199128204-Engineering-Chemical-Thermodynamics-Koretsky-Solutions-Manual.pdf

Chapter 1 Solutions Engineering and Chemical Thermodynamics

Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University [email protected]

1.2 An approximate solution can be found if we combine Equations 1.4 and 1.5:

1 2 mV  ekmolecular 2 3 kT  ekmolecular 2  3kT V  m

Assume the temperature is 22 ºC. The mass of a single oxygen molecule is m  5.14  10 26 kg . Substitute and solve:  V  487.6 m/s

The molecules are traveling really, fast (around the length of five football fields every second). Comment: We can get a better solution by using the Maxwell-Boltzmann distribution of speeds that is sketched in Figure 1.4. Looking up the quantitative expression for this expression, we have:  m  f (v)dv  4    2kT 

3/ 2

 m 2 2 v v dv exp  2kT 

where f(v) is the fraction of molecules within dv of the speed v. We can find the average speed by integrating the expression above 

 V 

 f (v)vdv 0 

 f (v)dv



8kT  449 m/s m

0

2

1.3 Derive the following expressions by combining Equations 1.4 and 1.5:

 3kT Va2  ma

 3kT Vb2  mb

Therefore,  Va2 mb   Vb2 ma

Since mb is larger than ma, the molecules of species A move faster on average.

3

1.4 We have the following two points that relate the Reamur temperature scale to the Celsius scale:

0 º C, 0 º Reamur  and 100 º C, 80 º Reamur  Create an equation using the two points: T º Reamur   0.8 T º Celsius 

At 22 ºC, T  17.6 º Reamur

4

1.5 (a) After a short time, the temperature gradient in the copper block is changing (unsteady state), so the system is not in equilibrium. (b) After a long time, the temperature gradient in the copper block will become constant (steady state), but because the temperature is not uniform everywhere, the system is not in equilibrium. (c) After a very long time, the temperature of the reservoirs will equilibrate; The system is then homogenous in temperature. The system is in thermal equilibrium.

5

1.6 We assume the temperature is constant at 0 ºC. The molecular weight of air is

MW  29 g/mol  0.029 kg/mol Find the pressure at the top of Mount Everest:      0.029 kg/mol9.81 m/s8848 m   P  1 atm exp      J    2 73.15 K   8.314      mol  K      P  0.330 atm  33.4 kPa

Interpolate steam table data:

T sat  71.4 º C

for P sat  33.4 kPa

Therefore, the liquid boils at 71.4 ºC. Note: the barometric relationship given assumes that the temperature remains constant. In reality the temperature decreases with height as we go up the mountain. However, a solution in which T and P vary with height is not as straight-forward.

6

1.7 To solve these problems, the steam tables were used. The values given for each part constrain the water to a certain state. In most cases we can look at the saturated table, to determine the state. (a)

(b)

(c)

(d)

Subcooled liquid Explanation: the saturation pressure at T = 170 [oC] is 0.79 [MPa] (see page 508); Since the pressure of this state, 10 [bar], is greater than the saturation pressure, water is a liquid. Saturated vapor-liquid mixture Explanation: the specific volume of the saturated vapor at T = 70 [oC] is 5.04 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 508); Since the volume of this state, 3 [m3/kg], is in between these values we have a saturated vaporliquid mixture. Superheated vapor Explanation: the specific volume of the saturated vapor at P = 60 [bar] = 6 [MPa], is 0.03244 [m3/kg] and the saturated liquid is 0.001 [m3/kg] (see page 511); Since the volume of this state, 0.05 [m3/kg], is greater than this value, it is a vapor. Superheated vapor Explanation: the specific entropy of the saturated vapor at P = 5 [bar] = 0.5 [MPa], is 6.8212 [kJ/(kg K)] (see page 510); Since the entropy of this state, 7.0592 [kJ/(kg K)], is greater than this value, it is a vapor. In fact, if we go to the superheated water vapor tables for P = 500 [kPa], we see the state is constrained to T = 200 [oC].

7

1.8 From the steam tables in Appendix B.1:

 m3  vˆ critical  0.003155    kg 

T  374.15 º C, P  22.089 MPa 

At 10 bar, we find in the steam tables vˆlsat

 m3   0.001127    kg 

 m3  vˆvsat  0.19444    kg  Because the total mass and volume of the closed, rigid system remain constant as the water condenses, we can develop the following expression: vˆ critical  1  x vˆlsat  xvˆvsat

where x is the quality of the water. Substituting values and solving for the quality, we obtain x  0.0105 or 1.05 %

A very small percentage of mass in the final state is vapor.

8

1.9 The calculation methods will be shown for part (a), but not parts (b) and (c) (a) Use the following equation to estimate the specific volume: vˆ1.9 MPa, 250 º C   vˆ1.8 MPa, 250 º C   0.5vˆ2.0 MPa, 250 º C   vˆ1.8 MPa, 250 º C  Substituting data from the steam tables,

 m3  vˆ1.9 MPa, 250 º C   0.11821    kg  From the NIST website:  m3  vˆ NIST 1.9 MPa, 250 º C   0.11791    kg  Therefore, assuming the result from NIST is more accurate  vˆ  vˆ NIST  % Difference    100 %  0.254 %  vˆ NIST  (b) Linear interpolation:

 m3  vˆ1.9 MPa, 300 º C   0.13284    kg  NIST website:  m3  vˆ NIST 1.9 MPa, 300 º C   0.13249    kg  Therefore, % Difference  0.264 %

(c) Linear interpolation:

 m3  vˆ1.9 MPa, 270 º C   0.12406    kg 

9

Note: Double interpolation is required to determine this value. First, find the molar volumes at 270 ºC and 1.8 MPa and 2.0 MPa using interpolation. Then, interpolate between the results from the previous step to find the molar volume at 270 ºC and 1.9 MPa. NIST website:  m3  vˆ NIST 1.9 MPa, 270 º C   0.12389    kg  Therefore, % Difference  0.137 %

With regards to parts (a), (b), and (c), the values found using interpolation and the NIST website agree very well. The discrepancies will not significantly affect the accuracy of any subsequent calculations.

10

1.10 For saturated temperature data at 25 ºC in the steam tables,  m3  L vˆlsat  0.001003    1.003    kg   kg 

Determine the mass: m

V  vˆlsat

1 L  0.997 kg L 1.003    kg 

Because molar volumes of liquids do not depend strongly on pressure, the mass of water at 25 ºC and atmospheric pressure in a one liter should approximately be equal to the mass calculated above unless the pressure is very, very large.

11

1.11 First, find the overall specific volume of the water in the container:

vˆ 

 m3  1 m3  0.2   5 kg  kg 

Examining the data in the saturated steam tables, we find at P sat  2 bar

vˆlsat  vˆ  vˆvsat

Therefore, the system contains saturated water and the temperature is T  T sat  120.23 º C Let ml represent the mass of the water in the container that is liquid, and mv represent the mass of the water in the container that is gas. These two masses are constrained as follows: ml  mv  5 kg We also have the extensive volume of the system equal the extensive volume of each phase ml vˆlsat  mv vˆvsat  V



 



ml 0.001061 m 3 / kg  mv 0.8857 m 3 /kg  1 m 3 Solving these two equations simultaneously, we obtain ml  3.88 kg mv  1.12 kg Thus, the quality is x

mv 1.12 kg   0.224 m 5 kg

The internal energy relative to the reference state in the saturated steam table is U  ml uˆlsat  mv uˆvsat

From the steam tables:

12

uˆlsat  504.47 kJ/kg 

uˆvsat  2529.5 kJ/kg  Therefore, U  3.88 kg 504.47 kJ/kg   1.12 kg 2529.5 kJ/kg  U  4790.4 kJ

13

1.12 First, determine the total mass of water in the container. Since we know that 10 % of the mass is vapor, we can write the following expression



V  m 0.9 vˆlsat  0.1vˆvsat



From the saturated steam tables for P sat  1 MPa  m3  sat vˆl  0.001127   kg 



 m3  vˆvsat  0.1944    kg  Therefore, m

V

0.9vˆlsat  0.1vˆvsat

m  48.9 kg



1 m3    m3    m3   0.9 0.001127      0.1 0.1944     kg kg        

and



Vv  0.1mvˆvsat  0.148.9 kg  0.1944 m 3 / kg Vv  0.950 m 3

14



1.13 In a spreadsheet, make two columns: one for the specific volume of water and another for the pressure. First, copy the specific volumes of liquid water from the steam tables and the corresponding saturation pressures. After you have finished tabulating pressure/volume data for liquid water, list the specific volumes and saturation pressures of water vapor. Every data point is not required, but be sure to include extra points near the critical values. The data when plotted on a logarithmic scale should look like the following plot. The Vapor-Liquid Dome for H2O 100.0000 Critical Point

Pressure, P (MPa)

10.0000 Liquid 1.0000 0.0001

Vapor-Liquid

0.001

0.01

Vapor

0.1

1

10

0.1000

0.0100

0.0010

Specific Volume, v (m 3/kg)

15

100

1000

1.14 The ideal gas law can be rewritten as

v

RT P

For each part in the problem, the appropriate values are substituted into this equation where  L  bar  R  0.08314   mol  K  is used. The values are then converted to the units used in the steam table. (a)

 L  v  30.7   mol   L  30.7  3 3 v  mol  10 3  m   1.70  m  vˆ    kg   L  MW  kg      0.018    mol  For part (a), the calculation of the percent error will be demonstrated. The percent error will be based on percent error from steam table data, which should be more accurate than using the ideal gas law. The following equation is used:  vˆ  vˆST   100 % %Error   IG vˆST   where v IG is the value calculated using the ideal gas law and vST is the value from the steam table. 3 3   1.70  m   1.6729  m    kg   kg  %Error     m3   1.6729   kg     

   %  1.62%    

This result suggests that you would introduce an error of around 1.6% if you characterize boiling water at atmospheric pressure as an ideal gas.

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(b)

 L  v  64.3   mol   m3  vˆ  3.57    kg  % Error  0.126% For a given pressure, when the temperature is raised the gas behaves more like an ideal gas. (c)

 L  v  0.643   mol   m3  vˆ  0.0357    kg  %Error = 8.87% (d)

 L  v  1.06   mol   m3  vˆ  0.0588    kg  %Error = 0.823%

The largest error occurs at high P and low T.

17

1.15 The room I am sitting in right now is approximately 16 ft long by 12 feet wide by 10 feet tall – your answer should vary. The volume of the room is

V  12 ft 16 ft 10 ft   1920 ft 3  54.4 m 3 The room is at atmospheric pressure and a temperature of 22 ºC. Calculate the number of moles using the ideal gas law:







1.01325  10 5 Pa 54.4 m 3 PV  RT   J  295.15 K   8.314   mol  K    n  2246 mol

n

Use the molecular weight of air to obtain the mass: m  nMW   2246.27 mol0.029 kg/mol  65.2 kg

That’s pretty heavy.

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1.16 First, find the total volume that one mole of gas occupies. Use the ideal gas law:

RT  v P

  J   8.314  293.15 K   mol  K   

1105 Pa 

 m3   0.0244    mol 

Now calculate the volume occupied by the molecules:  Number of molecules  Volume of  4    N A  πr 3     v occupied   per mole 3    one molcule   6.022  10 23 molecules   4 3  π 1.5  10 10 m  v occupied      3  1 mol     m3  v occupied  8.5110  6    mol 





Determine the percentage of the total volume occupied by the molecules:  v occupied  Percentage   100 %  0.0349 % v   A very small amount of space, indeed.

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1.17 Water condenses on your walls when the water is in liquid-vapor equilibrium. To find the maximum allowable density at 70 ºF, we need to find the smallest density of water vapor at 40 ºF which satisfies equilibrium conditions. In other words, we must find the saturation density of water vapor at 40 ºF. From the steam tables,

 m3  vˆvsat  153.74    kg 

(sat. water vapor at 40 ºF = 4.44 ºC)

We must correct the specific volume for the day-time temperature of 70 ºF (21.1 ºC ) with the ideal gas law.

vˆvsat ,277.6 K T277.6

vˆ  294.3 K T294.3





 m3   m3   T 294.3 K    vˆ294.3 K  294.3 vˆvsat  153 . 74 163     , 277.6 K T277.6 277.6 K   kg   kg   

Therefore,

ˆ 

1 vˆvsat

 kg   6.13  10  3    m3 





If the density at 70 ºF is greater than or equal to 6.13  10 3 kg/m 3 , the density at night when the temperature is 40 ºF will be greater than the saturation density, so water will condense onto the





wall. However, if the density is less than 6.13  10 3 kg/m 3 , then saturation conditions will not be obtained, and water will not condense onto the walls.

20

1.18 We consider the air inside the soccer ball as the system. We can answer this question by looking at the ideal gas law: Pv  RT

If we assume it is a closed system, it will have the same number of moles of air in the winter as the summer. However, it is colder in the winter (T is lower), so the ideal gas law tells us that Pv will be lower and the balls will be under inflated. Alternatively, we can argue that the higher pressure inside the ball causes air to leak out over time. Thus we have an open system and the number of moles decrease with time – leading to the under inflation.

21

1.19 First, write an equation for the volume of the system in its initial state: ml vˆlsat  mv vˆvsat  V

Substitute ml  1  x m and mv  xm :





m 1  x vˆlsat  xvˆvsat  V

At the critical point

mvˆ critical  V Since the mass and volume don’t change vˆ critical  1  x vˆlsat  xvˆvsat

From the steam tables in Appendix B.1:  m3  vˆ critical  0.003155    kg 

T  374.15 º C, P  22.089 MPa 

At 0.1 MPa, we find in the steam tables  m3  vˆlsat  0.001043    kg   m3  sat vˆv  1.6940    kg  Therefore, solving for the quality, we get x  0.00125 or 0.125 %

99.875% of the water is liquid.

22

1.20 As defined in the problem statement, the relative humidity can be calculated as follows

Relative Humidity 

mass of water mass of water at saturation

We can obtain the saturation pressure at each temperature from the steam tables. At 10 [oC], the saturation pressure of water is 1.22 [kPa]. This value is proportional to the mass of water in the vapor at saturation. For 90% relative humidity, the partial pressure of water in the vapor is:

pWater  0.9 1.22  1.10 [kPa] This value represents the vapor pressure of water in the air. At 30 [oC], the saturation pressure of water is 4.25 [kPa]. For 590% relative humidity, the partial pressure of water in the vapor is:

pWater  0.5  5.63  2.12 [kPa] Since the total pressure in each case is the same (atmospheric), the partial pressure is proportional to the mass of water in the vapor. We conclude there is about twice the amount of water in the air in the latter case.

23

1.21 (a) When you have extensive variables, you do not need to know how many moles of each substance are present. The volumes can simply be summed.

V1  Va  Vb (b) The molar volume, v1, is equal to the total volume divided by the total number of moles. v1 

V1 Va  Vb  ntot na  nb

We can rewrite the above equation to include molar volumes for species a and b. nb n v  nb vb na vb va  v1  a a  na  nb na  nb na  nb v1  xa va  xb vb

(c) The relationship developed in Part (a) holds true for all extensive variables.

K1  K a  K b (d)

k1  xa k a  xb kb

24



2.1 There are many possible solutions to this problem. Assumptions must be made to solve the problem. One solution is as follows. First, assume that half of a kilogram is absorbed by the towel when you dry yourself. In other words, let m H 2 O  0.5 kg 

Assume that the pressure is constant at 1.01 bar during the drying process. Performing an energy balance and neglecting potential and kinetic energy effects reveals

qˆ  hˆ Refer to the development of Equation 2.57 in the text to see how this result is achieved. To find the minimum energy required for drying the towel, assume that the temperature of the towel remains constant at T  25 º C  298.15 K . In the drying process, the absorbed water is vaporized into steam. Therefore, the expression for heat is v l qˆ  hˆH  hˆH 2O 2O

v is the specific enthalpy of water vapor at P  1.01 bar and T  298.15 K and where is hˆH O 2 hˆ l is the specific enthalpy of liquid water at P  1.01 bar and T  298.15 K . A hypothetical H 2O

path must be used to calculate the change in enthalpy. Refer to the diagram below P

hˆ

liquid 1 atm

vapor

P = 1 [atm] hˆ3

hˆ1 P

3.17 kPa

sat

hˆ2

liquid

vapor

By adding up each step of the hypothetical path, the expression for heat is qˆ  h1  h2  h3  hˆ l ,sat 25 º C   hˆ l



 25 º C, 1.01 bar  hˆHv,satO 25 º C  hˆHl ,satO 25 º C  hˆHv O 25 º C, 1.01 bar   hˆHv ,sat O 25 º C 

H 2O

H 2O

2

2

2

2

2

However, the calculation of heat can be simplified by treating the water vapor as an ideal gas, which is a reasonable assumption at low pressure. The enthalpies of ideal gases depend on temperature only. Therefore, the enthalpy of the vapor change due to the pressure change is zero. Furthermore, enthalpy is weakly dependent on pressure in liquids. The leg of the hypothetical path containing the pressure change of the liquid can be neglected. This leaves v 25 º C, 3.17 kPa   hˆHl 2 O 25 º C, 3.17 kPa  qˆ  hˆH O 2

From the steam tables:  kJ  v, sat  hˆH 2547 . 2  kg  O 2    kJ  l , sat hˆH  104 . 87  kg  O 2  

(sat. H2O vapor at 25 ºC) (sat. H2O liquid at 25 ºC)

which upon substitution gives  kJ  qˆ  2442.3    kg 

Therefore,   kJ   Q  0.5 kg  2442.3     1221.2 kJ   kg   

To find the efficiency of the drying process, assume the dryer draws 30 A at 208 V and takes 20 minutes (1200 s) to dry the towel. From the definition of electrical work, W  IVt  30 A 208 V 1200 s  7488 kJ 

Therefore, the efficiency is  1221.2 kJ   Q   100 %    100 %  16.3% W   7488 kJ  

 

There are a number of ways to improve the drying process. A few are listed below.  Dry the towel outside in the sun.  Use a smaller volume dryer so that less air needs to be heated.  Dry more than one towel at a time since one towel can’t absorb all of the available heat. With more towels, more of the heat will be utilized.

3

2.3 In answering this question, we must distinguish between potential energy and internal energy. The potential energy of a system is the energy the macroscopic system, as a whole, contains relative to position. The internal energy represents the energy of the individual atoms and molecules in the system, which can have contributions from both molecular kinetic energy and molecular potential energy. Consider the compression of a spring from an initial uncompressed state as shown below.

uncompressed

compressed x

State 1

M

State 2

Since it requires energy to compress the spring, we know that some kind of energy must be stored within the spring. Since this change in energy can be attributed to a change of the macroscopic position of the system and is not related to changes on the molecular scale, we determine the form of energy to be potential energy. In this case, the spring’s tendency to restore its original shape is the driving force that is analogous to the gravity for gravitational potential energy. This argument can be enhanced by the form of the expression that the increased energy takes. If we consider the spring as the system, the energy it acquires in a reversible, compression from its initial uncompressed state may be obtained from an energy balance. Assuming the process is adiabatic, we obtain:

E  Q  W  W We have left the energy in terms of the total energy, E. The work can be obtained by integrating the force over the distance of the compression:

W    F  dx   kxdx 

1 2 kx 2

Hence: E 

1 2 kx 2

We see that the increase in energy depends on macroscopic position through the term x. It should be noted that there is a school of thought that assigns this increased energy to internal energy. This approach is all right as long as it is consistently done throughout the energy balances on systems containing springs.

4

2.4 For the first situation, let the rubber band represent the system. In the second situation, the gas is the system. If heat transfer, potential and kinetic energy effects are assumed negligible, the energy balance becomes U  W

Since work must be done on the rubber band to stretch it, the value of the work is positive. From the energy balance, the change in internal energy is positive, which means that the temperature of the system rises. When a gas expands in a piston-cylinder assembly, the system must do work to expand against the piston and atmosphere. Therefore, the value of work is negative, so the change in internal energy is negative. Hence, the temperature decreases. In analogy to the spring in Problem 2.3, it can be argued that some of the work imparted into the rubber band goes to increase its potential energy; however, a part of it goes into stretching the polymer molecules which make up the rubber band, and the qualitative argument given above still is valid.

5

2.5 To explain this phenomenon, you must realize that the water droplet is heated from the bottom. At sufficiently high temperatures, a portion of the water droplet is instantly vaporized. The water vapor forms an insulation layer between the skillet and the water droplet. At low temperatures, the insulating layer of water vapor does not form. The transfer of heat is slower through a gas than a liquid, so it takes longer for the water to evaporate at higher temperatures.

6

2.6

Apartment

Surr. HOT System

Fridge

Q

+ W -

If the entire apartment is treated as the system, then only the energy flowing across the apartment boundaries (apartment walls) is of concern. In other words, the energy flowing into or out of the refrigerator is not explicitly accounted for in the energy balance because it is within the system. By neglecting kinetic and potential energy effects, the energy balance becomes

U  Q  W The Q term represents the heat from outside passing through the apartment’s walls. The W term represents the electrical energy that must be supplied to operate the refrigerator. To determine whether opening the refrigerator door is a good idea, the energy balance with the door open should be compared to the energy balance with the door closed. In both situations, Q is approximately the same. However, the values of W will be different. With the door open, more electrical energy must be supplied to the refrigerator to compensate for heat loss to the apartment interior. Therefore, Wajar  Wshut

where the subscript “ajar” refers the situation where the door is open and the subscript “shut” refers to the situation where the door is closed. Since, Qajar  Qshut U ajar  Qajar  Wajar  U shut  Qshut  Wshut  Tajar  Tshut

The refrigerator door should remain closed.

7

2.7 The two cases are depicted below. Heater off

Heater on

Surr.

Surr.

System

System

Heater

Q

Heater

Q

W

Let’s consider the property changes in your house between the following states. State 1, when you leave in the morning, and state, the state of your home after you have returned home and heated it to the same temperature as when you left. Since P and T are identical for states 1 and 2, the state of the system is the same and U must be zero, so

U  Q  W  0 or

Q W where -Q is the total heat that escaped between state 1 and state 2 and W is the total work that must be delivered to the heater. The case where more heat escapes will require more work and result in higher energy bills. When the heater is on during the day, the temperature in the system is greater than when it is left off. Since heat transfer is driven by difference in temperature, the heat transfer rate is greater, and W will be greater. Hence, it is cheaper to leave the heater off when you are gone.

8

2.8 The amount of work done at constant pressure can be calculated by applying Equation 2.57

H  Q Hence,

H  Q  mhˆ where the specific internal energy is used in anticipation of obtaining data from the steam tables. The mass can be found from the known volume, as follows:  3   L      1.0 kg    m3    0.0010     kg     

m

V  vˆ

1L 0.001 m

As in Example 2.2, we use values from the saturated steam tables at the same temperature for subcooled water at 1 atm. The specific enthalpy is found from values in Appendix B.1:  kJ   kJ   kJ  uˆ  uˆl ,2 at 100 o C  uˆl ,1 at 25 o C  419.02    104.87    314.15    kg   kg   kg 



 

  

Solve for heat:   kJ   Q  muˆ  1.0 kg  314.05     314.15 kJ   kg   

and heat rate:

Q Q   t

314.15 kJ   0.52 kW   60 s  10 min.   min 

This value is the equivalent of five strong light bulbs.

9

2.9 (a) From Steam Tables:

 kJ  uˆ1  2967.8    kg   kJ  uˆ 2  2659.8    kg 

(100 kPa, 400 ºC) (50 kPa, 200 ºC)

 kJ  uˆ  uˆ 2  uˆ1  308.0    kg  (b) From Equations 2.53 and 2.63 T2

T2

T1

T1

u  u 2  u1   cv dT 

 c P  R dT

From Appendix A.2 c P  R( A  BT  CT 2  DT 2  ET 3 ) T2





u  R  A  BT  CT 2  DT  2  ET 3  1 dT T1

Integrating   E 1 1 B C u  R ( A  1)(T2  T1 )  (T2 2  T12 )  (T23  T13 )  D(  )  (T2 4  T14 ) 2 3 4 T2 T1  

The following values were found in Table A.2.1 A  3.470 B  1.45  10  3 C 0 D  1.21 10 4 E 0 Substituting these values and using

10

 J  R  8.314   mol  K  T1  (400  273.15 K)  673.15 K T2  (200  273.15 K)  473.15 K

provides  J  u  5551   mol   kJ    J   1 [mol H 2 O]  1000 g  1 kJ      uˆ    5551  308 . 1       kg       mol   18.0148 [g H 2 O]  1 kg  1000 J    

The values in parts (a) and (b) agree very well. The answer from part (a) will serve as the basis for calculating the percent difference since steam table data should be more accurate. % Difference 

 308   308.1  100 %  0.03%  308.0

11

2.10 (a) Referring to the energy balance for closed systems where kinetic and potential energy are neglected, Equation 2.30 states

U  Q  W (b) Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal U  0

According to Equation 2.77 P W  nRT ln  2  P1

 P   n1 RT1 ln  2   P1

  

From the ideal gas law: n1RT1  P1V1 P  W  P1V1 ln 2   P1 

Substitution of the values from the problem statement yields







 5 bar  W  8  105 Pa 2.5  10  3 m 3 ln   8 bar  W  940 J  The energy balance is 0 J  Q W

 Q  940 J  (c) Since the process is adiabatic

Q0 The energy balance reduces to U  W

12

The system must do work on the surroundings to expand. Therefore, the work will be negative and U  0 T2

U  n  cv T  0 T1

 T2  T1 T2 will be less than 30 ºC

13

2.11 (a) (i). 1

Path B

Pa A th

P [bar]

3

2

2 1

0.01

0.02

0.03 3 v [m /mol]

(ii). Since internal energy is a function of temperature only for an ideal gas (Equation 2.4) and the process is isothermal u  0

Equation 2.48 states that enthalpy is a function of temperature only for an ideal gas. Therefore, h  0

Performing an energy balance and neglecting potential and kinetic energy produces u  q  w  0

For an isothermal, adiabatic process, Equation 2.77 states P  W  nRT ln 2   P1 

or w

P  W  RT ln 2  n  P1 

Substituting the values from the problem statement gives

14

  1 bar   J  w   8.314  (88  273.15) K ln    3 bar   mol  K     J  w  3299   mol 

Using the energy balance above

 J  q   w  3299   mol  (b) (i).

See path on diagram in part (a)

(ii). Since the overall process is isothermal and u and h are state functions u  0 h  0

The definition of work is

w    PE dv During the constant volume part of the process, no work is done. The work must be solved for the constant pressure step. Since it is constant pressure, the above equation simplifies to w   PE  dv   PE (v2  v1 ) The ideal gas law can be used to solve for v 2 and v1   J  mol    8.314  (88  273.15) K   m3  RT2   K     v2  0 . 030   P2 1 10 5 Pa  mol    J  mol    8.314  (88  273.15) K   m3  RT1   K    0 . 010 v1     P1 3  10 5 Pa  mol 

Substituting in these values and realizing that PE  P1 since the process is isobaric produces

15

  m3   m3   w  (3  10 5 Pa) 0.030    0.010     mol mol        J  w  6000   mol 

Performing an energy balance and neglecting potential and kinetic energy results in

u  q  w  0  J   q   w  6000   mol 

16

2.12 First, perform an energy balance. No work is done, and the kinetic and potential energies can be neglected. The energy balance reduces to

U  Q We can use Equation 2.53 to get T2

Q  n  cv dT T1

which can be rewritten as T2

Q  n  c P dT T1

since the aluminum is a solid. Using the atomic mass of aluminum we find n

5 kg  185.3 mol  kg  0.02698   mol 

Upon substitution of known values and heat capacity data from Table A.2.3, we get   J  Q  185.3 mol 8.314    2.486  1.49  10 3 T dT   mol K   294.15 K  Q  131.61 kJ 





323.15 K

17

2.13 First, start with the energy balance. Potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes

U  Q  W The value of the work will be used to obtain the final temperature. The definition of work (Equation 2.7) is V2

W    PE dV V1

Since the piston expands at constant pressure, the above relationship becomes W   PE V2  V1 

From the steam tables  m3  vˆ1  0.02641    kg 

(10 MPa, 400 ºC)

 

  m3   3  ˆ V1  m1v1  (3 kg) 0.02641     0.07923 m    kg   

Now V2 and v2 are found as follows V2  V1  V vˆ2  2 m2

 

 748740 J W  0.4536 m3  0.07923 m3  6 PE 2.0  10 Pa  m3  0.4536 m 3   0.1512   3 kg   kg 

 

Since vˆ2 and P2 are known, state 2 is constrained. From the steam tables: T2  400 º C

  3   20 bar, 0.1512  m      kg   

Now U will be evaluated, which is necessary for calculating Q . From the steam tables:

18

 kJ  uˆ 2  2945.2    kg 

  3   20 bar, 0.1512  m      kg   

 kJ  uˆ1  2832.4    kg 

100 bar, 400 º C

  kJ   kJ   U  m1 uˆ 2  uˆ1   3 kg  2945.2    2832.4     338.4 kJ   kg   kg   

Substituting the values of U and W into the energy equation allows calculation of Q

Q  U  W Q  338400 [J]   748740 J   1.09  10 6 J 

19

2.14 In a reversible process, the system is never out of equilibrium by more than an infinitesimal amount. In this process the gas is initially at 2 bar, and it expands against a constant pressure of 1 bar. Therefore, a finite mechanical driving force exists, and the process is irreversible.

To solve for the final temperature of the system, the energy balance will be written. The pistoncylinder assembly is well-insulated, so the process can be assumed adiabatic. Furthermore, potential and kinetic energy effects can be neglected. The energy balance simplifies to U  W

Conservation of mass requires n1  n2 Let n  n1  n2

The above energy balance can be rewritten as T2

V2

T1

V1

n  cv dT    PE dV

Since cv and PE are constant: ncv T2  T1    PE V2  V1  V2 and T1 can be rewritten using the ideal gas law

nRT2 P2 PV T1  1 1 nR

V2 

Substituting these expressions into the energy balance, realizing that PE  P2 , and simplifying the equation gives 5    P2  P1 V1 2  T2   7 nR 2 Using the following values

20

P1  2 bar  P2  1 bar 

V1  10 L

n  1.0 mol  L  bar  R  0.08314   mol  K 

results in T2  206 K 

To find the value for work, the energy balance can be used W  U  ncv T2  T1  Before the work can be calculated, T1 must be calculated PV T1  1 1  nR

2 bar 10 L 1 mol 0.08314  L  bar    mol  K   

 241 K 

Using the values shown above W  727 J 

21

2.15 The maximum work can be obtained through a reversible expansion of the gas in the piston. Refer to Section 2.3 for a discussion of reversible processes. The problem states that the piston assembly is well-insulated, so the heat transfer contribution to the energy balance can be neglected, in addition to potential and kinetic energy effects. The energy balance reduces to U  W

In this problem, the process is a reversible, adiabatic expansion. For this type of process, Equation 2.90 states W

1 P2V2  P1V1  k 1

From the problem statement (refer to problem 2.13), P1  2 bar 

V1  10 L

P2  1 bar 

To calculate W, V2 must be found. For adiabatic, reversible processes, the following relationship (Equation 2.89) holds: PV k  const where k is defined in the text. Therefore, 1 P k V2   1 V1k   P2 

c 7 Noting that k  P  and substituting the proper values provides cv 5 V2  16.4 L

Now all of the needed values are available for calculating the work. W  9 L  bar   900 J 

From the above energy balance, U  900 J 

22

The change in internal energy can also be written according to Equation 2.53: T2

U  n  cv dT T1

Since cv is constant, the integrated form of the above expression is 5  U  n R T2  T1  2  Using the ideal gas law and knowledge of P1 and V1 , T1  240.6 K 

and T2  197.3 K 

The temperature is lower because more work is performed during the reversible expansion. Review the energy balance. As more work is performed, the cooler the gas will become.

23

2.16 Since the vessel is insulated, the rate of heat transfer can be assumed to be negligible. Furthermore, no work is done on the system and potential and kinetic energy effects can be neglected. Therefore, the energy balance becomes

uˆ  0 or uˆ 2  uˆ1

From the steam tables  kJ  uˆ1  2619.2    kg   kJ   uˆ 2  2619.2    kg 

(200 bar, 400 ºC)

The values of uˆ 2 and P2 constrain the system. The temperature can be found from the steam tables using linear interpolation: T2  327.5 º C

  kJ   100 bar , uˆ 2  2619.2       kg   

Also at this state,  m3  vˆ2  0.02012    kg  Therefore,

 

  m3   3 Vvessel  mv2  1.0 kg  0.02012     0.020 m   kg    

24

2.17

Let the entire tank represent the system. Since no heat or work crosses the system boundaries, and potential and kinetic energies effects are neglected, the energy balance is u  0

Since the tank contains an ideal gas T2  T1  0 T2  T1  300 K The final pressure can be found using a combination of the ideal gas law and conservation of mass. T1 T  2 P1V1 P2V2 We also know V2  2V1

Therefore, P P2  1  5 bar 2

25

2.18 (a) First, as always, simplify the energy balance. Potential and kinetic energy effects can be neglected. Therefore, the energy balance is

U  Q  W Since, this system contains water, we can the use the steam tables. Enough thermodynamic properties are known to constrain the initial state, but only one thermodynamic property is known for the final state: the pressure. Therefore, the pressure-volume relationship will be used to find the specific volume of the final state. Since the specific volume is equal to the molar volume multiplied by the molecular weight and the molecular weight is constant, the given expression can be written Pvˆ1.5  const This equation can be used to solve for vˆ2 . 1

 P   1.5 vˆ2    1 vˆ11.5    P2  

Using P1  20 bar  vˆ1 

 

 m3  1.0 m 3  0.1   10 kg   kg 

P2  100 bar 

gives  m3  vˆ2  0.0342    kg  Now that the final state is constrained, the steam tables can be used to find the specific internal energy and temperature. T2  524.7 K   kJ  uˆ 2  3094.6    kg 

To solve for the work, refer to the definition (Equation 2.7). 26

V2

W    PE dV V1

or vˆ 2

wˆ    PE dvˆ vˆ1

Since the process is reversible, the external pressure must never differ from the internal pressure by more than an infinitesimal amount. Therefore, an expression for the pressure must be developed. From the relationship in the problem statement, Pvˆ1.5  P1vˆ11.5  const Therefore, the expression for work becomes vˆ 2

wˆ   

vˆ1

P1vˆ11.5 vˆ1.5

dvˆ   P1vˆ11.5

vˆ 2

1

 vˆ1.5 dvˆ

vˆ1

Integration and substitution of proper values provides  bar  m 3   kJ  wˆ  2.840    284    kg   kg    kJ   W  10 kg  284     2840 kJ   kg   

A graphical solution is given below:

P [bar]

2 100

60

1

20

Work 0.02

0.06

27

0.10 v [m 3/kg]

To solve for Q , U must first be found, then the energy balance can be used.   kJ   kJ   U  muˆ 2  uˆ1   10 kg  3094.6    2602.8     4918 kJ   kg   kg   

Now Q can be found, Q  U  W  4918 kJ   2840 kJ   2078 kJ 

(b) Since the final state is the same as in Part (a), U remains the same because it is a state function. The energy balance is also the same, but the calculation of work changes. The pressure from the weight of the large block and the piston must equal the final pressure of the system since mechanical equilibrium is reached. The calculation of work becomes: vˆ 2

W  mPE  dvˆ vˆ1

All of the values are known since they are the same as in Part (a), but the following relationship should be noted PE  P2

Substituting the appropriate values results in W  6580 kJ 

Again we can represent this process graphically:

P [bar]

2 100

Work 60

1

20

0.02

0.06

28

0.10 v [m 3/kg]

Now Q can be solved. Q  U  W  4918 kJ   6580 kJ   1662 kJ 

(c) This part asks us to design a process based on what we learned in Parts (a) and (b). Indeed, as is characteristic of design problems there are many possible alternative solutions. We first refer to the energy balance. The value of heat transfer will be zero when U  W

For the same initial and final states as in Parts (a) and (b), W  U  4918 kJ 

There are many processed we can construct that give this value of work. We show two alternatives which we could use: Design 1: If the answers to Part (a) and Part (b) are referred to, one can see that two steps can be used: a reversible compression followed by an irreversible compression. Let the subscript “i" represent the intermediate state where the process switches from a reversible process to an irreversible process. The equation for the work then becomes vî   1 1 . 5  W  m  P2 (vˆ2  vî )  P1vˆ1  dvˆ   4918000 J  1 . 5   vˆ1 vˆ  

Substituting in known values (be sure to use consistent units) allows calculation of vî :  m3  vî  0.0781    kg  The pressure can be calculated for this state using the expression from part (a) and substituting the necessary values. 1.5

v  Pi  P1  1   vi  Pi  29.0 bar 

29

Now that both Pi and vî are known, the process can be plotted on a P-v graph, as follows:

P [bar]

2 100

Work

60

i

1

20

0.02

0.06

0.10 v [m 3/kg]

Design 2: In an alternative design, we can use two irreversible processes. First we drop an intermediate weight on the piston to compress it to an intermediate state. This step is followed by a step similar to Part (b) where we drop the remaining mass to lead to 100 bar external pressure. In this case, we again must find the intermediate state. Writing the equation for work: W  m Pi (vî  vˆ1 )  P2 (vˆ2  vî )  4918000 J  However, we again have the relationship: v  Pi  P1  1   vi 

1.5

Substitution gives one equation with one unknown vi: 1.5    v1  W  m  P1   (vî  vˆ1 )  P2 (vˆ2  vî )  4918000 J     vi 

There are two possible values vi to the above equation. Solution A:  m3  vî  0.043    kg 

30

which gives Pi  70.8 bar 

This solution is graphically shown below:

P [bar]

2 100

i 60

20

Work

1

0.06

0.10 v [m 3/kg]

0.02

Solution B  m3  vî  0.0762    kg 

which gives Pi  30.0 bar 

This solution is graphically shown below:

P [bar]

2 100

60 i 20

Work

1

0.06

0.10 v [m 3/kg]

0.02

31

2.19 (a) Force balance to find k:

Fspring=kx

Fatm=PatmA

Piston

Fmass=mg

Fgas=PgasA

mg

Pgas  Patm  A  kx A

since V=Ax

Pgas  Patm  A  k2V A mg

since V is negative. Now solve: 2  10 5 Pa  1 10 5 Pa 

2040 kg 9.81 m/s 2   k 0.02 m 3 

0.1 m 2 2

0 .1 m 2

N  k  5.01 10 4   m

Work can be found graphically (see P-V plot) or analytically as follows: Substituting the expression in the force balance above: P k V  W A    atm  A  A 2 dV mg

V f

Vf

mg  kV  W A    Patm  d V  dV   2 A   A Vi Vi





2   5.01 10 4 N/m  0.02 m 3  0 2  W A  3.00  10 Pa 0.03  0.05 m  2   2 0 .1 m 2  



5



2





32



W A  5  103 J  -W

0.5kJ = 10squares square = 5 kJ

3 P [bar]

k V 2 A 2

mg A

2

Work

1 1 Patm

0.01

0.03

0.05 V [m3 ]

(b) You need to find how far the spring extends in the intermediate (int) position. Assume PVn=const (other assumptions are o.k, such as an isothermal process, and will change the answer slightly). Since you know P and V for each state in Part (a), you can calculate n. Pi Pf

 V n  V i  or f 

10 5 Pa 2 10 5 Pa

n

0.03 m 3    3   0.05 m

 n  1.35

Now using the force balance mg

Pgas,int  Patm  A  kV A2 with the above equation yields:

 P V

Pint  Pi V i int

n

mg k Vint  V i  atm  A  A2

This last equality represents 1 equation. and 1 unknown (we know k), which gives Vint  0.0385 m 3

Work can be found graphically (see P-V plot) or analytically using:

33

V int

Vf

Vi

Vint

W B   Pgas dV   PgasdV

expanding as in Part (a) V

int 5 N  W B    2  10 m2 dV 

Vi

Vint

V

V if

V i

Vint

Vint

f 6 N  5 V    5  10 d V   3  10 mN2  5 dV  m

6  5  10 N5 V d V  m

Therefore, WB  3.85 kJ 

just like we got graphically.

P [bar]

3

2 2

mg A mg A 1

1

Work Patm

0.01

0.03

0.05 V [m3 ]

(c) The least amount of work is required by adding differential amounts of mass to the piston. This is a reversible compression. For our assumption that PVn = const, we have the following expression:

WC , rev 

Vf

Vf

Vi

Vi

const

 Pgas dV   V 1.35 dV

Calculate the constant from the initial state

34





const.  1 105 Pa 0.05 m 3

1.35  1.75 103

Therefore, .03

WC,rev   1.75  103 dV 1.35 V

.05

10 V   1.750.35 3

0 .35 .03 .05

 2800 J 

P [bar]

3

2 2

1 1

Workrev Patm

0.01

0.03

0.05 V [m3 ]

35

2.20 Before this problem is solved, a few words must be said about the notation used. The system was initially broken up into two parts: the constant volume container and the constant pressure piston-cylinder assembly. The subscript “1” refers to the constant volume container, “2” refers the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f” denotes the final state.

To begin the solution, the mass of water present in each part of the system will be calculated. The mass will be conserved during the expansion process. Since the water in the rigid tank is saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water is constrained to a specific state. From the steam tables,  kJ  vˆ1l, i  0.001156    kg   kJ  vˆ1v, i  0.12736    kg   kJ  uˆ1l, i  850.64    kg   kJ  uˆ1v, i  2595.3    kg 

(Sat. water at 200 ºC)

P sat  1553.8 kPa  Knowledge of the quality of the water and the overall volume of the rigid container can be used to calculate the mass present in the container.

 

 

V1  0.05m1 vˆ1l, i  0.95m1 vˆ1v,i

 

Using the values from the steam table and V1  0.5 m 3 provides m1  4.13 kg 

Using the water quality specification, m1v  0.95m1  3.92 kg  m1l  0.05m1  0.207 kg 

For the piston-cylinder assembly, both P and T are known. From the steam tables

36

 m3  vˆ2, i  0.35202    kg   kJ  uˆ 2, i  2638.9    kg 

(600 kPa, 200 ºC)

Enough information is available to calculate the mass of water in the piston assembly. V m2  2  0.284 kg  vˆ2 Now that the initial state has been characterized, the final state of the system must be determined. It helps to consider what physically happens when the valve is opened. The initial pressure of the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final temperature and pressure of the entire system will match the surroundings’. In other words,  kJ  uˆ f  uˆ 2, i  2638.9    kg 

(600 kPa, 200 ºC)

Thus, the change in internal energy is given by

U  (m2  m1v, i  m1l, i )uˆ f  m2 uˆ 2, i  m1v, i uˆ1v, i  m1l, i uˆ1l, i Substituting the appropriate values reveals U  541.0 kJ 

To calculate the work, we realize the gas is expanding against a constant pressure of 600 kPa (weight of the piston was assumed negligible). From Equation 2.7, Vf

W   PE  dV   PE (V f  Vi ) Vi

where PE  600000 Pa 

 

V f  (m2  m1v, i  m1l,i )vˆ2, i  1.55 m 3

   

 

Vi  0.1 m 3  0.5 m 3  0.6 m 3

37

Note: vˆ2, i was used to calculate V f because the temperature and pressure are the same for the final state of the entire system and the initial state of the piston-cylinder assembly. The value of W can now be evaluated. W  570 kJ 

The energy balance is used to obtain Q. Q  U  W  541.0 kJ    570 kJ   1111 kJ 

38

2.21

A sketch of the process follows: system

boundary

system Process

B p 1,B = 20 bar T1,B = 250 o C

A p 1,A = 10 bar T1,A = 700 o C H2 O

boundary

H2 O

50 cm

A p2 T2 H2 O

B p2 T2 H2 O

10 cm

The initial states are constrained. Using the steam tables, we get the following:

p T v u V m

V v

State 1,A 10 [bar] 700 [oC] 0.44779 [m3/kg] 3475.35 [kJ/kg] 0.01 m3

State 1,B 20 [bar] 250 [oC] 0.11144 [m3/kg] 2679.58 [kJ/kg] 0.05 m3

0.11 [kg]

0.090 [kg]

All the properties in the final state are equal. We need two properties to constrain the system: We can find the specific volume since we know the total volume and the mass: v2 

V1, A  V1, B m1, A  m1, B



 

 m3  0.06 m 3  0.30   0.20 kg   kg 

We can also find the internal energy of state 2. Since the tank is well insulated, Q=0. Since it is rigid, W=0. An energy balance gives: U  Q  W  0

Thus, U 2  U1  m1,A u1, A  m1,Bu1,B or u2 

 kJ  U 2 m1, Au1, A  m1, B u1, B   3121   m2 m1, A  m1, B  kg 

We have constrained the system with u2 and v2, and can find the other properties from the steam Tables. Very close to T2 = 500 [oC] and

P2 = 1200 [kPa]

Thus,

39

  m3   V2, A  v2, A m2, A   0.30   0.09 kg   0.267 m and   kg     x  0.267  0.1  0.167 m

40

2.22 We start by defining the system as a bubble of vapor rising through the can. We assume the initial temperature of the soda is 5 oC. Soda is usually consumed cold; did you use a reasonable estimate for T1? A schematic of the process gives: State 2: T 2= ? P2= 1.01 bar

State 1: T 1 = 278 K P1 = 3 bar

where the initial state is labeled state 1, and the final state is labeled state 2. To find the final temperature, we perform an energy balance on the system, where the mass of the system (CO2 in the bubble) remains constant. Assuming the process is adiabatic and potential and kinetic energy effects are negligible, the energy balance is u  w

Expressions for work and internal energy can be substituted to provide cv T2  T1     PE dv   PE v 2  v1  where cv = cP – R. Since CO2 is assumed an ideal gas, the expression can be rewritten as T  PT  T  cv T2  T1    PE R  2  1    R T2  2 1  P1   P2 P1  

where the equation was simplified since the final pressure, P2, is equal to the external pressure, PE. Simplifying, we get:   RP  R T2 1    T1 1  2   c v P1   cv 

or  RP  c T2  T1 1  2  v  237 K  cv P1  c P

41

2.23 The required amount of work is calculated as follows: W   P V

The initial volume is zero, and the final volume is calculated as follows:

4 4 V  πr 3  π 0.5 ft 3  1.54 ft 3  0.0436 m 3 3 3 Assuming that the pressure is 1 atm, we calculate that







W  1.01325  105 Pa 0.0436 m 3  0 m 3  4417 J This doesn’t account for all of the work because work is required to stretch the rubber that the balloon is made of.

42

2.24 (a) Since the water is at its critical point, the system is constrained to a specific temperature, pressure, and molar volume. From Appendix B.1

 m3  vˆc  0.003155    kg  Therefore, m

V  vˆc

 

0.01 m 3  3.17 kg   m3  0.003155    kg 

(b) The quality of the water is defined as the percentage of the water that is vapor. The total volume of the vessel can be found using specific volumes as follows

V  m l vˆ l  m v vˆ v  1  x mvˆ l   xm vˆ v where x is the quality of the water. To solve for the quality, realize that starting with saturated water at a pressure of 1 bar constrains the water. From the steam tables,  m3  vˆ v  1.6940    kg   m3  ˆv l  0.001043    kg 

(sat. H2O at P = 1 bar)

Now the quality can be found x  0.00125

Thus, the quality of the water is 0.125%. (c) To determine the required heat input, perform an energy balance. Potential and kinetic energy effects can be neglected, and no work is done. Therefore,

U  Q

43

where



U  muˆ 2  1  x muˆ1l   xm uˆ1v



From the steam tables  kJ  uˆ 2  2029.58    kg   m3  l ˆ u1  417.33    kg   m3  uˆ1v  2506.1    kg 

(H2O at its critical point)

(sat. H2O at P=1 bar)

Evaluation of the expression reveals U  5102.6 kJ   5.10  10 6 J 

44

2.25 (a) Consider the air in ChE Hall to be the system. The system is constant volume, and potential and kinetic energy effects can be neglected. Furthermore, disregard the work. The energy balance is

du  q dt since the temperature of the system changes over time. Using the given expression for heat transfer and the definition of dU , the expression becomes cv dT  hT  Tsurr  dt We used a negative sign since heat transfer occurs from the system to the surroundings. If cv is assumed constant, integration provides cv ln T  Tsurr   ht  C where C is the integration constant. Therefore, T  Tsurr  C1e

 h    t   cv 

where C1 is a constant. Examining this equation reveals that the temperature is an exponential function of time. Since the temperature is decreasing, we know that the plot of temperature vs. time shows exponential decay.

Temperature

T0

Tsurr time

45

(b) Let time equal zero at 6 PM, when the steam is shut off. At 6 PM, the temperature of the hall is 22 ºC. Therefore, T  Tsurr  C1e

 h   t  c   v 

22 º C  2 º C  C1e (0) C1  20 º C After 10 PM, ( t  4 hr ), the temperature is 12 ºC. 12 º C  2 º C  20 º C e h   0.173 hr -1 cv

 h    ( 4 hr)   c   v 

At 6 AM, t  12 hr . Substitution of this value into the expression for temperature results in T  4 .5 º C

46

2.26 The gas leaving the tank does flow work as it exits the valve. This work decreases the internal energy of the gas – lowering the temperature. During this process, water from the atmosphere will become supersaturated and condense. When the temperature drops below the freezing point of water, the water forms a solid.

Attractive interactions between the compressed gas molecules can also contribute to this phenomena, i.e., it takes energy to pull the molecules apart as they escape; we will learn more of these interactions in Chapter 4.

47

2.27 Mass balance

dm  m in  m out  m in dt Separating variables and integrating: m2

t

m1

0

 dm   m in dt

or t

m2  m1   m in dt 0

Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is  dU      m out hout   m in hin  Q  W s  dt  sys out in

The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to  dU   dt

   m in hin  sys

The following math is performed U2

t

t

U1

0

0

 dU   m in hin dt  hin  m in dt

U 2  U 1  m2 uˆ 2  m1uˆ1  m2  m1 hîn where the results of the mass balance were used. Both m2 and m1 can be calculated by dividing the tank volume by the specific volume

48

m2 

V vˆ2

m1 

V vˆ1

Substitution of these relationships and simplification results in

uˆ 2  hin  uˆ1  hin  vˆ2



vˆ1

0

From the steam tables:  kJ  uˆ1  2583.6    kg   m3  vˆ1  0.19444    kg 

 kJ  hîn  3177.2    kg 

(sat. H2O vapor at 1 MPa)

(6 MPa, 400 ºC)

There are still two unknowns for this one equation, but the specific volume and internal energy are coupled to each other. To solve this problem, guess a temperature and then find the corresponding volume and internal energy values in the steam tables at 6 MPa. The correct temperature is the one where the above relationship holds. T  600 º C : Expression = 4427.6 T  500 º C : Expression = 1375.9 T  450 º C : Expression = -558.6

Interpolation between 500 ºC and 450 ºC reveals that the final temperature is T2  464.4 º C

49

2.28 We can pick room temperature to be 295 K

Tin  T1  295 K Mass balance dn  n in  n out  n in dt Separating variables and integrating: n2

t

n1

0

 dn   nin dt

or t

n2  n1   nin dt 0

Energy balance Neglecting ke and pe, he unsteady energy balance, written in molar units is written as:  dU    nin hin  nout hout  Q  W  dt  sys 

The terms associated with flow out, heat and work are zero.  dU    nin hin   dt  sys

Integrating both sides with respect to time from the initial state where the pressure is 10 bar to the final state when the tank is at a pressure of 50 bar gives: U2

t

t

U1

0

0

 dU   nin hin dt  hin  nin dt

since the enthalpy of the inlet stream remains constant throughout the process. Integrating and using the mass balance above: n2u2  n1u1  n2  n1 hin

50

Now we do some math: n2u 2  n1u1  n2  n1 hin n2 u 2  hin   n1 u1  hin  By the definition of h hin  uin  Pin vin  uin  RTin  u1  RT1 so n2 u 2  u1   n2 RT1  n1 u1  u1   n1R T1 n2 cv T2  T1   n2 RT1  n1RT1 Since cv  c P  R 

n2

3 R 2

3 T2  T1  n2T1  n1T1 2

or 3n2T2  5n2T1  2n1T1

dividing by n1: n n 3 2 T2  5 2 T1  2T1 n1 n1 Using the ideal gas law: n2 P2T1  n1 P1T2 so P T  P T  3 2 1 T2  5 2 1 T1  2T1  P1T2   P1T2 

or

51

P T  5 2 1   P1   434 [K] T2    P2   3   2   P1   (b) Closed system

u  q  w  q uˆ 

cv T2  T1   5R T2  T1   28.9  kJ  MW 2MW  kg 

 kJ  q  28.9    kg  (c)

P2T2  P3T3 PT P3  2 2  34 bar  T3

52

2.29 Mass balance

dn  n in  n out  n in dt Separating variables and integrating: n2

t

n1  0

0

 dn   nin dt

or t

n2   nin dt 0

Energy balance Neglecting ke and pe, the unsteady energy balance, in molar units, is written as:  dU     nin hin  nout hout  Q  W dt   sys The terms associated with flow out and heat are zero.

 dU     nin hin  W  dt  sys Integrating both sides with respect to time from the empty initial state to the final state gives: U2

t

t

t

U1

0

0

0

 dU   nin hin dt   W dt  hin  nin dt  W  hin n2  W

since the enthalpy of the inlet stream remains constant throughout the process. The work is given by:

W   n2 Pext (v2  v1 )   n2 Pext v2

53

n2u2  n2 hin  Pext v2  Rearranging,

u 2  hin  Pext v2  uin  Pin vin  Pext v2 u 2  uin  Pin vin  Pext v2 P T cv T2  Tin   RTin  R ext 2 P2

so T2 

cv  R    P  cv  ext R  P2  

Tin  333 K 

54

2.30 valve maintains pressure in system constant v T1 = 200 oC x1 = 0.4 V = 0.01 m3 l

Mass balance dm  m in  m out  m out dt Separating variables and integrating: m2

t

m1

0

 dm   m out dt

or t

m2  m1    m out dt 0

Energy balance  dU     m out hôut  Q  dt   sys Integrating m 2 uˆ 2

t

m1uˆ1

0





t

t

0

0

 dU    m out hôut  Q dt  hôut  m out dt   Q dt

Substituting in the mass balance and solving for Q Q  m2uˆ2  m1uˆ1  m2  m1 hôut

55

We can look up property data for state 1 and state 2 from the steam tables:  m3  vˆ1  (1  x)vˆ f  xvˆg  0.6  .001  0.4  0.1274  0.051    kg   m3  vˆ2  0.1274    kg 

So the mass in each state is:

 

m1 

0.01 m 3 V1  0.196 kg   vˆ1  m3  0.051    kg 

m2 

V2  vˆ2

 

0.01 m 3  0.0785 kg   m3  0.1274    kg 

m2  m1  0.1175 kg 

And for energy and enthalpy

 kJ  uˆ1  (1  x)uˆ f  xuˆ g  0.6  850.64  0.4  2597.5  1549    kg   kJ  uˆ 2  2595.3    kg   kJ  hôut  2793.2    kg  Solving for heat, we get Q  m2uˆ2  m1uˆ1  m2  m1 hôut  228 kJ 

56

2.31 Consider the tank as the system.

Mass balance dm  m in  m out  m in dt Separating variables and integrating: m2

t

m1

0


or t

m2  m1   m in dt 0

Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is  dU      m out hout   m in hin  Q  W s  dt  sys out in

The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to  dU     m in hin  dt  sys


t

U1  0

0

 dU   m

t

h dt  hin  m in dt

in in

0

U 2  m2 uˆ 2  m2 hîn where the results of the mass balance were used. Thus,

uˆ 2  hîn From the steam tables

57

 kJ  uˆ 2  3632.5    kg 

(9 MPa, 800 ºC)

so

 kJ  hîn  3632.5    kg  We can use the value of hin and the fact that the steam in the pipe is at 9 MPa to find the temperature.

Tin  600 º C

58

2.32 (a) First, the energy balance must be developed. Since the problem asks how much energy is stored in the battery after 10 hours of operation, the process is not steady-state. Let the battery be the system. Potential and kinetic energy effects can be neglected. Furthermore, heating of the battery as it is charged can be ignored. The energy balance is  dU     Q  W s dt   sys

No shaft work is performed, but electrical is supplied to the battery, which must be accounted for in W s . The value of Q is given explicitly in the problem statement. Both of these values remain constant over time, so integration provides





U  Q  W s t From the problem statement W s  5 kW  Q  1 kW  t  36000 s

Substituting these values allows the calculation of the amount of energy stored: U  144,000 kJ   144 MJ 

(b) To calculate the velocity of the falling water, an energy balance must be developed with the water passing through the electricity generator (probably a turbine) as the system, where the  water enters with a velocity V1 and leaves with a negligible velocity, which will be approximated as 0. Assume that potential energy changes can be neglected. Furthermore, assume that the temperature of the water does not change in the process, so the change in internal energy is zero. Also, view the process as adiabatic. The energy balance reduces to

E K  W river where W river is the power of the flowing water. The actual power being provided by the stream can be calculated using the efficiency information. Let  represent the efficiency.

59

W

  s Wriver

W 5 kW  W river  s   10 kW  0.5  The value of W river should be negative since the water is supplying work that is stored electrical energy. Therefore, the energy balance becomes

E K  10000 W  This expression can be rewritten as





1 2 2 m V2 V1  10000 W  2 From the problem statement and the assumptions made,

 kg  m  200    s   m V2  0   s Therefore,

 m V1  10   s There are a number of reasons for the low conversion efficiency. A possible potential energy loss inherent in the design of the energy conversion apparatus decreases the efficiency. Heat is lost to the surroundings during conversion. Some of the energy is also lost due to friction (drag) effects.

60

2.33 Considering the turbine to be the system, rearrangement of the steady-state, open system energy balance provides

 nout (h  ek  e p )out   nin (h  ek  e p )in  Q  W s out

in

Performing a mass balance reveals n in  n1  n out  n 2 Assuming the rate of heat transfer and potential energy effects are negligible and realizing that there is one inlet and one outlet allows the simplification of the above equation to W s  n 2 h2  h1   e K , 2  e K ,1 

h2  h1  can be rewritten using Equations 2.58 and Appendix A.2

h2  h1    c p dT  R  A  BT  CT 2  DT  2  ET 3 dT T2 T1





Since the quantity ek ,2  ek ,1 is multiplied by n , it is rewritten as follows for dimensional homogeneity

e

k ,2



 2 2 1  ek ,1   ( MW ) air V2  V1 2



To solve for n , the ideal gas law is used P2V2  n 2 RT21 P V n 2  2 2 RT2

To solve for the volumetric flow rate, the fluid velocity must be multiplied by the cross-sectional area

  D2 2V2    V2     4  The energy balance is now

61

   D 2V   T2  2 2  P  1 2 3 2 2 2 2     Ws  R A  BT  CT  DT  ET dT   ( MW ) air V2  V1   RT2  4     2  T1   Substituting values from Table A.2.1 and the problem statement results in





W  4.84  10 6 [W]  -4.84 [MW]

62





2.34 First, a sketch of the process is useful:

30 bar

20 bar

100 oC

150 oC

q

To find the heat in we will apply the 1st law. Assuming steady state, the open system energy balance with one stream in and one stream out can be written: 0  n h1  h2   Q which upon rearranging is: Q  h2  h1 n Thus this problem reduces to finding the change in the thermodynamic property, enthalpy from the inlet to the outlet. We know 2 intensive properties at both the inlet and outlet so the values for the other properties (like enthalpy!) are already constrained. From Table A.2.1, we have an expression for the ideal gas heat capacity: cp R

 1.424  14.394  10  3 T  4.392  10  6 T 2

with T in (K). Since this expression is limited to ideal gases any change in temperature must be under ideal conditions. From the definition of heat capacity:





T2 Q  h2  h1   1.424  14.394 10  3 T  4.392 10  6 T 2 dT n T1

By integrating and substituting the temperatures, we obtain: Q  J   5590  n  mol 

63

2.35 A schematic of the process follows: WÝs T1 = 350 oC

T1 = 308 oC P2= 1 atm

v1 = 600 cm3/mol

To solve for W s / n we need a first law balance. With negligible eK and eP, the 1st law for a steady state process becomes: 0  n h1  h2   Q  W s If heat transfer is negligible, W s  h n We can calculate the change in enthalpy from ideal gas heat capacity data provided in the Appendix.





T2 T2 W s  h   c p dT R  1.213  28.785  10  3 T  8.824  10  6 T 2 dT n T1

T1

Integrate and evaluate: W s  J   5358  n  mol 

64

2.36 (a) First start with the energy balance. Nothing is mentioned about shaft work, so the term can be eliminated from the energy balance. The potential and kinetic energy effects can also be neglected. Since there is one inlet and one outlet, the energy balance reduces to

Q  n 2 h2  n1h1 A mass balance shows n 2  n1

so the energy balance reduces to Q  n1 h2  h1  Using the expressions from Appendix A.1, the energy balance becomes T2





Q  n1R  A  BT  CT 2  DT  2  ET 3 dT T1

Using A  3.376 B  0.557  10  3 C 0 D  0.031 105 E 0  J  R  8.314   mol  K   mol  n1  20   s  T1  373.15 K  T2  773.15 K  gives Q  245063 W   245.1 kW 

65

(b) To answer this question, think about the structure of n-hexane and carbon monoxide. N-hexane is composed of 20 atoms, but carbon monoxide has two. One would expect the heat capacity to be greater for n-hexane since there are more modes for molecular kinetic energy (translational, kinetic, and vibrational). Because the heat capacity is greater and the rate of heat transfer is the same, the final temperature will be less.

66

2.37 First start with the energy balance around the nozzle. Assume that heat transfer and potential energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance reduces to n 2 (h  e K ) 2  n1 (h  e K )1  0

A mass balance shows n1  n 2

On a mass basis, the energy balance is



 1  hˆ2  hˆ1  eˆK ,1  eˆK ,2  V12  V2 2 2



Since the steam outlet velocity is much greater than the velocity of the inlet, the above expression is approximately equal to

 

1  hˆ2  hˆ1   V2 2 2

The change in enthalpy can be calculated using the steam tables.  J  h1  2827.9  103    kg   J  h2  2675.5  103    kg 

(10 bar, 200 ºC) (sat. H2O(v) at 100 kPa)

Therefore,

 m V2  552   s To solve for the area, the following relationship is used  AV2 m  vˆ2

From the steam tables

67

 m3  ˆv2  1.6940    kg  Now all but one variable is known.

 

A  3.07 10 3 m 2

68

2.38 First start with the energy balance around the nozzle. Assume that heat transfer and potential energy effects are negligible. The shaft work term is also zero. Therefore, the energy balance reduces to

n 2 (h  ek ) 2  n1 (h  ek )1  0 The molar flow rates can be eliminated from the expression since they are equal. Realizing that e K ,2  e K ,1 since the velocity of the exit stream is much larger than the velocity of the inlet stream simplifies the energy balance to h2  h1  ek ,2 Using Appendix A.2 and the definition of kinetic energy T2





 1 h2  h1  R  A  BT  CT 2  DT  2  ET 3 dT   ( MW ) C 3 H 8 V22 2 T1

From Table A.2.1 A  1.213 B  28.785  10  3 C  8.824  10  6 D0 E 0 It is also important that the units for the molecular weight and universal gas constant are consistent. The following values were used  J  R  8.314   mol  K   kg  ( MW ) C3 H 8  0.0441   mol  Integration of the above expression and then solving for T2 provides T2  419.2 K 

69

2.39 First start an energy balance around the diffuser. Assume that heat transfer and potential energy effects are negligible. The shaft work term is also zero. The energy balance reduces to

n 2 (h  ek ) 2  n1 (h  ek )1  0 A mass balance reveals n1  n 2

The molar flow rates can be eliminated from the expression. Using the definitions of enthalpy and the kinetic energy, the equation can be rewritten as



T2

 2 2 1 c dT   ( MW ) V 2  V1 P air  2



T1

The temperature and velocity of the outlet stream are unknown, so another equation is needed to solve this problem. From the conservation of mass,







  P1V1 P1 A1V1 P2 A2V2   T1 T1 T2



where A2, the cross-sectional area of the diffuser outlet, is twice the area of the inlet. Therefore,  1  P  T   V2   1  2 V1 2  P2  T1 

Using Appendix A.2 and the above expression, the energy balance becomes T2



R  A  BT  CT  DT T1

2

2

2    1  P1  T2    1   ET dT   ( MW ) air    V1   V12    2  P2  T1    2   3



Substituting values from the problem statement provides an equation with one unknown: T2  381 K

Therefore,  1  1 bar  381 K  m 300 m/s   111   V2    2  1.5 bar  343.15 K  s

70

2.40 To find the minimum power required for the compressor, one must look at a situation where all of the power is used to raise the internal energy of the air. None of the power is lost to the surroundings and the potential and kinetic energy effects must be neglected. Therefore, the energy balance becomes

0  n1h1  n 2 h2  W s Performing a mass balance reveals n1  n 2

The energy balance reduces to

W s  n1 h2  h1  Using Equation 2.58 and Appendix A.2, the equation becomes T2





W s  n1R  A  BT  CT 2  DT  2  ET 3 dT T1

Table A.2.1 and the problem statement provide the following values

A  3.355 B  0.575  10 3 C 0 D  1600 E 0  mol  n1  50   s  T1  300 K

To find the work, we still need T2. We need to pick a reasonable process to estimate T2. Since the heat flow is zero for this open system problem, we choose an adiabatic, reversible piston situation. For this situation, PV k  const.

Since we are assuming the air behaves ideally, we can rewrite the equation as

71

k

 n RT   n RT  P1  1 1   P2  2 2   P1   P2 

k

P11 k T1k  P21 k T2k

Substituting values from the problem statement, we obtain

1 7 / 5   7 / 5 1 bar  T2  300 K   10 bar 1 7 / 5  

5/7

 579 K

Substitute this value into the expression for the work and evaluate: W s  417.4 kW 

72

2.41 (a) Perform a mass balance:

n1  n 2  n out Apply the ideal gas law: P1V1 P2V2   n out RT1 RT2 Substitute values from the problem statement:

1105 5 10 3   2 105 2.5 10 3   nout

8.314373.15 8.314293.15 n out  0.366 mol/s

(b) No work is done on the system, and we can neglect potential and kinetic energy effects. We will assume the process is also adiabatic. The energy balance reduces to

0   nin hin   n out hout in

out

n out hout  n1h1  n 2 h2   n1 hout  h1   n 2 hout  h2   0 We can calculate the enthalpy difference from the given ideal heat capacity:

 3.267  5.324  10

Tout

n1 R

373.15

3



 3.267  5.324  10

Tout

T dT  n 2 R

3



T dT  0

293.15

Again, we must calculate the molar flow rates from the ideal gas law. Upon substitution and evaluation, we obtain Tout  329 K 

73

2.42 (a) Since the temperature, pressure, and volumetric flow rate are given, the molar flow rate is constrained by the ideal gas law.

  

Note: 1.67  10 8 m 3 /s  1 cm 3 / min n 

 



PV 1.0135  10 5 Pa 1.67  10 8 m 3 /s    7.45  10 7 mol/s 8.314 J/mol  K 273.15 K  RT

To recap, we have shown 1 SCCM   7.45  10 7 mol/s (b) Assumptions:

N2 is an ideal gas All power supplied by the power supply is transferred to the N2 Uniform temperature radially throughout sensor tube Kinetic and potential energy effects negligible in energy balance

Let x represent the fraction of N2 diverted to the sensor tube, and n s represent the molar flow rate through the sensor tube. Therefore, the total molar flow rate, ntotal , is n ntotal  s x We can use temperature and heat load information from the sensor tube to find the molar flow rate through the sensor tube. First, perform an energy balance for the sensor tube: H  n s hout  hin   Q The enthalpy can be calculated with heat capacity data. Therefore, n s 

Q T2

 cP dT

T1

Now, we can calculate the total molar flow rate.

74

Q

ntotal 

T2

x  c P dT T1

To find the flow rate in standard cubic centimeters per minute, apply the conversion factor found in Part (a) Q

vtotal (SCCM ) 

T2

x  c P dT

  1 SCCM     7.45 10  7 mol/s   

T1

(c) To find the correction factor for SiH4, re-derive the expression for flow rate for SiH4 and then divide it by the expression for N2 for the same power input, temperatures, and fraction of gas diverted to the sensor tube.

Q T2

Factor 

vtotal , SiH 4 vtotal , N 2

x  c P, SiH 4 dT 

T1

Q T2

x  c P, N 2 dT

  1 SCCM     7.45  10  7 mol/s   

  1 SCCM     7.45  10  7 mol/s   

T2

 c P, N



2

dT

T1 T2

 c P, SiH

4

dT

T1

T1

If we assume that heat capacities are constant, the conversion factor simplifies: Factor 

c P, N 2 c P, SiH 4

Using the values in Appendix A.2.2 at 298 K, we get Factor  0.67

75

2.43 (a) It takes more energy to raise the temperature of a gas in a constant pressure cylinder. In both cases the internal energy of the gas must be increased. In the constant pressure cylinder work, Pv work must also be supplied to expand the volume against the surrounding’s pressure. This is not required with a constant volume. (b) As you perspire, sweat evaporates from your body. This process requires latent heat which cools you. When the water content of the environment is greater, there is less evaporation; therefore, this effect is diminished and you do not feel as comfortable.

76

2.44 From the steam tables at 10 kPa:

T(K)

h

323.15

2592.6

373.15

2687.5

423.15

2783

473.15

2879.5

h vs. T 6000 y = 0.0003x

5000

523.15

2977.3

573.15

3076.5

4000

673.15

3279.5

3000

773.15

3489

873.15

3705.4

973.15

3928.7

1073.15

4159.1

1173.15

4396.4

1273.15

4640.6

1373.15

4891.2

1473.15

5147.8

1573.15

5409.7

+ 1.6241x + 2035.7 2

R =1

Series1

2000

Poly. (Series1) 1000 0 0

 dh  cP     1.6241  0.0006624T  dT  P

2

500

1000 T

 kJ   J   kg  K   3.516  0.001434T R  mol  K     

Now compare the above values to those in Appendix A.2. A

B

Steam Tables

3.516

0.001434

Appendix A

3.470

0.001450

% difference

1.3

1.4

77

1500

2000

2.45 For throttling devices, potential and kinetic energy effects can be neglected. Furthermore, the process is adiabatic and no shaft work is performed. Therefore, the energy balance for one inlet and one outlet is simplified to n1 h1  n 2 h2

which is equivalent to m 1 hˆ1  m 2 hˆ2 Since mass is conserved hˆ1  m 2 hˆ2 From the steam tables:  kJ  hˆ1  3398.3    kg   kJ   hˆ2  3398.3    kg 

(8 MPa, 500 ºC)

Now that we know uˆ 2 and P2 , T2 is constrained. Linear interpolation of steam table data gives T2  457 º C

78

2.46 (a) An expression for work in a reversible, isothermal process was developed in Section 2.7. Equation 2.77 is P  W  nRT ln 2   P1 

Therefore, P  w  RT ln 2   P1 

Evaluating the expression with  J  R  8.314   mol  K  T  300 K  P2  100 kPa  P1  500 kPa 

gives  J  w  4014    kg  (b) Equation 2.90 states

w

R T2  T1  k 1

Since the gas is monatomic 5 cP   R 2 3 cv    R 2 and

79

k

5 3

T2 can be calculated by applying the polytropic relation derived for adiabatic expansions. From Equation 2.89

PV k  const  Pv k  const By application of the ideal gas law RT P Since R is a constant, substitution of the expression for P into the polytropic relation results in v

P 1 k T k  const  P11 k T 1 k  P2 1 k T2 k This relation can be used to solve for T2 . T2  157.6 K 

Now that T2 is known, value of work can be solved.  J  w  1775.9    kg 

80

2.47 (a) The change in internal energy and enthalpy can be calculated using

hˆ  hˆ l 1 atm, 100 º C   hˆ l 1 atm, 0 º C  uˆ  uˆ l 1 atm, 100 º C   uˆ l 1 atm, 0 º C  We would like to calculate these values using the steam tables; however, the appendices don’t contain steam table data for liquid water at 0.0 ºC and 1 atm. However, information is provided for water at 0.01 ºC and 0.6113 kPa. Since the enthalpy and internal energy of liquid water is essentially independent of pressure in this pressure and temperature range, we use the steam table in the following way  kJ  hˆ l 1 atm, 100 º C   419.02    kg   kJ  hˆ l 1 atm, 0 º C   hˆ l 0.6113 kPa , 0.01 º C   0    kg   kJ  uˆ l 1 atm, 100 º C   418.91    kg   kJ  uˆ l 1 atm, 0 º C   uˆ l 0.6113 kPa , 0.01 º C   0    kg  Therefore,  kJ  hˆ  419.02    kg   kJ  uˆ  418.91    kg 

(b) The change in internal energy and enthalpy can be calculated using

hˆ  hˆ v 1 atm, 100 º C   hˆ l 1 atm, 100 º C  uˆ  uˆ v 1 atm, 100 º C   uˆ l 1 atm, 100 º C  From the steam tables

81

 kJ  hˆ v  2676.0    kg   kJ  hˆ l  419.02    kg   kJ  uˆ v  2506.5    kg   kJ  uˆ l  418.91    kg  Therefore,  kJ  hˆ  2256.99    kg   kJ  uˆ  2087.59    kg 

The change in internal energy for the process in Part (b) is 5.11 times greater than the change in internal energy calculated in Part (a). The change in enthalpy in Part (b) is 5.39 times greater than the change in enthalpy calculated in Part (a).

82

2.48 To calculate the heat capacity of Ar, O2, and NH3 the following expression, with tabulated values in Table A.2.1, will be used,

cP  A  BT  CT 2  DT  2  ET 3 R

where T is in Kelvin. From the problem statement T  300 K 

and from Table 1.1  J  R  8.314   mol  K  To find the A-E values, Table A.2.1 must be referred to. Formula Ar O2 NH3

A 3.639 3.5778

B  103 0.506 3.02

C  10 6 0 0

D  10 5 -0.227 -0.186

E 109 0 0

The values are not listed for Ar since argon can be treated as a monatomic ideal gas with a heat capacity independent of temperature. The expression for the heat capacity is 5 c P, Ar    R 2 Now that expressions exist for each heat capacity, evaluate the expressions for T  300 K  .  J  c P, Ar  20.785   mol  K   J  c P, O2  29.420   mol  K   J  c P, NH 3  35.560   mol  K  By examining the heat capacity for each molecule, it should be clear that the magnitude of the heat capacity is directly related to the structure of the molecule.

83

Ar

O2



Since argon is monatomic, translation is the only mode through which the atoms can exhibit kinetic energy.



Translation, rotation, and vibration modes are present. Since oxygen molecules are linear, the rotational mode of kinetic energy contributes RT per mol to the heat capacity.

NH3 

Translation, rotation, and vibration modes are present. Ammonia molecules are nonlinear, so the rotation mode contributes 3RT/2 per mole to the heat capacity.

The vibration contributions can also be analyzed for oxygen and ammonia, which reveals that the vibration contribution is greatest for ammonia. This is due to ammonia’s non-linearity.

84

2.49 For a constant pressure process where potential and kinetic energy effects are neglected, the energy balance is given by Equation 2.57:

Q  H

The change in enthalpy can be written as follows H  mh2  h1 

From the steam tables:  kJ  hˆ2  hsat. vapor at 10 kPa   2584.6    kg   kJ  hˆ1  hsat. liquid at 10 kPa   191.81    kg  Therefore,   kJ    kJ  Q  2 kg  2584.6    191.81     kg    kg   Q  4785.6 kJ 

We can find the work from its definition: Vf

W    PE dV Vi

The pressure is constant, and the above equation can be rewritten as follows W   PE mvˆ2  vˆ1 

From the steam tables:  m3  ˆv2  vˆsat. vapor at 10 kPa   14.674    kg   m3  vˆ1  vˆsat. liquid at 10 kPa   0.00101    kg  Therefore,

85

  m3     W  10000 Pa 2 kg 14.674    0.00101     293.5 kJ   kg   kg   

86

2.50 First, perform an energy balance on the system. Potential and kinetic energy effects can be neglected. Since nothing is mentioned about work in the problem statement, W can be set to zero. Therefore, the energy balance is

Q  U

Performing a mass balance reveals m2  m1

where m1  m1l  m1v

Now the energy balance can be written as



Q  m1 (uˆ 2  uˆ1 )  m1uˆ 2  m1l uˆ1l  m1v uˆ1v



Since two phases coexist initially (water is saturated) and P1 is known, state 1 is constrained. From the saturated steam tables  kJ  uˆ1l  191.79    kg   kJ  uˆ1v  2437.9  

(sat. H2O at 10 kPa)

 kg 

As heat is added to the system, the pressure does not remain constant, but saturation still exists. One thermodynamic property is required to constrain the system. Enough information is known about the initial state to find the volume of the container, which remains constant during heating, and this can be used to calculate the specific volume of state 2. l l v v V2 V1 m1 vˆ1  m1 vˆ1 v2    m2 m1 m1l  m1v

From the saturated steam tables

87

 m3  ˆv1l  0.001010    kg   m3  vˆ1l  14.674    kg 

(sat. H2O at 10 kPa)

Therefore,  m3  vˆ2  1.335    kg  The water vapor is now constrained. Interpolation of steam table data reveals  kJ  uˆ 2  2514.6    kg  Now that all of the required variables are known, evaluation of the expression for Q is possible. Q  11652 kJ 

88

2.51 Let the mixture of ice and water immediately after the ice has been added represent the system. Since the glass is adiabatic, no work is performed, and the potential and kinetic energies are neglected, the energy balance reduces H  0

We can split the system into two subsystems: the ice (subscript i) and the water (subscript w). Therefore, H  mi hi  mw hw  0 and mi hi   mw hw

We can get the moles of water and ice.

 

Vw 0.0004 m 3   0.399 kg  vˆw   m3    0.001003       kg    mw 0.399 kg   22.15 mol nw   MW H 2O 0.0180148  kg   mol  mi ni   5.55 mol MW H 2O mw 

Now, let’s assume that all of the ice melts in the process. (If the final answer is greater than 0 ºC, the assumption is correct.) The following expression mathematically represents the change in enthalpy.











ni c P, i 0   10 º C   h fus  c P, w T f  0 º C  nwc P, w T f  25 º C Note: Assumed the heat capacities are independent of temperature to obtain this expression.

From Appendix A.2.3 c P ,i  4.196 R c P ,w  9.069 R

and  kJ  h fus  6.0   mol 

89

Substitution of values into the above energy balance allows calculation of Tf. T f  3.12 º C

(Our assumption that all the ice melts is correct.) (b) To obtain the percentage of cooling achieved by latent heat, perform the following calculation

Fractionlatent 



ni  h fus





 nw c P, w T f  Tw, i



5.55 mol 6000 

J    mol    Fractionlatent   0.911  J  3.12 º C  25 º C  22.15 mol9.069  8.314   mol  K  Percentlatent  91.1%

90

2.52 A mass balance shows n2  n1l  n1v

To develop the energy balance, neglect kinetic and potential energy. Also, no shaft work is performed, so the energy balance becomes

U  Q The energy balance can be expanded to





Q  n1l  n1v u 2  n1l u1l  n1v u1v

If the reference state is set to be liquid propane at 0 ºC and 4.68 bar, the internal energies become u1l  0 u1v  u vap 0 º C  u 2  u vap 0 º C  

T2

 c P  R dT

273 K

Once the change in internal energy for vaporization and temperature of state 2 is determined, Q can be solved. As the liquid evaporates, the pressure increases. At state 2, where saturated propane vapor is present, the ideal gas law states P2 

RT2 v2

To find v2 , assume that v1v  v1l . The volume of the rigid container is V  n1v v1v  n1v

 m3  RT1  0.00485   P1  mol 

Therefore, v2 

 m3   0.00243   n1l  n1v  mol 



V



91

Also, since the propane is saturated, P2 and T2 are not independent of each other. They are related through the Antoine Equation,

 

B

ln P sat  A  T

sat

C

where P sat  P2 and T sat  T2

Substitution provides,  RT  B ln 2   A  T2  C  v2 

 bar  m 3  Using values from Table A.1.1 and R  8.314  10  5    mol  K  T2  301.7 K

To find u vap , refer to the definition of enthalpy. hvap  h v  h l  (u  Pv) v  u  Pv l Since v v  v l , the above the change in internal energy of vaporization can be written as u vap  h vap  Pv v  h vap  RT Therefore,  kJ  u vap 0 º C   14.39   mol  Evaluation of the following equation after the proper values have been substituted from Table A.2.1





301.7 K   Q  n1l  n1v u vap 0 º C    c P  R dT   n1l 0  n1v uvap 0 º C    273 K

92

gives Q  18.1 kJ 

93

2.53 The equation used for calculating the heat of reaction is given in Equation 2.72. It states

 i

 hrxn   vi h f

This equation will be used for parts (a)-(e). Since the heat of reaction at 298 K is desired, values from Appendix A.3 can be used. (a) First the stoichiometric coefficient must be determined for each species in the reaction.

vCH 4 ( g )  1 vO2 ( g )  1 vCO2 ( g )  1 v H 2 O( g )  1

From Tables A.3.1 and A.3.2 kJ  hf ,298 CH ( g )  74.81  mol  kJ  hf ,298 O ( g )  0  mol  kJ  hf ,298 CO ( g )  393.51  mol  kJ  hf ,298 H O( g )  241.82  mol  4

2

2

2

From Equation 2.72, the equation for the heat of reaction is

















     hrxn ,298  vCH 4 ( g ) h f ,298 CH ( g )  vO2 ( g ) h f ,298 O ( g )  vCO2 ( g ) h f ,298 CO ( g )  v H 2 O ( g ) h f ,298 H O ( g ) 4 2 2 2          kJ   kJ   kJ   kJ   hrxn    0      393.51      241.82   , 298    74.81      mol     mol     mol     mol   

 kJ   hrxn , 298  560.52   mol  Now that a sample calculation has been performed, only the answers will be given for the remaining parts since the calculation process is the same.

94

(b)

 kJ   hrxn , 298  604.53   mol  (c)

 kJ   hrxn , 298  206.12   mol  (d)

 kJ   hrxn , 298  41.15   mol  (e)

 kJ   hrxn , 298  905.38   mol 

95

2.54 The acetylene reacts according to the following equation

C2H2(g) + (5/2)O2(g)  2CO2(g) + H2O(g) (a) First, choose a basis for the calculations. nC 2 H 2  1 mol

Calculate the heat of reaction at 298 K using Equation 2.72 and Appendix A.3

 i  hrxn  h f  C H  hrxn   vi h f 2

 O  2hf CO  hf H O

 2.5 h f

2

2

2

2

 J   hrxn  1.255 10 6   mol 





 H rxn,298  nC 2 H 2 hrxn  1.255 10 6 J 

The required amount of oxygen is calculated as follows

nO 1  2.5(nC H 2

2

2

)1  2.5 mol

The compositions for both streams are Streams 1 (Inlet) 2 (Outlet)

nC 2 H 2

nO2

nN 2

nCO2

nH 2O

1 0

2.5 0

0 0

0 2

0 1

D  10 5 -1.299 -0.227 -1.157 0.121

E  109 0 0 0 0

From Table A.2.2 Species C2H2 O2 CO2 H2O

A 6.132 3.639 5.457 3.470

B  103 1.952 0.506 1.045 1.45

C  10 6 0 0 0 0

Integration of the following equation provides an algebraic expression where only T2 is unknown.

96

T2

H rxn,298 

  ni 2 c P i dT  0

298 i

Substituting the proper values into the expression gives T2  6169 K

(b) The calculations follow the procedure used in Part (a), but now nitrogen is present. The basis is nC 2 H 2  1 mol

The heat of reaction is the same as in Part (a), but the gas composition is different. Since stoichiometric amount of air is used,

nO 1  2.5 mol 2

 yN   n N 2  nO2  2   9.40 mol 1 1 y   O2  air

   

The composition of the streams are summarized below Streams

nC 2 H 2

nO2

nN 2

nCO2

nH 2O

1 2

1 0

2.5 0

9.40 9.40

0 2

0 1

D  10 5 -1.299 -0.227 -1.157 0.121 0.04

E  109 0 0 0 0 0

From Appendix A.2 Species

A

C2H2 O2 CO2 H2O N2

6.132 3.639 5.457 3.470 3.280

B  103 1.952 0.506 1.045 1.45 0.593

C  10 6 0 0 0 0 0

Therefore, T2  2792 K

97

(c) Now excess air is present, so not all of the oxygen reacts. The heat of reaction remains the same because only 1 mole of acetylene reacts. Since the amount of air is twice the stoichiometric amount

nO 1  5 mol 2

 yN   n N 2  nO2  2   18.80 mol 1 1 y   O2  air

   

The compositions of the streams are summarized below Streams

nC 2 H 2

nO2

nN 2

nCO2

nH 2O

1 2

1 0

5 2.5

18.8 18.8

0 2

0 1

The table of heat capacity data in Part (b) will be used for this calculation. Using the expression shown in Part (a) T2  1787 K

98

2.55 (a) The combustion reaction for propane is

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) For all subsequent calculations, the basis is one mole of propane. The heat of reaction is calculated as follows

 C H  5hf O  3hf CO  4hf H O

 hrxn   h f

3

8

2

2

2

 J   hrxn  2.044  10 6   mol 





 H rxn,298  nC 3 H 8 hrxn  2.044 10 6 J 

The required amount of oxygen for complete combustion of propane is

nO 1  5(nC H 2

3

8

)1  5 mol

 yN   n N 2  nO2  2   18.81 mol 1 1 y   O2  air The stream compositions are listed below

    Streams 1 (Inlet) 2 (Outlet)

nC 3 H 8

nO2

nN 2

nCO2

nH 2O

1 0

5 0

18.8 18.8

0 3

0 4

D  10 5 0.04 -1.157 0.121

E  109 0 0 0

From Table (a)2.2 Species N2 CO2 H2O

A 3.280 5.457 3.470

B  103 0.593 1.045 1.45

C  10 6 0 0 0

Now all of the necessary variables for the following equation are known, except T2 . T2

H rxn,298 

  ni 2 c P i dT  0

298 i

Solving the resulting expression provides T2  2374 K

99

(b) The combustion reaction for butane is

C4H10(g) + (13/2)O2(g)  4CO2(g) + 5H2O(g) For all subsequent calculations, the basis is one mole of butane. The heat of reaction is calculated as shown in Part (a) H rxn,298  2.657  10 6 J 

The moles of nitrogen and oxygen in the feed stream are calculated according to the method in Part (a). The compositions are Streams 1 (Inlet) 2 (Outlet)

nC 4 H 10

nO2

nN 2

nCO2

nH 2O

1 0

6.5 0

24.5 24.5

0 4

0 5

The c P data listed in Part (a) can also be used for this reaction since there is no remaining butane. T2  2376 K

(c) The combustion reaction for pentane is

C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(g) The basis is one mole of pentane. The heat of reaction is calculated as shown in Part (a). H rxn,298  3.272  10 6 J 

The moles of nitrogen and oxygen in the feed stream are calculated according to the method in Part (a). The compositions are listed below Streams 1 2

nC 5 H 12

nO2

nN 2

nCO2

nH 2O

1 0

8 0

30.1 30.1

0 5

0 6

Substitution of the values into the expression used to find T2 and subsequent evaluation results in T2  2382 K

The adiabatic flame temperatures are nearly identical in all three cases. 100

2.56 The equation for the combustion of methane is

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Using Equation 2.72  J   hrxn  8.02  105   mol  The basis for this problem is

nCH 1  1 mol 4

Also, let  represent the fractional conversion of methane. Therefore, the composition of the product gas leaving the reactor is

nCH 2  11    mol nO 2  21    mol nN 2  7.52 mol nCO 2   mol nH O 2  2 mol 4

2

2

2

2

Furthermore, the heat of reaction is calculated as follows





H rxn,298    8.02 105 J 

From Table A.2.2 Species

A

CH4 O2 CO2 H2O N2

1.702 3.639 5.457 3.470 3.280

B  103 9.081 0.506 1.045 1.45 0.593

C  10 6 -2.164 0 0 0 0

D  10 5 0 -0.227 -1.157 0.121 0.04

E  109 0 0 0 0 0

After substitution of the outlet composition values, heat capacity data, and the heat of reaction into the following equation

101

1273

H rxn, 298 

  n  c  dT  0 i 2

P i

298 i

integration provides an equation with one unknown:  . Solving the equation gives

 =0.42 Since the fractional conversion is 0.42, 58% of the methane passed through the reactor unburned.

102

2.57 For the entire cycle,

U11  U12  U 23  U 31  0

 U12  50 kJ  From state 1 to state 2 U12  Q12  W12  Q12  400 kJ  From state 2 to state 3

U 23  Q23  W23  Q23  0 kJ  From state 3 to state 1 U 31  Q31  W31

W31  250 kJ 

Hence, the completed table is Process State 1 to 2 State 2 to 3 State 3 to 1

U kJ  -50 800 -750

W kJ  -400 800 -250

Q kJ  350 0 -500

To determine if this is a power cycle or refrigeration cycle, look at the overall heat and work, W11 and Q11 . W11  W12  W23  W31  150 kJ 

Q11  Q12  Q23  Q31  150 kJ 

Since work is done on the system to obtain a negative value of heat, which means that heat is leaving the system, this is a refrigeration cycle.

103

2.58 Refer to the graph of the Carnot cycle in Figure E2.20. From this graph and the description of Carnot cycles in Section 2.9, it should be clear that state 3 has the lowest pressure of all 4 states, and state 1 has the highest pressure. States 1 and 2 are at the higher temperature. States 3 and 4 have the lower temperature. Since both the temperature and pressure are known for states 1 and 3, the molar volume can be calculated using

v

RT P

The table below summarizes the known thermodynamic properties.



State

T K 

P bar 

1 2 3 4

1073 1073 298 298

60

v m 3 /mol 0.00149

0.2

0.124



For each step of the process, potential and kinetic energy effects can be neglected. The step from state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in internal energy is 0, and the energy balance becomes Q12  W12

From Equation 2.77, P  W12  nRT1 ln 2   P1 

where W12 is the work done from state 1 to state 2. The value of P2 is not known, but recognizing that the process from state 2 to 3 is an adiabatic expansion provides an additional equation. The polytropic relationship can be employed to find P2 . A slight modification of Equation 2.89 provides Pv k  const From the ideal gas law P

RT v

Combining this result with the polytropic expression and noting that R is constant, allows the expression to be written as

104

Tv k 1  const Therefore, 1

T  k 1 v2   3 v3k 1   T2  Substituting the appropriate values (k=1.4) gives  m3  v2  0.00504    mol  Applying the ideal gas law P2 

RT2  17.7 bar  v2

Now, W12 can be calculated. W12  10.89 kJ 

Calculation of W34 follows a completely analogous routine as calculation for W12 . The following equations were used to find the necessary properties 1  m3  T  k 1  0.0367  v4   1 v1k 1    T4   mol 

P4 

RT4  0.675 bar  v4

Now the following equation can be used P  W34  nRT3 ln 4   P3 

which gives W34  3.01 kJ 

105

For an adiabatic, reversible process, Equation 2.90 states W

nR T2  T1  k 1

This equation will be used to calculate the work for the remaining processes. nR T3  T2   16.11 kJ k 1 nR T1  T4   16.11 kJ  W41  k 1

W23 

To find the work produced for the overall process, the following equation is used Wnet  W12  W23  W34  W41 Evaluating this expression with the values found above reveals Wnet  7.88 kJ  Therefore, 7.88 kilojoules of work is obtained from the cycle. The efficiency of the process can be calculated using Equation 2.98:



Wnet 7.88   0.72 QH 10.89

since QH  W12  10.89 kJ  . Alternatively, if we use Equation E2.20D.

  1

TC TH

where TC  T3  T4 and TH  T1  T2 . Upon substitution of the appropriate values

  0.72

106

2.59 Since this is a refrigeration cycle, the direction of the cycle described in Figure 2.17 reverses. Such a process is illustrated below: Constant T H QH

Isothermal compression Adiabatic compression State 2 Well - insulated

T2=T1 P2

T1 P1

State 3 State 1 T1 P1

Adiabatic expansion

State 1

T3 P3

State 4 T4 P4

State 4 T4=T3 P4

State 2 State 3 T3 T2 P3 P2

Isothermal expansion QC Constant T C

States 1 and 2 are at the higher temperature. States 3 and 4 have the lower temperature. Since the both the temperature and pressure are known for states 2 and 4, the molar volume can be calculated using v

RT P

The following table can be made State

T K 

1 2 3 4

1073 1073 298 298



P bar 

v m 3 /mol

60

0.00149

0.2

0.124



For each step of the process, potential and kinetic energy effects can be neglected. The process from state 1 to state 2 is a reversible, isothermal expansion. Since it is isothermal, the change in internal energy is 0, and the energy balance becomes Q12  W12

107

From Equation 2.77, P  W12  nRTH ln 2   P1 

where W12 is the work done from state 1 to state 2. The value of P1 is not known, but recognizing that the process from state 4 to 1 is an adiabatic compression provides an additional relation. The polytropic relationship can be employed to find P1 . A slight modification of Equation 2.89 provides Pv k  const From the ideal gas law P

RT v

Combining this result with the polytropic expression and noting that R is constant allows the expression to be written as Tv k 1  const Therefore, 1

T  k 1 v1   4 v 4k 1   T1  Substituting the appropriate values (k=1.4) gives 1 k 1

T   m3  v1   4 v 4k 1   0.00504    mol   T1  RT P1  1  17.7 bar  v1

Now, W12 can be calculated. W12  10.9 kJ 

Calculation of W34 follows a completely analogous routine as the calculation for W12 . The following equations were used to find the necessary properties: 108

 m3  v3  0.0367    mol 

Applying the ideal gas law P3 

RT3  0.675 bar  v3

Now the following equation can be used P  W34  nRTC ln 4   P3 

which gives W41  3.0 kJ 

Equation 2.90 can be used to determine the work for adiabatic, reversible processes. This equation will be used to calculate the work for the remaining processes. nR TC  TH   16.11 kJ  k 1 nR TH  TC   16.11 kJ  W41  k 1 W23 

To find the work produced for the overall process, the following equation is used Wnet  W12  W23  W34  W41 Evaluating this expression with the values found above reveals Wnet  7.88 kJ  Therefore, 7.88 kJ of work is obtained from the cycle. The coefficient of performance is defined in Equation 2.99 as follows COP 

QC Wnet

where QC is the equal to Q34 . From the energy balance developed for the process from state 3 to state 4

109

Q34  W34  3.01 kJ  Therefore, COP 

3.01 kJ   0.382 7.88 kJ 

110

2.60 (a) The Pv path is plotted on log scale so that the wide range of values fits (see Problem 1.13)

logP 100

0.075

4

1

3 2

log v v (b) The work required to compress the liquid is the area under the Pv curve from state 3 to state 4. Its sign is positive. The power obtained from the turbine is the area under the curve from state 1 to 2. Its sign is negative. The area under the latter curve is much larger (remember the log scale); thus the net power is negative. (c) First, perform a mass balance for the entire system:

m 1  m 2  m 3  m 4  m Since no work is done by or on the boiler, the energy balance for the boiler is m hˆ1  m hˆ4  Q H

Similarly, the energy balance for the condenser is m hˆ3  m hˆ2  Q C To find the necessary enthalpies for the above energy balances, we can use the steam tables:  kJ  hˆ1  3424.5    kg 

(520 ºC, 100 bar)

111

   kJ    kJ    kJ  hˆ2  0.10168.77     0.90 2574.8     2334.2    kg    kg    kg    (sat. liq at 7.5 kPa) (sat. vap. at 7.5 kPa)

 kJ  hˆ3  168.77    kg 

(sat. liquid at 0.075 bar)

 kJ  hˆ4  342.81    kg 

(subcooled liquid at 80 ºC, 100 bar)

Now, we can calculate the heat loads:   kJ    kJ  Q H  100 kg/s  3424.5    342.81     308169 kW   kg    kg     kJ    kJ  Q C  100 kg/s 168.77    2334.2     216543 kW   kg    kg  

(d) Use Equation 2.96:

W net  Q net  0 From Part (c), we know Q net  308169 kW   216543 kW   91626 kW  Therefore, W net  91626 kW  (e) Using the results from Parts (c) and (d): η

91626 kW   0.297 308169 kW 

112



3.1 Since entropy is a state function s sys  s sys, step1  s sys, step 2

Step 1 is a constant volume process. Therefore, no work is done. After neglecting potential and kinetic energy effects, the energy balance for a reversible process becomes

u  q  du  q cv dT  q Using the definition of entropy, final



s sys ,step1 

initial

q rev T



T2

cv dT  T T1

Step 2 is an isothermal process. For and ideal gas u is zero and the energy balance is (PE and KE neglected)

q  w  dq  w For an ideal gas, the following can be shown

wrev   Pdv  

RT dv v

Therefore, final

s sys ,step 2 



initial

qrev T



v2

R

 v2   1 

 v dv  R ln v

v1

Combination of both steps yields T2

s sys 



T1

v  cv dT  R ln 2  T  v1 

2

3.2 Equation 3.62 states T2

s sys 

cP

T

T1

P  dT  R ln 2   P1 

Substituting the equation for c P yields T2

s sys 



T1

P  A  BT  CT 2 dT  R ln 2  T  P1 





P  T  C s sys  A ln 2   BT2  T1   T22  T12  R ln 2  2  P1   T1 

3

3.3 Equation 3.3 states S univ  S sys  S surr

where S sys  m( sˆ2  sˆ1 )

S surr 

Qsurr Tsurr

(the temperature of the surroundings is constant)

First, we will calculate S sys . From the steam tables:  kJ  sˆ1  7.1228    kg  K 

(300 ºC, 10 bar)

Since the container is rigid and mass is conserved in the process, vˆ2  vˆ1

 m3  vˆ1  0.25794    kg 

(300 ºC, 10 bar)

 m3   vˆ2  0.25794    kg  To find the number of phases present in state 2, compare the specific volume of state 2 to the specific volume for saturated water and saturated water vapor at 1 bar. From the steam tables,

 m3  vˆ2l , sat  0.001043    kg 

(sat. H2O(l) at 1 bar)

 m3  vˆ2v, sat  1.6940    kg 

(sat. H2O(v) at 1 bar)

Since vˆ2l , sat  vˆ2  vˆ2v, sat , two phases are present. The quality of the water can be calculated as follows vˆ2  1  x vˆ2l , sat  xvˆ2v, sat

Therefore,

4

x  0.152


 kJ  sˆ2l , sat  1.3025    kg  K   kJ  sˆ2v, sat  7.3593    kg  K 

(sat. H2O(l) at 1 bar) (sat. H2O(v) at 1 bar)

The entropy of state 2 can be calculated using the following equation: sˆ2  1  x sˆ2l , sat  xsˆ2v, sat  kJ  sˆ2  2.2231    kg  K 

Now the entropy change of the system can be calculated.   kJ   kJ   kJ   49.0   S sys  10 kg  2.2231   7.1228       K  kg  K   kg  K   

To find the change in entropy of the surroundings, an energy balance will be useful. Since no work is done, the energy balance is

U  Q We also know

Qsurr  Q To following expression is used to solve for the internal energy:



 

U  m 1  x uˆ 2l , sat  xu 2v, sat  uˆ1

From the steam tables

 kJ  uˆ1  2793.2    kg   kJ  uˆ 2l , sat  417.33    kg 

(300 ºC, 10 bar) (sat. H2O(l) at 1 bar)

5

 kJ  uˆ 2v, sat  2506.1    kg 

(sat. H2O(v) at 1 bar)

The expression for internal energy yields U  20583.77 kJ 

Therefore, Q  20583.77 kJ 

and S surr 

Qsurr 20583.77 kJ   kJ    70.25   Tsurr 293 K  K

Now the change in entropy of the universe can be calculated

 kJ   kJ   kJ  S univ  S sys  S surr  49.0    70.25    21.25   K K K Since Suniv > 0, this process is possible.

6

3.4 Entropy Balance: S univ  S sys  S surr

Considering the copper block to be the system, no work is done on the system; thus, the energy balance is

du  q cv dT  q

(neglecting PE and KE)

This can be used in the following expression for entropy T2

s sys 

q rev



T

T1 T2

s sys 

cv

T

dT

T1

From Table A.2.3 c P  2.723R cv  c P  2.723R

(for liquids and solids)

Therefore, T2

s sys 



T1

T   2.723R  J    280 K    ln dT  2.723R ln 2   2.723 8.314   T  mol  K    373.15 K     T1 

 J  s sys  6.50   mol  K      10 kg    6.50  J    1024  J   S sys   mol  K    K    kg    0 . 063465   mol     Since the temperature of the lake remains constant, the change in entropy of the surroundings can be calculated as follows S surr 

Qsurr  nCu c P T2  T1   2.723nCu RT2  T1  Q    Tsurr Tsurr Tsurr Tsurr

7

   J    280 K   373.15 K    2.723 8.314    10 kg    mol  K     S surr       kg 280 K      0.063546   mol       J S surr  1184   K  From the definition of entropy: J J J S univ  S sys  S surr  1024    1184    160   K  K  K 

8

3.5 (a) Since the process is adiabatic and reversible, S sys  0

(b) The change in entropy is calculated by

liquid



S sys  n

qrev T

vapor



 nh Tb

vap

  kJ    1 mol 8.2    mol   J    73.9   111 K K 

The sign is negative because there is less randomness in the liquid phase. (c) For this situation δq  c P dT

Therefore, Tf

S sys  m  Ti

q rev T

  J    273.15 K  cP    dT 18 . 0148 g 4 . 2   g  K   ln 373.15 K    T     373.15 K  273.15 K



J S sys  23.6   K  The sign is negative because as the water cools, less translational energy states are occupied by the molecules. Therefore, the randomness decreases. (d) First, calculate the temperature at which the blocks (block A and block B) equilibrate. The energy balance for the process is n Ac P T2  373.15 K   n B c P T2  473.15 K   0

Since the heat capacities number of moles of A and B are equal, we find that T2  423.15 K

Now, calculate the change in entropy:

9

S sys  S sys, A  S sys, B

 T   T  S sys  n Ac P ln 2   n B c P ln 2   T1, A   T1, B      Substituting values, we obtain J S sys  0.337   K  The sign is positive because two objects at different temperatures will spontaneously equilibrate to the same temperature when placed together.

10

3.6

The solution below compares problems 2.14 and 2.15, the calculation of 2.13 was erroneously included in the problem statement of the first printing and is shown at the end of this problem. Problem 2.14 Since the system is well-insulated no heat is transferred with the surroundings. Therefore, the entropy change of the surroundings is zero and S univ  S sys

The gas in the piston-cylinder system is ideal and c P is constant, so  T   P  S sys  n c P ln 2   R ln 2   T1   P1   From the problem statement, we know P1  2 bar  P2  1 bar 

The ideal gas law can be used to solve T1 . PV 2 bar 10 L  T1  1 1   240.6 K  nR   L  bar    0.08314  1 mol  mol  K    Solving for T2 is slightly more involved. The energy balance for this system where potential and kinetic energy effects are neglected is U  W

Conservation of mass requires n1  n2 Let n  n1  n2

The energy balance can be rewritten as T2

V2

T1

V1

n  cv dT    PE dV

11

Since cv and PE are constant ncv T2  T1    PE V2  V1  V2 and T1 can be rewritten using the ideal gas law

nRT2 P2 PV T1  1 1 nR

V2 

Substituting these expressions into the energy balance, realizing that PE  P2 , and simplifying the equation gives 5    P2  P1 V1 2  T2   7 nR 2 Using the following values P1  2 bar  P2  1 bar 

V1  10 L

n  1.0 mol  L  bar  R  0.08314   mol  K 

results in T2  206 K 

Since both states are constrained, the entropy can be calculated from Equation 3.63:   J    206 K    J   1 bar   S univ  S sys  1.0 mol 3.5  8.314   ln ln   8.314     mol  K    240.6 K    mol  K   2 bar    J Suniv  S sys  1.24   K 

12

Problem 2.15 Since the initial conditions in the piston-cylinder assembly are equal to the initial conditions of Problem 2.14, T1 and P1 are known. Moreover, P2 is known, so we only need to find T2 in order to calculate the entropy change. For adiabatic, reversible processes, the following relationship (Equation 2.89) holds:

PV k  const This can be used to find V2 . 1 k P V2   1 V1k    P2

c 7 Noting that k  P  and substituting the proper values results in cv 5 V2  16.4 L

Th polytropic expression can also be manipulated to yield

TV k 1  const. Therefore, T2 

T1V1k 1 V2k 1

Substitution of the appropriate variables provides T2  197.4 K 

Now the entropy can be calculated,   J   1 bar    J    197.4 K   S univ  S sys  1.0 mol 3.5  8.314  ln    8.314   ln   mol  K   2 bar    mol  K    240.6 K   

J S univ  S sys  0.004   K  The value of the entropy shown represents round-off error. Since the process is reversible and adiabatic, we know from Table 3.1 and the related discussion in Section 3.3 that the entropy changes of the system, surroundings, and universe will be zero. 13

Problem 2.13

Equation 3.3 states S univ  S sys  S surr  m( sˆ2  sˆ1 ) 

Qsurr Tsurr

where the temperature of the surroundings is constant. First, we will determine s 2 . The first law can be applied to constrain state 2. With potential and kinetic energy effects neglected, the energy balance becomes U  Q  W The value of the work will be used to obtain the final temperature. The definition of work (Equation 2.7) is V2

W    PE dV V1

Since the piston expands at constant pressure, the above relationship becomes W   PE V2  V1 

From the steam tables  kJ  sˆ1  6.2119   (10 MPa, 400 ºC)  kg  K   m3  vˆ1  0.02641    kg 

(10 MPa, 400 ºC)

 

  m3   3 V1  m1vˆ1  (3 kg) 0.02641     0.07923 m    kg   

Now V2 and v2 are found as follows

 

 748740 J W  0.4536 m3  0.07923 m3  6 PE 2.0  10 Pa 3  m3  V 0.4536 m ˆv2  2   0.1512   m2 3 kg   kg 

V2  V1 

 

14

Since vˆ2 and P2 are known, state 2 is constrained. From the steam tables:  kJ  sˆ2  7.1270    kg  K 

  3   20 bar, 0.1512  m     kg   

Now U will be evaluated, which is necessary for calculating Qsurr . From the steam tables:  kJ  uˆ 2  2945.2    kg 

  3   20 bar, 0.1512  m     kg   

 kJ  uˆ1  2832.4    kg 

100 bar, 400 º C

  kJ    kJ  U  m1 uˆ 2  uˆ1   3 kg  2945.2    2832.4     338.4 kJ   kg    kg  

Substituting the values of U and W into the energy equation allows calculation of Q Q  U  W Q  338400 [J]   748740 J   1.09  10 6 J   Qsurr

so S univ  m( sˆ2  sˆ1 ) 

Qsurr Tsurr

  kJ   1.09  103 kJ   kJ   kJ    1.126     3 kg  7.1270  6 . 2119     673.15 K  K  kg K    kg K  

Therefore, this process is irreversible.

15

3.7 (a)

From Equation 3.63: T  P  s sys  c P ln 2   R ln 2   T1   P1    0.5 bar    J    7  500 K    J  s sys   8.314    ln   20.63    ln   mol  K    2  300 K    mol  K   1 bar   

(b) From Equation 3.65: T  v  s sys  cv ln 2   R ln 2   T1   v1     3    0.025  m        J    5  500 K    mol    4.85  J   ln s sys   8.314    ln         mol  K   m 3     mol  K    2  300 K       0.05    mol      

(c) First, we can find the molar volume of state 1 using the ideal gas law.

v1 

 m3  RT1  0.025   P1  mol 

Now, we can use Equation 3.65 T  v  s sys  cv ln 2   R ln 2   T1   v1     3    0.025  m      mol    J   J    5  500 K      10.62   ln s sys   8.314  ln          m 3     mol  K    2  300 K    mol  K      0.025     mol    

16

3.8 (i). We wish to use the steam tables to calculate the entropy change of liquid water as it goes from its freezing point to its boiling point. The steam tables in Appendices B.1 – B.5 do not have data for subcooled water at 1 atm. However, there is data for saturated water at 0.01 ºC and a pressure of 0.6113 kPa. If we believe that the entropy of water is weakly affected by pressure, then we can say that the entropy of water at 0.01 ºC and 0.6113 kPa is approximately equal to the entropy at 0 ºC and 1 atm. The molar volumes of most liquids do not change much with pressure at constant temperature. Thus, the molecular configurations over space available to the water molecules do not change, and the entropy essentially remains constant. We do not need to consider the molecular configurations over energy since the temperature difference is so slight. So, from the steam tables:

 kJ  sˆ0 º C , 1 atm   sˆ0.01 º C , 0.6113 kPa   0    kg  K   kJ  sˆ100 º C , 1 atm   1.3068    kg  K  Therefore,  kJ   kJ  sˆ  sˆ100 º C , 1 atm   sˆ0 º C , 1 atm   1.3068  0    kg  K   kg  K   kJ  sˆ  1.3068    kg  K  (ii). From the steam tables:  kJ  sˆl 100 º C , 1 atm   1.3068    kg  K   kJ  sˆ v 100 º C , 1 atm   7.3548    kg  K  Therefore,  kJ   kJ  sˆ  sˆ v 100 º C , 1 atm   sˆ l 100 º C , 1 atm   7.3548   1.3068     kg  K   kg  K   kJ  sˆ  6.048    kg  K  The change in entropy for process (ii) is 4.63 times the change in entropy for process (i). There are many ways to reconcile this difference, but think about it from a molecular point of view. In process (i), the available molecular configurations over energy are increased as the temperature increases. As the temperature increases, the molar volume also increases slightly, so the 17

available molecular configurations over space also increase. Now consider process (ii), where the molecules are being vaporized and entering the vapor phase. The molecular configuration over space contribution to entropy is drastically increased in this process. In the liquid state, the molecules are linked to each other through intermolecular interactions and their motion is limited. In the vapor state, the molecules can move freely. Refer to Section 3.10 for a discussion of entropy from a molecular view.

18

3.9 Before calculating the change in entropy, we need to determine the final state of the system. Let the mixture of ice and water immediately after the ice has been added represent the system. Since the glass is adiabatic, no work is performed, and the potential and kinetic energies are neglected, the energy balance reduces H  0

We can split the system into two subsystems: the ice (subscript i) and the water (subscript w). Therefore, H  mi hi  mw hw  0 and mi hi   mw hw

We can get the moles of water and ice.

 

Vw 0.0004 m 3   0.399 kg  vˆw   m3    0.001003       kg    mw 0.399 kg   22.15 mol  nw  MW H 2O 0.0180148  kg   mol  mi  5.55 mol ni  MW H 2O

mw 

Now, let’s assume that all of the ice melts in the process. (If the final answer is greater than 0 ºC, the assumption is correct.) The following expression mathematically represents the change in internal energies (cP=cv).





ni c P ,i 0   10 º C   h fus  c P ,w T f  0 º C   nw c P ,w T f  25 º C  Note: Assumed the heat capacities are independent of temperature to obtain this expression. From Appendix A.2.3 c P ,i  4.196 R c P ,w  9.069 R and  kJ  h fus  6.0   mol 

19

Substitution of values into the above energy balance allows calculation of Tf. T f  3.12 º C (Our assumption that all the ice melts is correct.) Now, we can calculate the change in entropy. From Equation 3.3 suniv  s sys  s surr Since the glass is considered adiabatic, s surr  0 suniv  s sys We will again break the system into two subsystems: the ice and the water. The change in entropy of the universe can be calculated as follows S univ  ni si  nw s w The definition of change in entropy is final

s 



initial

δqrev T

Assuming the heat capacities of ice and water are independent of temperature, the expressions for the change in entropy of the subsystems are h fus  276.27 K   273.15 K   c P, w ln si  c P, i ln    273.15 K   263.15 K  273.15 K  276.27 K  s w  c P, w ln   298.15 K  Therefore h fus    273.15 K   276.27 K   276.27 K  S univ  ni c P, i ln  c P, w ln    n w c P, w ln   263.15 K  273.15 K  273.15 K   298.15 K    Substituting the values used before, we obtain

20

 J  Suniv  6.59   mol 

21

3.10 (a) The maximum amount of work is obtained in a reversible process. We also know the entropy change for the universe is zero for reversible processes. From the steam tables

 kJ  sˆ1  6.2119    kg  K   kJ  sˆ2  8.5434    kg  K 

(400 ºC, 100 bar) (400 ºC, 1 bar)

Using the entropy criterion, Suniv  0  msˆ2  sˆ1   S surr   kJ   kJ   S surr  msˆ2  sˆ1   0.5 kg  8.5434   6.2119      kg  K   kg  K     kJ  S surr  1.1658   K

Since the process is isothermal, we know that the temperature of the surroundings is constant at 400 ºC. Therefore, Qsurr  kJ   S surr  1.1658   Tsurr K Qsurr  784.76 kJ 

To find the work obtained, perform an energy balance. An energy balance where potential and kinetic energy effects are neglected is

U  Q  W We can calculate the change in internal energy from the steam tables, and we also know that Q  Qsurr . From the steam tables:  kJ  uˆ1  2832.4    kg   kJ  uˆ 2  2967.8    kg 

(400 ºC, 100 bar) (400 ºC, 1 bar)

Therefore,

22

W  muˆ 2  uˆ1   Q   kJ    kJ   2832 . 4  W  0 . 5 kg  2967 . 8     784 . 76 kJ    kg    kg   W  717.06 kJ 

(b) If steam is modeled as an ideal gas, the change in internal energy is zero. Therefore, the energy balance is

Q  W The work can be found using Equation 2.77 which is developed for reversible, isothermal processes. P  W  nRT ln 2   P1   0.5 kg   J   1 bar  W  8.3145  673.15 K ln     mol  K    100 bar   kg     0.0180148    mol    W  715332 J   715.3 kJ 

Therefore, Q  715.3 kJ 

and

Qsurr  715.3 kJ 

Since the temperature of the surroundings is constant,  kJ  S surr  1.06   K

23

3.11 (a) The initial pressure is calculated as follows

P1  Psurr 



25000 kg  9.81 m/s 2 0.05 m

2

  24.6 105 Pa

Similarly, the final pressure is P2  Psurr 



35000 kg  9.81 m/s 2 0.05 m 2

  34.4  105 Pa

(b) The temperature should rise, which can be understood by considering an energy balance. Because the system is insulated, the work done by the mass being added to the piston is transformed into molecular kinetic energy. (c) The final temperature can be calculated with an energy balance: U  W ncv T2  T1   W

Since the pressure of the surroundings is constant after the block is added to the piston, the work is calculated as follows: W   P2 V2  V1 

Assume ideal gas behavior: V2 

nRT2 P2

V1 

nRT1  0.00169 m 3 P1

Now, we can create one equation with one unknown:  nRT2  n5 / 2 R T2  T1    P2   V1   P2 

Substitute values and solve for T2: T2  557 K

24

(d) Since the system is well-insulated

s surr  0 Use Equation 3.65 to calculate the change in entropy: T  V  T  T P  suniv  s sys  cv ln 2   R ln 2   5 / 2 R ln 2   R ln 2 1   T1   V1   T1   T1P2 

Substituting values, we obtain,  J  s sys  0.354   mol  K  (e) Because the change in entropy of the universe is equal to change in entropy of the system, which is positive in this situation, the second law is not violated.

25

3.12 (a) To calculate ssys, we need to take the gas from state 1 to state 2 using a reversible process. The process in part a can be drawn as follows: 24.6 bar 500 K

State 1

34.4 bar 557 K rev. adiabatic

State 2 rev. isothermal

PI T I = 557 K State I

We need to find the intermediate pressure, PI where we end up at the temperature in state 2, T2. To find it, we can use the results from pages 78-79 of the text: T1P11 k  / k  TI PI1 k  / k  T2 PI1 k  / k Therefore, k

 T  1k  P1  35.9 [bar] PI   1   T2 

(I)

If we draw the process on a PT diagram, we get: 40 P I 2 Adiabatic compression

Isothermal expansion

1 20 490

560 T

26

The change in entropy can be represented as follows: s sys  s adiabatic  sisothermal

For the reversible adiabatic process, the change in entropy is zero. Therefore, s sys  sisothermal

For an isothermal process,

q rev  wrev  

RT dP P

Therefore, final



s sys 

q

initial

T

P2

  PI

P  R  J  dP   R ln 2   0.354  P  mol  K   PI 

If we substitute relation (I) in the expression above, we get the same expression in the book: P s sys   R ln 2  P1

 T  T  P  k   R ln 2   c P ln 2   R ln 2   1  k   T1   T1   P1 

(b) For this construction, we use the path diagrammed on the following PT diagram: 40 P

2 Isothermal compression

Isobaric heating 1 20 490

560 T

We write

27

s sys  sisobar  sisothermal

First, find an expression for the isobaric heating

q rev  dh  c P dT Therefore, T2

sisobar 

T 

c

 TP dT  c P ln T12 

T1

Now, we need an expression for the isothermal, reversible expansion

q rev  wrev  Pdv Therefore, v 2  RT2 / P2

sisothermal 



v I  RT2 / P1

P dv  T

v 2  RT2 / P2



v1  RT2 / P1

P  P  R dv  R ln 1    R ln 2  v  P2   P1 

Combine the two steps: T  P   J  s sys  c P ln 2   R ln 2   0.354   mol  K   T1   P1 

(c) For this construction, we use the path diagrammed on the following PT diagram:

28

40 P

2 Isothermal compression I

Isochoric heating

1 20 490

560 T

For isochoric heating followed by an isothermal expansion, the entropy can be expressed as follows: s sys  sisochoric  sisothermal

For the reversible, isothermal expansion, we obtain (refer to Parts (a) and (b) to see how this is derived) P  sisothermal   R ln 2   PI  However, we can relate the intermediate pressure to the initial pressure through the ideal gas law:

PI P1  T2 T1 or PI  27.4 [bar]

     P2  T  P  P2    sisothermal   R ln    R ln  R ln 2   R ln 2   T2   PI   T1   P1  P1    T1 

For the isochoric process:

q rev  cv dT Therefore,

29

T2

sisochoric 

cv

T 

 T dT  cv ln T12 

T1

Combining these results: T s sys  cv  R  ln 2  T1

 P   R ln 2   P1

 T  P   J    c P ln 2   R ln 2   0.354   mol  K    T1   P1 

Using any of these three reversible paths, we get the same answer!

30

3.13 (a) Since the process is reversible and adiabatic, the entropy change for the process is zero.

s sys  0 Furthermore, since the process is adiabatic, the changes in entropy of the surroundings and the universe are also zero. (b) Since this process is isentropic (s=0), we can apply an expression for the entropy change of an ideal gas. T2 c P s  0   P dT  Rln 2 P1 T T 1

Insert the expression for cP from Appendix A

P  3.639  0.506 10 3 T  0.227 10 6 T 2 dT  R ln 2  T  P1  T1

T2

s  0  R 

which upon substituting

T1  250 K  P1  1 bar 

P2  12.06 bar  yields T2  482 K 

(c) An energy balance gives T2

T2

T1

T1

w  u   cv dT 

 c

P

 1dT

31

 J  w  R  2.639  0.506  10 3 T  0.227  10 6 T 2 dT  5390   mol  T1 T2

(d) Reversible processes represent the situation where the minimum amount of energy is required for compression. If the process were irreversible, more work is required, and since the process is adiabatic, the change in internal energy is greater. Since the change in internal energy is greater, so too is the change in temperature. Therefore, the final temperature would be higher than the temperature calculated in Part (b).

32

3.14 The problem statement states that the vessel is insulated, so we can assume that heat transfer to the surroundings is negligible. Therefore, the expression for the entropy change of the universe is suniv  s sys

An energy balance will help us solve for the entropy change. Neglecting potential and kinetic energy effects, the energy balance is

U  Q  W Since the vessel is insulated, the heat term is zero. Furthermore, no work is done, so the energy balance is U  muˆ 2  uˆ1   0  u1  u 2

The values for the initial pressure and temperature constrain the value of specific energy at state 1. From the steam tables,

 kJ  uˆ1  2619.2    kg   kJ   uˆ 2  2619.2    kg 

(400 ºC, 200 bar)

The problem statement provides the pressure of state 2 (100 bar). Since we know the pressure and internal energy at state 2, the entropy is constrained. Information given in the problem statement also constrains state 1. From the steam tables,

 kJ  sˆ1  5.5539    kg  K   kJ  sˆ2  5.7754    kg  K 

(400 ºC, 200 bar)   kJ   100 bar, uˆ 2  2619.2       kg   

Therefore,

 kJ  sûniv  sˆsys  sˆ2  sˆ1  0.2215    kg  K   kJ  S univ  S sys  0.2215   K

33

3.15 The subscript “2” refers to the final state of the system. “1” refers to the gas initially on one side of the partition. If you take the system to be the entire tank (both sides of the partition), then no net work is performed as the gas leaks through the hole. Furthermore, the tank is well insulated, and kinetic and potential energy effects can be neglected. Thus, the energy balance is u  0

and T2  T1

(ideal gas)

Initially, the partition is divided into two equal parts. The gas fills the entire volume for the final state ( V2  2V1 ). The ideal gas law can be used to calculate the moles of gas present.

n

 

P1V1 10 105 Pa 0.5 m 3   200.5 mol  RT1   J  300 K   8.314   mol  K   

Now, the change in entropy can be solved using a slight modification of Equation 3.65:

3   nv  T   v  T  S sys  n cv ln 2   R ln 2   n  R ln 2   R ln 2   nv1   T1   v1   T1  2    J   3  S sys  200.5 mol 8.314    ln1  ln2   mol  K    2   J S sys  1155   K 

34

3.16 (a) A schematic of the process is shown below: 1 bar, 298 K

N2

N2 N2

N2

P2, T2

1 bar, 298 K

O2

Mixing Process

O2

N2

N2

N2 N2

N2

N2

N2

N2

O2

O2

N2 N2

N2 N2

The tank is insulated. The change in entropy of the universe can be rewritten as S univ  S sys, N 2  S sys, O2

since the tank is well-insulated. After the partition ruptures, the pressure and temperature will remain constant at 1 bar and 298 K, respectively. This can be shown by employing mass and energy balances. Therefore, the partial pressures in the system are

p2, N 2  0.79 bar p2,O2  0.21 bar Use Equation 3.62 to determine the entropies: 298 K

s sys, N 2 

 P2  cP   J    0.79 bar      8.314    ln ln dT R   T P1  N mol  K    1 bar     298 K 2

 J  s sys, N 2  1.96   mol  K  298 K

s sys, O2 

P   cP  J    0.21 bar  dT  R ln 2    8.314   ln  T P1 O mol  K    1 bar     298 K 2



 J  s sys ,O2  12.98   mol  K  Therefore,

   J   J  S univ  0.79 mol1.96    0.21 mol12.98     mol  K    mol  K    

35

J S univ  4.27   K  (b) A schematic of the process is shown below: 2 bar, 298 K

N2

N2

N2

N2

N2

N2

O2

N2

N2 N2

N2

N2

N2 N2

P2, T 2

1 bar, 298 K

N2

N2

Mixing Process

N2

N2

N2

N2 N2

N2 N2

O2

N2

N2

O2 N2

N2

N2 N2

N2 O2 N2

N2

N2

Before calculating the change in entropy, we need to find how the temperature and pressure change during the process. The energy balance simplifies to U  0

which can be rewritten as

nO

2



 n N 2 cvT2  n N 2 cv 298 K   nO2 cv 298 K   0

Assuming the heat capacities are equal, we can show that T2  298 K

We can find the pressure after the rupture by recognizing that the tank is rigid. Therefore, VO2  V N 2  Vtot

By employing the ideal gas law, we get the following equation (it has been simplified):

nN 2 P1, N 2



nO2 P1, O2



nO

2

 nN 2



P2

Substitute values and solve to obtain P2  1.65 bar

Now, use Equation 3.62 to calculate the entropies as was done in Part (a):

36

298 K

s sys, N 2 

P   cP  J    1.30 bar  dT  R ln 2    8.314   ln  T P1  N mol  K    2 bar     298 K 2



 J  s sys, N 2  3.58   mol  K  P   cP  J    0.347 bar  dT  R ln 2    8.314   ln  T  mol  K    1 bar    P1 O2 298 K

298 K

ssys ,O2 



 J  s sys ,O2  8.8   mol  K  Therefore,    J   J  S univ  0.79 mol 3.58    0.21mol 8.8     mol  K    mol  K     J Suniv  4.68   K

37

3.17 First start with the energy balance for the throttle. Potential and kinetic energy effects can be neglected. During the throttling process, no shaft work is performed and the rate of heat transfer is negligible. Therefore, m 2 hˆ2  m 1hˆ1

A mass balance allows the energy balance to be simplified to hˆ2  hˆ1 From the steam tables:  kJ  hˆ1  3398.3    kg   kJ   hˆ2  3398.3    kg 

(500 ºC, 8 MPa)

Now calculate entropy: sûniv  sˆsys  sˆsurr

Since the process is adiabatic,  kJ  sˆsurr  0    kg  K  The steam tables can be used to calculate the change in entropy of the system.  kJ  sˆ2  8.7098    kg  K   kJ  sˆ1  6.7239    kg  K 

  kJ   100 kPa, hˆ2  3398.3       kg   

(8 MPa, 500 ºC)

Thus,  kJ  sûniv  sˆsys  sˆ2  sˆ1  1.9859    kg  K 

38

3.18 For this process to work, conservation of mass and the first and second laws of thermodynamics must hold. The subscript “1” refers to the inlet stream, “2” refers to the cold outlet, and “3” refers to the hot outlet. To test the conservation of mass, perform a mass balance

m 1  m 2  m 3  kg   kg   kg  2    0.5    1.5   s   s  s  Clearly, the conservation of mass holds. Now test the first law of thermodynamics by writing an energy balance. Since there is no heat transfer or work, the energy balance becomes m 2 hˆ2  m 3hˆ3  m 1hˆ1  0

(PE and KE effects neglected)

Using the conservation of mass we can rewrite the mass flow rate of stream 1 in terms of streams 2 and 3:









m 2 hˆ2  hˆ1  m 3 hˆ3  hˆ1  0

If we assume that the heat capacity of the ideal gas is constant, the equation can be written as follows: m 2 cˆP T2  T1   m 3cˆP T3  T1   0 Therefore,  kg   kg  0.5   cˆP (60 K)  1.5   cˆP (20 K)  0 s  s  This proves that the first law holds for this system. For the second law to be valid, the rate of change in entropy of the universe must greater than or equal to 0, i.e., dS  0 dt univ Assuming the process is adiabatic, we can write the rate of entropy change of the universe for this steady-state process using Equations 3.48-3.50:  dS     m 2 sˆ2  m 3 sˆ3  m 1sˆ1  m 2 sˆ2  sˆ1   m 3 sˆ3  sˆ1   dt univ

39

where a mass balance was used. For constant heat capacity, we can calculate the entropy differences using Equation 3.63: 5  T   P  P  T  s2  s1  cP ln 2   R ln 2   R  ln 2   ln 2   P1   P1   T1   2  T1  5  T   P  P  T  s3  s1  cP ln 3   R ln 3   R  ln 3   ln 3   P1   P1   T1   2  T1  Applying these relations, we get: 1  dS  2.77 R  0.057cP      dt univ MW

Therefore, the second law holds if cP  48.6 R ; this value is far in excess of heat capacity for gases, so this process is possible.

40

3.19 (a) T1 = 640 oC P2 = 100 kPa

P1 = 4 MPa V1= 20 m/s

Nozzle

V2=?

(b) Since the process is reversible and adiabatic,

J S sys  0   K (c) The steam tables can be used to determine the final temperature. From the interpolation of steam table data

 kJ  sˆ2  sˆ1 640 º C, 4 MPa   7.4692    kg  K  Now two thermodynamic properties are known for state 2. From the steam tables:   kJ    P2  0.1 MPa, sˆ2  7.4692      kg K   

T2  121.4 º C

(d) An energy balance is required to calculate the exit velocity. The energy balance for the nozzle is









0  m 2 hˆ  eˆK 2  m 1 hˆ  eˆ K 1 Realizing that m 1  m 2 allows the energy balance to be written as





eˆK ,2  hˆ  eˆ K 1  hˆ2 which is equivalent to





  V2  2 hˆ  0.5V 2 1  hˆ2



Substituting the following values

41

 m V1  20   s

 J  hˆ2  2719100    kg   J  hˆ1  3767000    kg 

(0.1 MPa, 121.4 ºC) (4 MPa, 640 ºC)

yields

 m V2  1448   s Note: the velocity obtained is supersonic; however, the solution does not account for this type of flow.

42

3.20 A schematic of the process follows:

Propane in v1 = 600 cm3/mol T1= 350 oC

Turbine

ws Propane P = 1 atm 2 out

Since this process is isentropic (s=0), we can apply an expression for the entropy change of an ideal gas. We must be careful, however, to select an expression which does not assume a constant heat capacity. T2 c P s  0   P dT  Rln 2 P1 T T 1

Insert the expression for cP from Appendix A

P  1.213  28.785  10 3 T  8.824  10 6 T 2 s  0  R  dT  R ln 2  T  P1  T1 T2

We can also relate P1 to T1 and v1 (which are known) through the ideal gas law: P1 

RT1 v1

This leaves us with 1 equation and 1 unknown (T2). Integrating: T  Pv  s  0  1.213 ln 2   28.785  10 3 (T2  T1 )  4.412 10 6 T22  T12  ln 2 1   T1   RT1 



Using T1 = 623 K, v1 = 600 cm3/mol, P2 = 1 atm

43



R=82.06 cm3 atm/mol K, we get: T2 = 454 K

To solve for

W S we need a first law balance, with negligible ke and pe, the 1st law for a steady n

state process becomes: 0  n h1  h2   Q  W S If heat transfer is negligible (isentropic), T2 T2 2 W S  h2  h1   c p dT  R  1.213  28.785  10 3 T  8.824  10 6 T dT n T1 T1





integrating:









W S  R 1.213T2  T1   14.393 10  3 T22  T12  2.94110  6 T23  T13 n Substituting  J  R  8.314   mol  K  T2  454 K T1  623 K

results in W S  J   19860  n  mol 

44



3.21 A reversible process will require the minimum amount of work. For a reversible process,

suniv  0  s sys  s surr For this process, the change in entropy of the system can be calculated as follows s sys  sO2  s N 2

where

sO2 

T2 c



P O2

T1

s N 2 

T2 c



T1

T

P  dT  R ln 2   P1 

P N 2

T

O2

P  dT  R ln 2   P1 

N2

Since the temperature of the streams do not change, the expressions reduce to

P  sO2   R ln 2   P1  O2 P  s N 2   R ln 2   P1  N 2 Therefore,    s surr  R  ln    

P2 P1

 P    ln  2  O2  P1  N 2

   

Since we are assuming the temperature of the surroundings remain constant at 20 ºC, q s surr  surr Tsurr where

 qsurr  q

45

Therefore,

 P   P2   2    q surr  Tsurr s surr  RTsurr ln   ln     P1  O  P1  N 2  2    1 bar    J   1 bar        q surr   8.314  293 . 15 K ln ln        mol  K    0.79 bar  N 2    0.21 bar  O2   J  q surr  4378.2   mol  and

 J  q  4378.2   mol  Energy Balance:

 n

h  nout hout  Q  W S  0

in in

in

out

Because the temperature of the oxygen and nitrogen doesn’t change in the process, the energy balance per mole of feed becomes wS   q

 J  wS  4378.2   mol 

46

3.22 (a) Take the entire container to be the system and assume no heat or work crosses the system boundary.

Energy Balance: U  0

After the oscillation cease, the temperature and pressure on both sides of the piston will be the same assuming the metallic piston is very thin and the thermal conductivity coefficient is large. Now, let’s rewrite the energy balance.

U  n1, A ugas A  n1,B ugas B  0 or n1, A ugas A  ugas B  0 n1,B n1, A PA,1VA,1TB ,1 P T   A,1 B ,1  2.41 n1,B PB ,1VB ,1TA,1 2 PB ,1TA,1 since 2V A,1  VB,1 so T





2 3  2.41 RT2  773.15 K   R  5/2  1.5  10-3 T dT  0 2  373.15 K

T2  585 K

Mass Balance: n A,1  n B,1  n A,2  n B,2 Using the ideal gas law, we get

47

PA,1VA,1 PB ,1VB ,1 PA, 2VA, 2 PB , 2VB , 2 P2Vtot     RTA,1 RTB ,1 RTA, 2 RTB , 2 RT2 since the pressure and temperature of state 2 are equal on both sides. But Vtot  V A,1  VB,1  3V A,1 Using the volume relationships and simplifying, we get P2 

T2  PA,1 2 PB ,1  585 K   10 bar  1 bar       2   3  TA,1 TB ,1  3  773.15 K  373.15 K  

P2  3.56 bar 

(b) The container is well-insulated, so suniv  s sys

The entropy change of the system can be split into two subsystems:

ssys 

n1, A n1,B s gas A  s gas B n1, A  n1,B n1, A  n1,B

Using Equation 3.63 and realizing that c P  cv  R ,

sgas A  sgas B

 P  T  2.41   J  2.5 R ln 2   R ln 2   1.96  3.41   mol  K   PA,1   TA,1 

T  P  1  2 3.5  1.5  10 3 T  J  R   dT  R ln 2   1.51  3.41  TB ,1 T  mol  K   PB ,1  

Therefore,

 J   J   J  suniv  ssys  1.96   1.51   3.47     mol  K   mol  K   mol  K  The process is possible.

48

3.23

Equation 3.3 can be modified to show S univ  S sys  S surr

Let’s first calculate S sys . Before this problem is solved, a few words must be said about the notation used. The system was initially broken up into two parts: the constant volume container and the constant pressure piston-cylinder assembly. The subscript “1” refers to the constant volume container; “2” refers the piston-cylinder assembly. “i" denotes the initial state before the valve is opened, and “f” denotes the final state. First, the mass of water present in each part of the system will be calculated. The mass will be conserved during the expansion process. Since the water in the rigid tank is saturated and is in equilibrium with the constant temperature surroundings (200 ºC), the water’s entropy is constrained. From the steam tables,

 kJ  vˆ1l, i  0.001156    kg   kJ  vˆ1v, i  0.12736    kg   kJ  sˆ1l, i  2.3308    kg   kJ  sˆ1v, i  6.4322    kg 

(Sat. water at 200 ºC)

P sat  1553.8 kPa  Knowledge of the quality of the water and the overall volume of the rigid container can be used to calculate the mass present in the container.

 

 

V1  0.05m1 vˆ1l, i  0.95m1 vˆ1v,i

 

Using the values from the steam table and V1  0.5 m 3 provides m1  4.13 kg 

Using the water quality specification,

49

m1v  0.95m1  3.92 kg  m1l  0.05m1  0.207 kg 

For the piston-cylinder assembly, both P and T are known. From the steam tables  m3  vˆ2, i  0.35202    kg   kJ  sˆ2, i  6.9665    kg  K 

(600 kPa, 200 ºC)

Now enough information is available to calculate the mass of water in the piston assembly. V m2  2  0.284 kg  vˆ2 Now the final state of the system must be determined. It helps to consider what physically happens when the valve is opened. The initial pressure of the rigid tank is 1553.8 kPa. When the valve is opened, the water will rush out of the rigid tank and into the cylinder until equilibrium is reached. Since the pressure of the surroundings is constant at 600 kPa and the surroundings represent a large temperature bath at 200 ºC, the final temperature and pressure of the entire system will match the surrounding’s. In other words,  kJ  sˆ f  sˆ2, i  6.9665    kg 

(600 kPa, 200 ºC)

Thus, the change in entropy is given by S sys  (m2  m1v, i  m1l, i ) sˆ f  m2 sˆ2, i  m1v, i sˆ1v, i  m1l, i sˆ1l, i

Substituting the appropriate values reveals  kJ  S sys  3.05   K Now we calculate the change in entropy of the surroundings. Since the temperature of the surroundings is constant, S surr 

Qsurr Q  Tsurr Tsurr

After neglecting potential and kinetic energy effects, the energy balance becomes 50

U  Q  W The change in internal energy and work will be calculated in order to solve for Q. The following equation shows how the change in internal energy can be calculated. U  (m2  m1v, i  m1l, i )uˆ f  m2 uˆ 2, i  m1v, i uˆ1v, i  m1l, i uˆ1l, i

From the steam tables  kJ  uˆ1l, i  850.64    kg   kJ  uˆ1v, i  2595.3    kg 

 kJ  uˆ f  uˆ 2, i  2638.9    kg 

(sat. H2O at 200 ºC)

(600 kPa, 200 ºC)

Using these values and the values of mass calculated above, U  541.0 kJ 

Calculating the work is relatively easy since the gas is expanding against a constant pressure of 600 kPa (weight of the piston was assumed negligible). From Equation 2.7, Vf

W   PE  dV   PE (V f  Vi ) Vi

where PE  600000 Pa 

 

V f  (m2  m1v, i  m1l,i )vˆ2, i  1.55 m 3

   

 

Vi  0.1 m 3  0.5 m 3  0.6 m 3

Note: vˆ2, i was used to calculate V f because the temperature and pressure are the same for the final state of the entire system and the initial state of the piston-cylinder assembly. The value of W can now be evaluated. W  570 kJ 

51

The energy balance can be used to obtain Q. Q  U  W  541.0 kJ    570 kJ   1111 kJ 

Therefore, S surr 

 1111 kJ   kJ   2.35   473.15 K  K

and  kJ   kJ   kJ  Suniv  S sys  S surr  3.05    2.35    0.70   K K K

52

3.24 Let the subscript “1” represent the state of the system before the partition is removed. It has two components: component a and component b. Subscript “2” represents the system after the partition is removed.

Mass balance: n1, a  n1, b  n2 Energy balance for the adiabatic process: U sys  W













 n1, a cv, a T2  T1, a  n1, b cv, b T2  T1, b   P V2  V1, a  V1, b



The external pressure, P, for the above energy balance is equal to P1,a. To find the final temperature, first find the volumes using the ideal gas law.

V1, a 

V2 

2 mol 8.314 

J  300 K   mol  K     0.05 m 3  1000 kg  9.8 m  s - 2    0.098 m 2  

 

    

4 mol 8.314 

J  T2  mol  K     3.3256  10  4 T2 m 3 2  1000 kg  9.8 m  s    0.098 m 2  

 

    

Substitute the volumes into the energy balance and solve for T2 :

2 mol 3 R T2  300 K   2 mol 3 R T2  300 K   1 105 Pa 3.3256  10  4 T2  0.15 m 3 

2  T2  360.4 K

2 

To calculate the change in entropy, we can use the following relationship   T   P  T   P  S sys  na c P ln 2   R ln 2   nb c P ln 2   R ln 2   T1   P1  b  T1   P1  a  

The only unknown in the above equation is P1,b, so we can calculate it with the ideal gas law:

53

P1, b 

J  300 K   mol  K  

2 mol 8.314  

0 .1 m

3

 0.5  10 5 Pa

Substitute values into the expression for entropy and solve:  5  360.4 K   5  360.4 K   1 bar   1 bar    ln S sys  2 molR  ln    2 molR  ln   ln  0.5 bar  b  1 bar  a  2  300 K   2  300 K 

J S sys  3.72   K 

54

3.25 First perform an energy balance on the process: U sys  W

The change in internal energy can be written: U  n A, f u A, f  nB , f u B , f  n A,i u A mass balance gives: n A,i  n A , f  nB , f These two expressions can be substituted to give: U  n A, f cv TA, f  Ti   nB , f cv TB , f  Ti   W

5 R , we obtain 2 5 5 5 PB, f VB, f  PA, f V A, f  PA, iV A, i  W 2 2 2

Using the ideal gas law and cv 

Now, find an expression for the work Vf

W    PE dV Vi

The pressure is not constant; its value is given by k VB  VB ,i  mg kx mg   Patm   A2 A A A 5 5 PE  1.06  10  8.284  10 VB Pa  PE  Patm 

Integrating W  1.06 105VB, f  4.142 105VB2, f

Substituting this expression into the energy balance gives: 5 5 5 PB, f VB, f  PA, f V A, f  PA,iV A, i  1.06  105VB, f  4.142  105VB2, f 2 2 2

55

This expression can be simplified by recognizing that PB, f  PA, f and V A, f  V A, i :





5 5 PB, f VB, f  V A, i  PA, iV A, i  1.06 105VB, f  4.142  105VB2, f 2 2 We still have two unknowns: PB, f and VB, f . We can eliminate PB, f by writing a force balance for the final state: PB, f  1.06 105  8.284 105VB, f Pa 

Substituting this into the energy balance, we obtain







5 5 1.06  105  8.284 105VB, f VB, f  V A, i  PA, iV A, i  1.06 105VB, f  4.142 105VB2, f 2 2 Solving for the final volume: VB, f  0.268 m 3

and for the final pressure. PB, f  PA, f  Pf  1.06  105  8.284  105 0.268  3.28  105 Pa

Since the gas in A has undergone an adiabatic, reversible expansion S sys, A  0

Therefore, T  P  0  c P ln 2   R ln 2   T1   P1   3.28  10 5 Pa     J    7  T2  0   8.314    ln   ln  7.0  10 5 Pa   mol  K    2  313.15 K    

Solve for T2: T2  252 K

56

3.26 A schematic of the process is drawn: Q surr=69.86 kW Steam in P1 = 60 bar T1 = 500 oC

W

Turbine Steam out

P2 = 10 bar

Tsurr=300 K

(a) and (b) The maximum work occurs for a reversible process. Applying the second law, we get:

 dS   dS   dS        0  dt  univ  dt  sys  dt  surr

or m sˆ2  sˆ1  

Q surr 0 Tsurr

Calculate the change in entropy of the surroundings:  kJ  We can look up property values of state 1 from the steam tables sˆ1  6.8802   and  kgK   kJ  hˆ1  3,422.13   . Converting the units of mass flow rate gives:  kg    kg   1 hr    kg  m   4,500      1.25    hr   3,600 s   s  

so sˆ2  sˆ1 

 kJ  Q surr  6.6939   m Tsurr  kgK 

We now know two properties of steam ( sˆ2 and P2) From the steam tables: T2  200 oC

Using the steam tables for superheated steam at 1 MPa, we find that when water has this value of entropy

57

 kJ  hˆ2  2827.9    kg 

Energy balance:





m hˆ2  hˆ1  Q  W s

Calculate Ws:





  kJ   kJ   W s  m hˆ2  hˆ1  Q  1.25 kg/s  2827.9    3422.1      69.86 kW   kg   kg    W s  673 kW 

(c) The isentropic efficiency is defined as follows: η

W S actual W S reversible

For this situation,

W S actual  0.665 673 kW   447.5 kW  (d) The real temperature should be higher since not as much energy is converted into work. (e) Use the energy balance:





m hˆ2  hˆ1  Q  W s  kJ   kJ  Q  W s ˆ  69.86 kW  -447.5 kW  h1   3422.1    3008.2   hˆ2  1.25 kg/s m  kg   kg 

At 1 MPa, water has this value of enthalpy when

T2 actual  280.2 º C

58

3.27 Since the compressor is adiabatic, the energy balance after neglecting potential and kinetic energy becomes

n h2  h1   W S

Using the ideal gas law and Appendix A.2, the above equation becomes





T P1V1 2  WS  A  BT  CT 2  DT  2  ET 3 dT T1  T1

Substituting the following values P1  1 bar   1 105 Pa   m3  V1  1    s  T1  293.15 K  T2  473.15 K  A  3.355; B  0.575 10 3 ; C  0; D  0.016 105 ; E  0 (Table A.2.2)

gives W S  218.8 kW 

This is the work of our real turbine (with 80% isentropic efficiency). We can use the isentropic efficiency to calculate the work of an equivalent 100% efficient process.

W 

compressor   S reversible WS 

compressor

W S reversible  W S compressor compressor  0.8218.8 kW  W S reversible  175.4 kW 

Since the process is adiabatic, we can use the following equation again to calculate what the final temperature would be in a reversible process.



T P1V1 2  WS  A  BT  CT 2  DT  2  ET 3 dT  T1 T1

59



P V 175400 W   1 1 T1

 A  BT  CT

T2

2

 DT  2  ET 3 dT



T1

Substituting the values from above and solving for T2 , we obtain T2, rev  164.8 K

To obtain the pressure the final pressure for the real compressor, we can calculate the final pressure for the reversible process because the final pressure is the same in both cases. For the isentropic expansion T2

s  0 

P  dT  R ln 2  T  P1  T1



cp

 A  BT  CT 2  DT  2  ET 3  P  s  0  R    dT  R ln 2  T  P1   T1  T2





 T2  438 K    P2  3.355  0.575  10  3 T  1600T  2  dT ln s  0  R       1 bar      T     293 . 15 K T   1 

Solve for P2: P2  4.16 bar 

60

3.28 Isentropic efficiency for a turbine is defined as



ws actual ws rev

If the rate of heat transfer is assumed negligible, the energy balance for this process is





0  m hˆ1  hˆ2  W s

For a reversible, adiabatic process, s sys  0

sˆ2,rev  sˆ1 500 º C, 10 MPa 

From the steam tables  kJ  sˆ2,rev  sˆ1  6.5965    kg  K 

Since sˆ2l , sat  sˆ2  sˆ2v, sat , a mixture of liquid and vapor exists. The quality of the water can be calculated as follows sˆ2,rev  1  x sˆ2l ,sat  xsˆ2v ,sat

  kJ    kJ   kJ      x 7.3593  6.5965   1  x 1.3025        kg K kg K      kg  K       x  0.874

Therefore, hˆ2,rev  1  x hˆ2l ,sat  xhˆ2v ,sat    kJ    kJ   hˆ2,rev  1  0.874  417.44     0.874 2675.5     kg    kg      kJ  hˆ2,rev  2391.0    kg 

Also, from the steam tables,  kJ  hˆ1  3373.6    kg 

(500 ºC, 10 MPa)

61

The reversible work is calculated as:

W S rev  hˆ m 1

2

 kJ   kJ   kJ   hˆ1  2391.0    3373.6    982.6    kg   kg   kg 

and the actual work as

W S actual m 1



W S rev  835  kJ  m 1

 kg   

The exit temperature is calculated by determining the enthalpy of the actual exit state:

 W S actual  kJ  ˆh ˆ  2538.4   2, actual  h1  m 1  kg  This state is still saturated (although the quality is higher), so the temperature is T2 = 99.6 oC

62

3.29 The subscripts “2” and “3” represent the two outline streams, and “1” represents the inlet stream. First, perform a mass balance:

n1  n 2  n3 

2 1 n1  n1 3 3

Now write the energy balance: 0  n1h1  n2 h2  n3 h3   Q  W S Since the system is insulated and there is no shaft work, the energy balance can be rewritten as: 2 1 n1h2  n1h3  n1h1  0 3 3 2 h2  h1   1 h3  h1   0 3 3 Substituting expressions for heat capacity A.2.2, we obtain the following expression: 400 K





T





3   4000  1 4000  3 3  3.280  0.593 10 T  T 2  dT  3 R  3.280  0.593  10 T  T 2  dT  0 300 K 300 K Solving, we find

2 R 3

T3  100.6 K Now set up an entropy balance for the process. The minimum pressure is required for a reversible process.

2 1 n1s2  n1s3  n1s1  0 3 3 2 s2  s1   1 s3  s1   0 3 3 Assuming ideal gas behavior, we can express the changes in entropies using Equation 3.62: 400 K  3.280  1bar  4000    ln s 2  s1  R    dT  0.593  10  3   3  T P  T   1   300 K 100.6 K  3.280  1bar  4000    ln s3  s1  R    dT  0.593 10  3   3  T P  T   1    300 K









63

Substitute these expressions into the entropy balance and solve for P1: P1  1.85 bar

64

3.30 A schematic of the process is illustrated below:

reversible

Tsurr = 25 oC Psurr = 1 bar Pure H 2O P2 = 20 bar

Pure H2O

T 2 = ? oC

P1 = 60 bar T1 = 540 oC

m = 5 kg

m = 5 kg

well-insulated

(a) Since the process is adiabatic and reversible, suniv  0 s sys  0 s surr  0

(b) We can obtain the final temperature using the steam tables.

 kJ  sˆ1  6.999    kg  K   kJ   sˆ2  6.999    kg  K 

(540 ºC, 60 bar) (20 bar)

The pressure and entropy of state 2 can be used to back out T2 from the steam tables. T2  362.5 º C

(c) To obtain the value of work, perform an energy balance. The process is adiabatic, and potential and kinetic energy effects can be neglected. Therefore, the energy balance is U  muˆ 2  uˆ1   W

From the steam tables,  kJ  uˆ2  2881.15    kg 

(362.5 ºC, 20 bar)

65

 kJ  uˆ1  3156.12    kg 

(540 ºC, 60 bar)

Hence,   kJ   kJ   W  5 kg  2881.15    3156.12     1373 kJ   kg   kg   

(d) The specific volume can be found in the steam tables

 m3  vˆ2  0.14173    kg 

Therefore,

 

  m3   3 V2  5 kg  0.14173     0.709 m    kg   

66

3.31 A schematic of the process is shown below: very tiny leak hole

Tsurr = 25 oC Psurr = 1 bar

Pure H2O

Pure H2O P1 = 60 bar

P2 = 20 bar

T1 = 540 oC

T ' 2 = ? oC

m = 5 kg

well-insulated

In order to leave the system, the gas must do flow work on the surroundings. The initial state is the same as for Problem 3.30 and the final pressures are the same. Since the water only expands against 1 bar, the work is lower than that for the differential process described in Problem 3.30. Thus, this adiabatic process looses less energy, leading to a higher final temperature. Another way to view this argument is to look at this process as a closed system. This depiction is the expansion analog of the compression process depicted for Example 2.5 in Figure E2.5B (page 57). We can represent this process in terms of two latches, one that keeps the process in its initial state at 60 bar and one that stops the expansion after the pressure has reduced to 20 bar. The process is initiated by removal of the first latch and ends when the piston comes to rest against the second latch. Such a process is depicted as “Problem 3.31’ below. The corresponding reversible process of Problem 3.30 is shown next to it for comparison. Clearly the process on the left does less work, resulting in a greater final temperature.

Problem 3.31

Problem 3.30

T surr = 25 oC Psurr = 1 bar latches Pure H2O P1 = 60 bar T1 = 540 oC

irreversible

reversible

Tsurr = 25 oC Psurr = 1 bar

Pure H2O

Pure H 2O

P2 = 20 bar

P2 = 20 bar

Pure H2O

T ' 2 = ? oC

T 2 = ? oC

P1 = 60 bar

m = 5 kg

m = 5 kg

T1 = 540 oC

m = 5 kg

m = 5 kg

well-insulated

67

3.32 (a) Consider the tank as the system.

Mass balance dm  m in  m out  m in dt

Separating variables and integrating: m2

t

m1

0


or t

m2  m1   m in dt 0

Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is  dU      m out hout   m in hin  Q  W s dt   sys out in

The process is adiabatic and no shaft work is done. Furthermore, there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to  dU     m in hin  dt  sys


t

U1 0

0

 dU   m

t

h dt  hin  m in dt

in in

0

U 2   m2uˆ 2  m2 hîn

where the results of the mass balance were used. Thus, uˆ 2  hîn

68

From the steam tables,  kJ  hîn  3456.5    kg 

(3 MPa, 773 K)

Now the water in the tank is constrained. From the steam tables:  kJ  sˆ2  7.743    kg  K  3

m vˆ2  0.14749    kg 

  kJ    3 MPa, uˆ 2  3456.5       kg   

Compute the change in entropy. An entropy balance gives:  dS   dS        m in sin  dt  univ  dt  sys

Integrating with sin constant t

S univ  m2 sˆ2  sîn  m in dt  m2 sˆ2  sîn  0

From the steam tables:  kJ  sîn  7.2337    kg  K 

(3 MPa, 773 K)

Therefore,       0.05 m 3  kJ    kJ  kJ  7.743    0.173    7.2337  S univ           m 3   K  kg  K    kg  K   0.14749    kg     

(b) If it tank sits in storage for a long time and equilibrates to a final temperature of 20 ºC, some or all of the vapor will condense and exchange heat with the surroundings. Let the subscript “3” designate the final state of the water when it has reached a temperature of 20 ºC. The change in entropy between state 2 and state 3 is given by the following equation 69

S univ  m2 sˆ3  sˆ2  

Qsurr T

Let the subscript “3” designate the final state of the water when it has reached a temperature of 20 ºC. First, find quality of the water. From the steam tables:  m3  ˆv3sat , l  0.001002  kg   

(sat. water at 20 ºC)

 m3  vˆ3sat  57 . 79   ,v  kg 

Calculate the quality as follows: ˆ sat vˆ3  vˆ2  1  x vˆ3sat , l  xv3, v

  m3   m3     m3   0.14749    1  x  0.001002     x 57.79     kg   kg     kg    x  0.0025

Now calculate the entropy of state 3 ˆ sat sˆ3  1  x sˆ3sat , l  xs3, v

Substitute values from the steam tables:  sˆ3  1  0.0025 0.2966   kJ  sˆ3  0.3175    kg  K 

  kJ    kJ     0 . 0025 8 . 6671   kg  K    kg  K        

Now calculate the amount of heat transferred to the surroundings: Q surr  Q  m2 uˆ 3  uˆ 2 

Calculate the internal energy of state 3: ˆ sat uˆ 3  1  x uˆ 3sat ,l  xu 3,v

Substituting values from the steam tables: 70

 kJ  uˆ 3  89.74    kg 

Therefore,       0.05 m 3  kJ    kJ   89.74    3456.5     1141 kJ  Q surr      m 3    kg    kg  0 . 14749       kg   

Now calculate the change in entropy of the universe

S univ

S univ

    3  kJ   kJ   1141 kJ    0.05 m  0 .3175 7 . 743    kg  K   kg  K    293 K 3    m      0.14749      kg     kJ   1.38   K

The entropy change for both processes can be fount by adding together the entropy change from Part (a) and Part (b):

S univ (a) and (b)  1.55 kJ  K

71

3.33 A schematic is given below valve maintains pressure in system constant v T1 = 200 oC x1 = 0.4 V = 0.01 m3 l

Mass balance dm  m in  m out  m out dt

Separating variables and integrating: m2

t

m1

0

 dm   m

out

dt

or t

m2  m1    m out dt 0

Entropy Balance: Q Q  dS   dS  Suniv     m out sout  surr     m out sout  Tsurr  dt  sys Tsurr  dt  sys

Integrating with sout constant Suniv  m2 s2  m1s1  m2  m1 sout 

Q Tsurr

(1)

From steam tables:

72

 kJ  s1  (1  x)u f  xu g  0.6  2.3309  0.4  6.4323  3.9715    kg K   kJ   kJ  s 2  6.4323  is the same as sout  6.4323     s2  kg K   kg K  Thus Equation 1 simplifies to

Suniv  m1 ( s 2  s1 ) 

Q

(2)

Tsurr

Energy Balance to find Q :  dU     m out hôut  Q  dt  sys

Integrating

 dU   

m 2 uˆ 2

t

m1uˆ1

0



t

t

0

0

 m out hôut  Q dt  hôut  m out dt   Q dt

Substituting in the mass balance and solving for Q Q  m 2 uˆ 2  m 1 uˆ 1  m 2  m 1 hˆ out

To find the mass in each state:  m3  v1  (1  x)v f  xv g  0.6  .001  0.4  0.1274  0.051    kg   m3  v2  0.1274    kg 

 

V 0.01 m 3  0.196 kg  m1  1  v1  m3  0.051    kg 

V and m2  2  v2

From the seam tables:

73

 

0.01 m 3  0.0785 kg   m3  0.1274    kg 

 kJ  u1  (1  x)u f  xu g  0.6  850.64  0.4  2597.5  1549    kg   kJ  u 2  2595.3    kg 

and

 kJ  hout  2793.2    kg 

Plugging in: Q  m2uˆ2  m1uˆ1  m2  m1 hôut  228 kJ 

Back into Equation 2  kJ  228    kJ    Q  kg    s  0  S univ  m1 ( s 2  s1 )    0.196    6.4323  3.9715    Tsurr   s    kg K   473  K  s   Note: Vaporization is reversible if Tsurr = Tsys.

74

3.34 The process can be represented as:

V1 = 1 m3 P1 = 10 bar

Process

V2 = 10 m3 P2 = ?

T1 = 1000 K

Work out Q=0

T2 = ?

State 1

State 2

Solving for the number of moles: PV n  1 1  120.3 mol RT1

The maximum work is given by a reversible process. Since it is also adiabatic, the entropy change of the system is zero: T  P  s  0  cP ln 2  R ln 2  T1  P1  5 Since cP  R 2  T2     T1 

5/ 2



P2 nRT2  P1 V2 P1

Solving for T2 gives  nRT 5 / 2  1 T2     V2 P1 

2/3

 215 K

An energy balance on this closed system gives: T2

T2

T1

T1

U = Q  W  W  n  c v dT  n  cP  R dT

Solving for work, we get 3 6 W  n RT2  T1   1.18 10 J 2

75

3.35 The maximum efficiency is obtained from a Carnot cycle. From Equation 3.32 T n  1 C TH

where temperature is in Kelvin. Hence, n  1

298.15 K  0.614 773.15 K

76

3.36 The labeling described in Figure 3.8 will be used for this solution. First, a summary of known variables is provided.

State 1 2 3 4

P (bar) 30 0.1

T (ºC) 500

From our knowledge of ideal Rankine cycles, the table can be expanded as follows State 1 2 3 4

Thermodynamic State Superheated Steam Sat. Liq. and Vapor Sat. Liquid Subcooled Liquid

P (bar)

T (ºC)

30 0.1 0.1 30

500

The saturation condition constrains state 3. First, start with the turbine. Since the rate of heat transfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle, the following is known sˆ1  sˆ2

and



W s  m hˆ2  hˆ1



From the steam tables:  kJ  sˆ1  7.2337    kg  K   kJ  sˆ2l  0.6492    kg  K   kJ  sˆ2v  8.1501  

 kg  K   kJ  hˆ1  3456.5    kg   kJ  hˆ2l  191.81    kg  hˆ2v

 kJ   2584.6    kg 

(500 ºC, 30 bar)

(sat. H2O at 0.1 bar)

(500 ºC, 30 bar)

(sat. H2O at 0.1 bar)

77

The quality of the water can be calculated as follows sˆ1  (1  x) sˆ2l  xsˆ2v

By substituting the steam table data, we find x  0.878

Now the quality can be used for the following expression hˆ2  (1  x)hˆ2l  xhˆ2v

Substituting the steam table data and the quality calculated above, we get  kJ  hˆ2  2292.7    kg 

Therefore,  kJ   kJ     kg   W s  100    2292.7    3456.5     116.38 MW   s     kg   kg  

At state 3,  kJ  hˆ3  191.81    kg   m3  ˆv3  0.001010    kg 

(sat. H2O(l) at 0.1 bar)

Therefore,





Q C  m hˆ3  hˆ2  210.09 MW 

Since the molar volume of water does not change noticeably with pressure and the compressor is adiabatic, the work required for compressing the fluid can be calculated as follows W c  m vˆ3 P4  P3   m3    kg  W c  100   0.00101    30  105  0.1  105 Pa   3.02  10 5 W   0.302 MW   s   kg  





78

Now we find the rate of heat transfer for the evaporator. For the entire Rankine cycle, Equation 3.84 gives Q net  W net  Q C  Q H  W s  WC  0

We have Q H  W s  WC  Q C

Using the values calculated above: Q H  326.2 MW 

Equation 3.84: W net  W S  WC W net  116.08 MW 

The efficiency can be calculated with Equation 3.82 W

116.08 MW 

  net   0.356 326.02 MW  QH

79

3.37 To make the Rankine cycle more efficient, we need to increase the area that represents the net work in Figure 3.8. This can be done in a variety of ways:

1. Increase the degree of superheating of steam in the boiler. This process is sketched in the upper left hand Ts diagram below. This change reduces the moisture content of steam leaving the turbine. This effect is desirable since it will prolong the life of the turbine; however, if the steam is heated too high, materials limitations of the turbine may need to be considered. 2. Lower the condenser pressure. Lowering the pressure in the condenser will lower the corresponding saturation temperature. This change will enlarge the area on the Ts diagram, as shown on the upper right below. Thus, we may want to consider lowering the pressure below atmospheric pressure. We can achieve this since the fluid operates in a closed loop. However, we are limited by the low temperature off the heat sink that is available. increase the superheating of steam in the boiler

T

lower the condenser pressure

T

1new

1

wnet

wnet 4new

4 2new

3

3new

2new

s

s

2 stage turbine with reheat

increase the boiler pressure

Turbine 1 Turbine 2

1

T

T 1

3

6 2 wnet

wnet

4 4 3

5

2

s

s

80

3. Increase the boiler pressure. This will increase the boiler temperature which will increase the area as shown on the bottom left Ts diagram. 4. We can be more creative about how we use the energy available in the boiler. One way is to divide the turbine into two stages, Turbine 1 and Turbine 2, and reheat the water in the boiler between the two stages. This process is illustrated below. Rankine cycle with reheat Ws

1 Fuel Air

Ws Turbine 2

turbine 1 2

4 QH

3

Boiler

QC

cooling water

6 5

condenser

Wc compressor

The pressure in Turbine 1 will be higher than the pressure in turbine 2. This process is schematically shown on the bottom right Ts diagram. This process leads to less moisture content at the turbine exit (desirable) and limits the temperature of the superheat (desirable).

81

3.38 The labeling described in Figure 3.8 will be used for this solution. First, a summary of known variables is provided.

State 1 2 3 4

P (bar) 100 1

T (ºC) 500

From our knowledge of ideal Rankine cycles, the table can be expanded as follows

State 1 2 3 4

Thermodynamic State Superheated Steam Sat. Liq. and Vapor Sat. Liquid Subcooled Liquid

P (bar)

T (ºC)

100 1 1 100

500

The saturation condition constrains state 3. First start with the turbine (state 2). Since the rate of heat transfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle, the following is known sˆ1  sˆ2

From the steam tables,  kJ  sˆ1  6.5965    kg  K   kJ  sˆ2l  1.3025    kg  K  sˆ2v

 kJ   7.3593    kg  K 

(500 ºC, 100 bar)

(sat. H2O at 1 bar)

The quality of the water can be calculated as follows sˆ1  (1  x) sˆ2l  xsˆ2v

By substituting the steam table data, we find x  0.874

82

States 1, 2, and 3 are constrained; therefore, the enthalpies can be determined using the steam tables. The enthalpy of state 4 can be calculated from Equation 3.80: hˆ4  hˆ3  vˆl P4  P3   m3  ˆ ˆ where vl  v3  0.001043   since specific volume of liquids are insensitive to pressure  kg  changes. From the steam tables:

State

1

 kJ  3240.8    kg  Note: hˆ2  (1  x)hˆ2l  xhˆ2v hˆ

2

3

4

 kJ  2391.0    kg 

 kJ  417.44    kg 

 kJ  427.8    kg 

Equation 3.84:

W net  Q H  Q C Therefore,



 

 

 

W net  m hˆ1  hˆ4  m hˆ3  hˆ2  m hˆ1  hˆ4  hˆ3  hˆ2



and

m 

W net

hˆ1  hˆ4  hˆ3  hˆ2 



100 103 kW 

  kJ   kJ     kJ   kJ    3240.8    417.5      427.8    2391.0      kg   kg     kg   kg  

 kg  m  116.3   s  Now consider the non-ideal turbine and compressor. Use the definition of isentropic efficiencies:

 W s actual turbine   Ws reversible  W S actual  turbine W S reversible  turbine m hˆ2  hˆ1 reversible

83

WC reversible WC actual  WC reversible m hˆ4  hˆ3 reversible  

compressor   WC actual

compressor

compressor

Equation 3.84 is used to find the mass flow rate

W net  W s  WC





 hˆ4  hˆ3 reversible   W net  m turbine hˆ2  hˆ1  reversible compressor  

Substituting the enthalpy values from the table shown above and the efficiency values gives

 kg  m  147    s  A higher flow rate is needed as compared to the reversible process.

84

3.39 The labeling shown in Figure 3.8 will be used for this solution. First, a summary of known variables is provided.

State

Thermodynamic State Sat. Vapor

1 2 3 4

P (MPa) 1.7 0.7

From our knowledge of ideal Rankine cycles, the table can be expanded as follows State 1 2 3 4

Thermodynamic State Sat. Vapor Sat. Liq. And Vapor Sat. Liquid Subcooled Liquid

P (MPa) 1.7 0.7 0.7 1.7

The enthalpy of states 1 and 3 can be found using the NIST website. For states 2 and 4, use the following expressions to find the enthalpies

hˆ2  (1  x)hˆ2l  xhˆ2v hˆ4  hˆ3  vˆl P4  P3   hˆ3  vˆ3 P4  P3  To find x, use the fact that the turbine is isentropic. sˆ1  sˆ2 sˆ1  (1  x) sˆ2l  xsˆ2v

From the NIST website:

 J  s1  146.38   mol  K   J  s2l  90.45   mol  K   J  s2v  150.07   mol  K 

(sat. vapor at 1.7 MPa)

(sat. mixture at 0.7 MPa)

By substituting the above values into the entropy relationship, we find

85

x  0.938

Also, from the NIST website:  J  h1  36046   mol   J  h2l  18416   mol   J  h2v  35353   mol   J  h3  18416   mol   3 5 m v3  6.954 10    mol 


(sat. mixture at 0.7 MPa)

(sat. liquid at 0.7 MPa)

Substituting these values into the expression for h4 and h2 yields  J  h2  34302.9    mol   J  h4  18485.5   mol  Equation 3.82 states

 rankine 

h2  h1  h4  h3  h1  h4

 J   J    J   J   36046   18485.5   18416   34302.9      mol   mol    mol   mol    rankine   J   J   18485.5  36046    mol   mol   rankine  0.095 This efficiency is significantly lower than a conventional power plant.

86

3.40 (a) Using the information in the problem statement and our knowledge of ideal vapor compression cycles, the following table can be created

State

Thermodynamic P (MPa) State 1 Sat. Mixture 0.12 2 Sat. Vapor 0.12 3 Superheated Vapor 0.7 4 Sat. Liquid 0.7 Note: Refer to Figure 3.9 to review labeling convention. Equation 3.85 states Q C  n h2  h1  which upon combination with Equation 3.89 gives Q C  n h2  h4  From the NIST website:  kJ  h2  39.295   mol   kJ  h4  24.181    mol 

(sat. vapor at 0.12 MPa) (sat. liquid at 0.7 MPa)

Therefore,   mol    kJ   kJ    Q C   0.5  39 . 295 24 . 181      mol   mol    7.557 kW   s     (b) The power input to the compressor can be calculated with Equation 3.86:

WC  n h3  h2  Equation 3.87 states s 2  s3

87

From the NIST website:  J  s3  s2 sat. vapor at 0.12 MPa   177.89   mol  K  Now, state 3 is constrained.  kJ  h3  43.031   mol 

  J   0.7 MPa, s3  177.89    mol  K   

Therefore,   mol    kJ   kJ    39.295  WC   0.5   43.031    1.87 kW     s    mol   mol    (c) Equation 3.90 states:

 kJ   kJ   24 . 181 39.295   mol  h h h h  mol  COP  2 1  2 4   4.05 h3  h2 h3  h2  kJ   kJ   39.295  43.031   mol   mol 

88

3.41 Using the information in the problem statement and our knowledge of ideal vapor compression cycles, the following table can be created

State

Thermodynamic P (MPa) State 1 Sat. Mixture 0.12 2 Sat. Vapor 0.12 3 Superheated Vapor 0.7 4 Sat. Liquid 0.7 Note: Refer to Figure 3.9 to review labeling convention. From Equation 3.90 COP 

Q C h2  h1  W C h3  h2

The enthalpy of state 2 can be found directly from the NIST website, but the enthalpies of states 1 and 3 require the use of additional information.  kJ  h2  39.295   mol 


For the process between state 2 and state 3, s 2  s3 From the NIST website:  J  s3  s 2  177.89   mol  K 


Now, state 3 is constrained.  kJ  h3  43.031   mol 

  J   0.7 MPa, s3  177.89     mol  K   

The process between state 4 and state 1 is also isentropic.  J  s1  s 4  115.09   mol  K 

(sat. liquid at 0.7 MPa)

The quality of the R134a can be calculated as follows

89

s1  1  x s1l  xs1v

where  J  s1l  90.649   mol  K   J  s1v  177.89    mol  K 

(sat. R134a(l) at 0.12 MPa) (sat. R134a(v) at 0.12 MPa)

Therefore,   J   J    J   1  x  90.649  115.09    x177.89      mol  K   mol  K     mol  K     x  0.280 Using the quality of R134a, the enthalpy of state 1 can be calculated as follows h1  1  x h1l  xh1v

From the NIST website:  kJ  h1l  17.412   mol   kJ  h1v  39.295    mol 

(sat. R134a(l) at 0.12 MPa) (sat. R134a(v) at 0.12 MPa)

Therefore,    kJ    kJ   h1  1  0.2817.412    0.28 39.295     mol    mol      kJ  h1  20.74   mol  Now, everything needed to calculate COP is available. Using Equation 3.90,  kJ   kJ  39.295   20.74   h h  mol   mol   4.97 COP  2 1  h3  h2  kJ   kJ  43.031   39.295    mol   mol 

90

Is this modification practical? No. An isentropic turbine adds significant level of complexity to the cycle. Turbines are expensive and wear over time. Furthermore, the real turbine added to the cycle will not be 100% efficient, so the COP will not increase as much. The cost of the turbine is not justified by the increase in COP.

91

3.42 The subscript “h” refers to the hotter cycle, while “c” refers to the cooler cycle. The (a) The flow rate of the cooler cycle can be found by performing an energy balance on the condenser/evaporator shared between the two cycles. An energy balance shows: Q H , c  Q C , h  Q

Written in terms of enthalpy, the equation is n c h8  h7    n h h2  h1  We can find the enthalpies for positions 2 and 8 directly from the thermodynamic tables because the fluid is saturated in at these positions. To find the other enthalpies we must use the following relationships: s 6  s7 h1  h4 Using the NIST website:  J  h1  24181   mol   J  h2  40967   mol   J  h7  41510   mol   J  h8  21099   mol 

(sat. R134 liquid at 0.7 MPa) (sat. R134 vapor at 0.35 MPa)  J  ) (R134 vapor at 0.35 MPa with s  177.89   mol  K  (sat. R134 liquid at 0.35 MPa)

Now the flow rate can be calculated:

n c

nc

  J   J   0.5 mol/s 40967   24181     mol   mol       J   J   41510   21099     mol   mol     mol   0.411   s 

92

(b) Performing an energy balance we find

Q C  n c h6  h5  We can also use the following relationship  J  h5  h8  21099   mol  From the NIST website:  J  h6  39295   mol 

(sat. R134 vapor at 0.12 MPa)

Therefore,   mol    J   J  Q C   0.411   21099   39295    7.48 kW     s    mol   mol    (c) The power input is calculated as follows: WC , total  WC , c  WC , h

where WC , c  nc h7  h6  WC , h  n h h3  h2 

We have all of the required enthalpies except h3. State 3 is constrained because  J  s3  s 2  175.95  mol  K  P3  0.7 MPa

(sat. R134a vapor at 0.35 MPa)

From the NIST website:  J  h3  42428   mol 

 J  ) (R134a vapor at 0.7 MPa with s  175.95  mol  K 

Now compute the power for each unit:

93

   J   J   J   J   39295   40967  WC , total  0.411 mol/s 41510    0.5 mol/s 42428       mol   mol    mol   mol     W C , total  1.64 kW 

(d) The coefficient of performance is calculated using the following equation: COP 

Q C

WC , total



7.48 kW  4.56 1.64 kW

(e) The COP for the cascade is 4.56, while the COP is 4.05 in Problem 3.40. The cascade system’s COP is 12.6% greater.

94

3.43 One design follows; your design may differ:

In order to cool a system to -5 ºC, the temperature of the fluid must be colder than -5 ºC so that heat transfer will occur. We arbitrarily specify that the working fluid evaporates at -10 ºC. Similarly, in order to eject heat to the 20 ºC reservoir, the fluid must condense at a temperature greater than 20 ºC. Arbitrarily, we choose 25 ºC. One possible refrigeration cycle is presented below. Ý valve 3

T 1 QH

evaporator refrigerated unit at -5 C

4

4 QC 20 C reservoir

QH 2

1 3

condenser

Wc

QC

2

s compressor

A number of fluids will work sufficiently for this system, but the design process will be illustrated using R134a. For states 3 and 4, the pressure is constant, and for states 1 and 2, the pressure is constant at a different value. From the NIST website, we find P1  P2  0.201 MPa (Tsat = -10 ºC) P3  P4  0.665 MPa (Tsat = 25 ºC) (Note: The temperatures of each state are not constant. The listed saturation temperatures are the temperatures at which the fluid evaporates and condenses.)

Now that the pressures are known, we can compute the required flow rate required in order to provide 20 kW of cooling. Q C  n h2  h1  We can get h2 from the thermodynamic tables for saturated R134a. In order to find h1, we can use the following relationship: h1  h4

From the NIST website:  J  h2  40064   mol 

(sat. R134 vapor at 0.201 MPa)

95

 J  h4  23931   mol 

(sat. R134 liquid at 0.665 MPa)

Now, calculate the required flow rate of R134a. Q C 20,000 J/s  h2  h1    J   J   23931   40064     mol   mol     mol  n  1.24   s 

n 

96

3.44 The schematic of the process and the corresponding Ts is presented below T = 5 oC valve

Fridge 2 valve

6

T

QC,R

1 QH

1

evaporator

Wc

6

3

5 oC

T = -15 oC Freezer evaporator QC,F 4

QH 5

high T reservoir

-15 oC

QC,R 2

3 QC,F 4

5

condenser

Wc

s compressor

A number of refrigerants will work for this system, but the design process will be illustrated for R134a only. To define each state, we need to thermodynamically constrain each state. You should note that the problem doesn’t state what temperature the fluid condenses at. Therefore, we can assume the temperature is 25 ºC. By using information in the Ts diagram and from the NIST website, we find P1  P2  0.34966 MPa P3  P4  0.16394 MPa

P5  P6  0.66538 MPa

(T1=T2=Tsat = 5 ºC) (T3=T4=Tsat = -15 ºC) (T6 = 25 ºC)

Before solving for additional pressures and temperatures, we will list the known temperatures and pressures. State 1 2 3 4 5 6

Temperature (ºC) 5 5 -15 -15 25 25

Pressure (MPa) 0.24334 0.24334 0.16394 0.16394 0.66538 0.66538

States 4, 5, and 6 are completely constrained as confirmed by Gibbs phase rule. Now, we need to find the liquid and vapor compositions of states 1, 2, and 3 to completely constrain the states. Since the heat duties are equal for the refrigerator and the freezer, we have the following relationship: Q C , F  Q C , R

97

which upon performing an energy balance becomes h4  h3  h2  h1 Energy balances around the valves provide: h1  h6 h2  h3 Substituting these relationships into the expression that equates the heat loads results in h  h6 h2  h3  4 2

From the NIST website:  J  h4  39755   mol   J  h6  23931   mol  Now constrain states 1, 2, and 3 by determining the enthalpies:  J  h1  h6  23931   mol   J   J   23931  39755    mol   mol   31843  J  h2  h3   mol  2 Technically, the states are now all constrained, but we would like also like to know the vapor and liquid compositions of each state. The compositions of states 4, 5, and 6 are already known; we can calculate the compositions of states 1, 2, and 3 as follows: sat h1  1  x1 h1sat , l  x1h1, v sat h2  1  x2 h2sat , l  x2 h2, v sat h3  1  x3 h3sat , l  x3 h3, v

From the NIST website,

98

 J  h1sat , l  21095  mol     J  h2sat , l  21095  mol     J  h3sat , l  18380  mol   

 J  h1sat , v  40965  mol     J  h2sat , v  40965  mol     J  h3sat , v  39755   mol 

Now, we can solve the composition of vapor for each state: x1  0.143

x2  0.541

x3  0.630

The following table presents a summary of our results: State

Temperature (ºC)

Pressure (MPa)

1

5

0.24334

2

5

0.24334

3

-15

0.16394

4

-15

0.16394

5

25

6

25

Phases

Saturated Liquid and Vapor Saturated Liquid and Vapor Saturated Liquid and Vapor

Liquid Vapor Composition Composition

0.857

0.143

0.459

0.541

0.37

0.630

Saturated Vapor

0

1

0.66538

Superheated Vapor

0

1

0.66538

Saturated Liquid

1

0

99

3.45 In this case, the working material is a solid. The four states of the magnetic material are shown of the sT diagram below. Note the axis are shifted from the usual manner.

3 2

QH

QC 1

4

(a) The heat expelled by the cold reservoir can be approximated by:

qC  TC s 2  s1  where T is the average temperature between states 1 and 2, which is approximately 1 K. Taking values of S/R from the sT diagram, we get:  J  qC  R1.9  1.3  5.0  mol  K  (b) Similarly, the heat absorbed by the hot reservoir can be approximated by:

 J  q H  TH s 4  s 4   8.8R1.2  1.9  51  mol  K  (b) The coefficient of performance is given by:

COP 

qC wC ,total



qC  0.11 q H  qC

(d) The value of COP is much lower than a typical refrigeration process (COP= 4-6); in general, refrigeration processes at these low temperatures are much less efficient.

F. Work is supplied to magnetize the material and to spin the wheel. 100

3.46 When the polymer is unstretched it is in a more entangled state. When stretched the polymer chains tend to align. The alignment decreases the spatial configurations the polymer can have, and therefore, reduces that component of entropy. If the process is adiabatic, the entropy of the system cannot decrease. Consequently, its thermal entropy must increase. The only way this can be accomplished is by increased temperature.

101

3.47 Assume the temperature is 298 K. The following data was taken from Table A.3.2:

Species

h ºf (kJ/mol)

g ºf (kJ/mol)

Cu2O (s) O2 (g) CuO (s)

-170.71 0 -156.06

-147.88 0 -128.29

The data listed above were used to create the next table using s ºf 

h ºf  g ºf T Species

s ºf (kJ/mol K)

Cu2O (s) O2 (g) CuO (s)

-0.0766 0 -0.0932

Now the change in entropy of the reaction can be calculated in a method analogous to Equation 2.72

 

  kJ     kJ   º s rxn   vi s ºf  2 - 0.0766    4 - 0.0932    i  mol  K     mol  K     kJ  º s rxn  0.22   mol  Does this violate the second law of thermodynamics? This problem shows that the entropy change of the system is negative, but nothing has been said about the entropy change of the universe. We must look at the change in entropy of the surroundings to determine if the second law is violated. By looking at the enthalpies, we see that the reaction is exothermic, which means that heat is transferred from the system to the surroundings. Therefore, the entropy of the surroundings will increase during this reaction.

102

3.48 The temperature and pressure terms in the equation for entropy do not contribute anything to the entropy change because they are constant. Therefore, the only remaining entropy contribution, the randomness of the atomic arrangements, must be considered. The randomness does not increase when CdTe forms. In pure crystals of Cd and Te, the location of each atom is known because the crystal lattice constrains the atomic locations. In CdTe, the crystal lattice still defines the location of each atom, so the randomness has not increased. Therefore, the change in entropy is zero. 3.49 This argument is not scientifically sound. Morris is arguing that since evolution results in more order, the second law of thermodynamics is violated, so evolution must be impossible. However, the flaw in this argument is caused by ignoring the entropy change of the entire universe. The second law states that the entropy of the universe will remain constant or increase for any process. Morris’ argument was based on the entropy of the system undergoing evolution – not the entropy of the entire universe. A system can decrease in entropy if the entropy of the surroundings increases by at least that much.

103

3.50 Entropy is related to the number of configurations that a state can have. The greater the number of configurations, the more probable the state is and the greater the entropy. We can qualitatively relate this concept to the possible hands in a game of poker, but it is more interesting to quantify the results using some basic concepts of probability.

We consider a hand of poker containing 5 cards randomly draw from a deck of 52 cards.. There are a finite number of permutations in which we can arrange a 52 card deck in 5 cards. For the first card in the hand, we select from 52 cards, the second card can be any other card so we select from 51 cards, the third card has 50 cards, and so on. Thus the number of permutations of 5 cards is:

P  52  51 50  49  48  311,875,200 However, we do not care the order in which the cards are dealt, so we must divide this number by the number of ways we can come up with the same hand. We do this math in a similar way. For a given hand there are five cards we can pick first, four we can pick second, …. So the number of ways we can make the same hand is: N  5  4  3  2  1 120

The number of unique configurations can be found by dividing P by N . Thus, the cards can display C

P  2,598,960 N

or 2,598,960 unique configurations. To find the “entropy” of a given hand, we need to find out how many of these unique configurations belong to the hand Consider a four of a kind. There are 13 different possible ranks of four of a kind, one for each number A, 2, 3, 4, 5, 6, 7,8, 9, 10, J, Q, K. The fifth card in the hand could be any of the other 48 cards. Therefore, the number of combinations of four of a kind is: 13 x 48 = 624 The probability of a four of a kind is: 624 / 2598960 = 1 / 4165 which is very unlikely. Thus the “entropy” of this hand is low. In contrast, there are 1,098,240 to have a hand that has one pair; therefore the probability of getting this hand is much greater, 1/2.4, and its “entropy” is high. The probability of having nothing is 1/2 which represents the most likely hand in poker, or the hand with the highest “entropy”. In fact, the rules of poker are defined so the hand of lower “entropy” always beats the hand of higher “entropy.”

104



1

4.1. (a) Yes, the form of the equation is reasonable. This can be rewritten: PA 

RT a  n v A v A

It is equivalent to Pv  RT , except the pressure term has been corrected to account for the ions’ intermolecular forces. The coulombic forces between the gas molecules affect the system pressure. This modification is similar to the van der Waals equation. Since we are limited to 1 parameter, we need to choose the most important interaction. Since net electric point charges exert very strong forces, this effect will be more important than size. (b) The sign should be negative for a because the positively charged gas molecules repel each other due to coulombic forces. Therefore, the system pressure increases, i.e.,:

PA  Pideal Coulombic forces are much stronger than van der Waals interactions, so a will be large – much larger than a values for the van der Waals EOS. (c) Coulombic repulsion is the primary intermolecular force present in the gas. Coulombic potential energy is proportional to r-1. v is proportional to r3, so the coulombic potential goes as v-1/3. Therefore,

n

1 3

a must have must have the following units to maintain dimensional homogeneity with the pressure term:  a  N  J   v1/ 3    m 2    m 3  so

a  

J  1/3   m  mol  2

2

4.2 The attractive interactions are described by van der Waals forces. For both O2 and propane, the dipole moments are zero. Therefore, the expression for the interactions reduce to 

3  i j 2 r6

 Ii I j   Ii  I j 

   

From Table 4.1: I a  12.07 eV   1.933  10 11 erg 

I b  10.94 eV  1.753  10 11 erg Now obtain the requested expressions:

   1.933 10



3 16  10  25 cm 3 aa   2 r6 aa 

 3.71  10 59 r

ab   ab 

6

cm

6

2

r

6

cm

6



   1.753 10



bb 

   1.933 10

 erg

3 62.9  10  25 cm 3 bb   2 r6

 5.20  10 58 r

6

cm

6





 erg

3 16  10  25 cm 3 62.9  10  25 cm 3 2 r6

 1.39  10 58



erg 1.933  10 11 erg   11 erg  1.933  10 11 erg  1.933  10

 



11

2

 erg

11

11





erg 1.753  10 11 erg    1.933  10 11 erg  1.753  10 11 erg 





erg 1.753  10 11 erg   1.753  10 11 erg  1.753  10 11 erg 



(b) Calculation:









   5.20  10  58   3.71  10  59 6  cm  erg  cm 6  erg  aa bb     r6 r6   

3

aa bb 

 1.39  10 58

cm

6



 erg r Note: Disregarded the positive value. The value of

6

aa bb is equal to ab . The values are equal because the ionization energies are

similar. (c) An expression for the average intermolecular attraction in the mixture can be found using the mixing rules mix  y a2 aa  2 y a yb ab  yb2 bb  3.71  10 59 y a2  1.39  10 58 y a yb  5.20  10 58 yb2 mix  cm 6  erg 6 r





4



4.3 (a) 300 K, 10 atm. The intermolecular distance of molecules is greater at lower pressures. Therefore, fewer intermolecular interactions exist, which cause less deviation from ideality. (b) 1000 K, 20 atm. At higher temperatures, the kinetic energy of the molecules (speed) is greater. The molecules interact less; thus, the compressibility factor is closer to unity. (c) Let subscript “1” denote BClH2 and “2” denote H2. For the mixture, we calculate the compressibility factor as follows B z  1  mix v

where

Bmix  y12 B1  2 y1 y2 B12  y 22 B2 Since BClH2 is polar and H2 is non-polar and small B1  Bmix  B2 Therefore, the plot may look like the following. quadradic, not linear

zH2 close to 1 1

z

0.9

zBClH2 lowest

0.8

0.7

0.6 0

0.2

0.4

0.6

yBClH2

5

0.8

1

4.4 (a) The intermolecular attractions and volume occupied by the styrene monomers will contribute to the deviations from ideality. Since styrene monomers are essentially non-polar, the order of importance is as follows

dispersion  dipole - dipole  induction The van der Waals EOS is appropriate since it accounts for the occupied molar volume and intermolecular forces, but it should be noted that more modern EOSs will give more accurate results. (b) Use critical data to calculate the a and b parameters: 2

  J    8.314  647.15 K   2  3  mol  K   27 RTc  27     3.13  J  m   a   2 64 Pc 64 39  105 Pa   mol  b

RTc  8 Pc

  J  647.15 K   8.314   mol  K   





8 39  105 Pa 

 m3   1.72  10  4    mol 

Now we can use the van der Waals EOS to solve for temperature.

P

RT a  v  b v2

  J   8.314  T  mol  K    5 10  10 Pa   3 4  m  3.0  10   1.72  10 4   mol  T  550.8 K  277.65 º C

 m3   mol   



 J  m3  3.13  2   mol    3.0  10 4  

 m3    mol    

2

The styrene will not decompose at this temperature. (c) The a parameter is related to attractive intermolecular forces. Dispersion is the controlling intermolecular force in this system, and its magnitude is directly related to the size of the molecules (polarizability component of dispersion). For the 5-monomer long polymer chain, a is 5 times the a value in Part (b). The b parameter is also related to the size of the molecule since it accounts for the volume occupied by the molecules. Again, the b parameter for the reduced polymer chain is 5 times the b value from Part (b).

6

 J  m3  a  15.65  2   mol   m3  b  8.6  10  4    mol 

(d) We must realize that if we initially believed there were 100 moles of styrene in the reactor, then there can only be 20 moles of the 5-monomer long polymer chain. Therefore,

 m3  v  1.5  10  3    mol  and  J  m3    J  15 . 65  8.314  T  mol 2   mol  K      5  10  10 Pa   2 3 3 3     m m   4 3  m    1.5  10 3  8 . 6 10   mol  1.5  10     mol      mol   T  612.41K  339.26 º C

Decomposition will occur.

7

4.5

There are many ways to solve this problem, and the level of complexity varies for each method. To illustrate the principles in Chapter 4, two of the simplest solution methods will be illustrated. Method 1. Polarizability of each atom The polarizability of a molecule scales with the number of atoms; the polarizabilities of individual atoms are additive. Using the first two molecules, solution of the following system of equations 1 C  4 H  1 CH 4

2 C  6 H  1 C2 H 6 gives

  cm 

 C  11.4  10 25 cm 3  H  3.65  10  25

3

Now, the polarizability of the chlorine atom can be found. For chloroform, 1 C  4 Cl  1 CCl 4

Therefore,

 

 Cl  23.4  10 25 cm 3

The values of C3 H 8 , CH 3Cl , CH 2Cl2 , and CHCl3 are calculated with the polarizabilities found above as follows, and compared to the values given. Species C3H8 CH3Cl CH2Cl2 CHCl3

 calculated (x 1025 cm3) 63.4 45.8 65.5 85.3

 reported (x 1025 cm3) 62.9 45.6 64.8 82.3

% Difference 0.8 0.3 1.1 3.6

All agree reasonably well. Now, the polarizabilities of C 4 H10 and C 2 H 5Cl will be calculated.

C

4

H 10



  

 

 

 4 11.4  10  25 cm 3  10 3.65  10  25 cm 3  82.1  10  25 cm 3

8

C

2

H 5 Cl

  



  

 

 

 2 11.4  10  25 cm 3  5 3.65  10  25 cm 3  1 23.4  10  25 cm 3  64.45  10  25 cm 3

Method 2. Bond Polarizability For this method, we calculate the molecules’ polarizabilities by adding the polarizability of each bond, instead of the atoms. For the methane molecule 4 C  H  1 CH 4

 

 

26  10  25 cm 3  6.5  10  25 cm 3 4

 C  H 

To calculate the polarizability of a C-C bond, use ethane as follows: 6 C  H   C  C  1 C 2 H 6

  

 

 

  C  C  44.7  10  25 cm 3  6 6.5  10  25 cm 3  5.7  10  25 cm 3

The C-C and C-H polarizability calculated above predict the given polarizability of propane well. The polarizability of C-Cl bonds is calculable with the polarizability of chloroform. 4 C  Cl  1 CCl 4   C  Cl 

 

 

105  10  25 cm 3  26.25  10  25 cm 3 4

The values of C3 H 8 , CH 3Cl , CH 2Cl2 , and CHCl3 are calculated with the polarizabilities found above as follows, and compared to the values given. Species C3H8 CH3Cl CH2Cl2 CHCl3

 calculated (x 1025 cm3) 63.4 45.8 65.5 85.3

 reported (x 1025 cm3) 62.9 45.6 64.8 82.3

% Difference 0.8 0.3 1.1 3.6

All agree reasonably well. This value predicts the polarizabilities of the other species in the table reasonably well. Now, the polarizabilities of C 4 H10 and C 2 H 5Cl will be calculated. 1 C 4 H 10  3 C  C  10 C  H

C

4

H 10



  

 

 

 3 5.7  10  25 cm 3  10 6.5  10  25 cm 3  82.1  10  25 cm 3

9

1 C 2 H 5 Cl  1 C  C  5 C  H  1 C  Cl

C

2

H 5 Cl



  

  

 

 

 1 5.7  10  25 cm 3  5 6.5  10  25 cm 3  1 26.25  10  25 cm 3  64.45  10  25 cm 3

Note both the atom method and the bond method yield identical results. More accurate values for the polarizabilities can be calculated using more of the data given in the problem.

10

4.6 (a)  increases with molecular size. Therefore,

 I  S  O 2

2

2

 scales with magnitude of van der Waals interactions. Because these are non-polar, diatomic molecules, only dispersion forces are present. Dispersion forces depend on the first ionization potential and polarizability. Ionization energy is approximately equal for each molecule. The polarizability scales with molecular size, so

 I   S  O 2

2

2

(b)  increases with molecular size. Diethylether and n-butanol have the same atomic formula and similar spatial conformations. Therefore, they should be about equal in size. Methyl ethyl ketone has fewer atoms, but has two exposed electron pairs on the double-bonded oxygen. The size of the molecular electron orbital of methyl ethyl ketone is approximately equal to the sizes of diethyl ether and n-butanol, so

 n - butanol   diethyl ether   methyl ethyl ketone Methyl ethyl ketone and n-butanol are much more polar than diethyl ether due to their greater asymmetry, so their  values are greater than diethyl ether’s. Now we must determine if there is greater attraction in n-butanol or methyl ethyl ketone.  scales with the magnitude of van der Waals interactions. Since induction and dispersion forces are similar in these molecules, we must consider the strength of dipole-dipole forces. There is greater charge separation in the double bond of ketone, so

 methyl ethyl ketone   n - butanol   diethyl ether

11

4.7 (a) At 30 bar, the water molecules are in closer proximity than they are at 20 bar. Intermolecular attractions are greater, so the magnitude of molecular potential energy is greater. The potential energy has a negative value for attractive interactions. The molecular kinetic energy is identical since the temperature is the same. Hence, the internal energy, the sum of kinetic and potential energies, is less at 30 bar. (b) The key to this phenomenon is hydrogen bonding. At 300 K and 30 bar, isopropanol and npentane are both liquids. The hydrogen bonding and dipole-dipole interactions are present in isopropanol, and dispersion is present in n-pentane. The intermolecular forces are greater in the isopropanol, so the compressibility factor is smaller for isopropanol.

At 500 K and 30 bar, both species are gases. In the gas phase, hydrogen bonding does not play a significant role. The dispersion forces in n-pentane are stronger than the dipole-dipole forces of isopropanol. Therefore, the compressibility factor is smaller for n-pentane.

12

4.8 (a) Ideal: For ideal NH3, the compressibility factor is equal to one. For real NH3, the strong intermolecular forces (dipole-dipole and dispersion) cause the molar volume to decrease. They outweigh the volume displaced by the physical size of NH3; thus, z will be less than one. (b) Internal energy will be greater for the ideal gas. In the real gas, intermolecular attractions are present. Internal energy value is the sum of potential and kinetic energies of the molecules. The absolute values of the kinetic energies are identical at identical temperature: however, the potential energy decreases for real NH3 due to attractive interactions - so the internal energy is less for the real NH3. (c) The entropy will be greater for the ideal gas. Entropy is a measure of possible molecular configurations or “randomness”. Ammonia has an electric dipole in which positive and negative charge are separated. The intermolecular forces in the real gas cause the molecules to align so that the positive charge in one molecule is adjacent to a negative charge in a neighboring molecule to reduce potential energy. Therefore, fewer possible configurations exist, which creates less randomness and lower entropy.

13

4.9 (a) We can determine which case has the higher compressibility factor by comparing the molar volumes at constant T and P. With Ne, very weak intermolecular attractions are present, so volume displacement becomes important. The compressibility factor will be slightly greater than one. In NH3, the strong intermolecular attractions decrease the molar volume, so z is less than one. The compressibility factor is greater for Ne. (b) Entropy is a measure of randomness. Both species are gases at these conditions. The intermolecular attractions present in NH3 reduce the number of possible configurations. The weak forces present in Ne have a much smaller effect. However, NH3 is asymmetrical, while Ne is symmetrical. The asymmetry of NH3 results in more possible configurations that NH3 can have. Therefore, it is difficult to qualitatively show for which case the entropy is greater. Since both species are gases, intermolecular interactions are relatively weak, and we can guess that entropy is greater in NH3.

14

4.10 (a) To find the average distance between the two atoms of Ar, we can find the volume that each atom occupies. The molar volume can be found from the compressibility factor. At 300 K and 25 bar,

Tr  1.99 Pr  0.513 From the generalized compressibility charts: z  0.9859

Therefore,

v  0.9859

  J   8.314  300 K   mol  K    25  105 Pa 

 9.84  10

4

 m3     mol 

 m3  v  1.63 10  27    atom  The second number was found by dividing the molar volume by Avogadro's number. To estimate the distance between each atom, we note that the distance between molecules can be related to the volume by:

r3  v Consider the geometry shown below: 3

v

r

Ar

Ar

A rough estimate of r is



r  1.63  10  27 m 3

1 / 3  1.18 109 m 15

r  11.8 A 

(b) The potential energy due to gravity can be calculated as follows G 

 G  m 2Ar r

  m2   where G is the gravitation constant  G  6.67  10 11    and mAr is the mass of an argon 2    kg  s    atom. The mass of an argon atom is calculated as follows

 1 kg    39.948 g Ar  1 mol  kg     6.633  10 - 26  m Ar     23    atom   1 mol Ar   6.023  10 atoms  1000 g   Using the distance calculated in Part (a), we get

G  2.49 10 -52 J  (c) Equation 4.13 quantifies the potential energy due to London interactions Ar  Ar  

3  Ar Ar 2 r6

 I Ar I Ar   I Ar  I Ar

  

From Table 4.1

I Ar  15.76 eV  2.52  10 -4 erg

 

 Ar  16.6  10 25 cm 3





Using the distance calculated in Part (a) r  1.18  10  7 cm , we get

Ar  Ar  1.93  10 10 erg  1.93  10 17 J  (d) The potential energy due to London interactions is around 10 35 times greater than the potential energy due to gravity. Clearly, London interactions are much more important, and gravitational effects can be neglected.

16

4.11 We want to compare the Lennard-Jones potential to one with an exponential repulsion term. As provided in the text, Equation 4.19, the Lennard-Jones potential is

  12    6    4        r    r  To simplify further analysis, we can rewrite the equation in dimensionless quantities:

* 

  1 1     4  r *12 r *6 

r* 

r σ

where

We want to compare this to an exponential repulsive function * exp

   c2  1    c1 exp     r *  r * 12   

 

We have two adjustable parameters, c1 and c2, to match the first (LJ) potential to the second (exp) potential. We need to choose reasonable criteria to specify. For this solution we choose equal well depths and equal values at =1. Other choices may be just as valid; you should realize that you have two parameters to fit and so must specify two features. Using the above criteria, we iterate on a spreadsheet, to get the solution: c1 = 143,000 and

c2 = 11.8

This solution is shown at two magnifications in the plots on the following page:

17

Potential functions 0.3 0.2

Lennard-Jones exponential

0.1 0 0.9

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2

2.1

-0.1 -0.2 -0.3

Potential functions 20

Lennard-Jones

15

exponential 10

5

0 0.7

1

1.3

1.6

1.9

-5

We draw the following conclusions: 1. The most stable configuration (the bottom of the well) occurs at a greater separation for the exp model. 2. The Lennard-Jones potential increases more steeply at small radii, i.e., it behaves more like the hard-sphere potential. 3. The two models are in reasonable qualitative agreement

18

4.12 (a) The bond strength of a sodium ion can be viewed as the amount of energy it would take to remove the sodium ion from the crystal lattice. The interaction between the chlorine and sodium ions is Coulombic attraction.







QNa QCl 4.803  10 10 esu  4.803  10 10 esu 6  5.01  10 11 erg  r 2.76  10 -8 cm   31.3 eV 

6





(b)

6

QNa QCl Q Q  12 Na Na r r

  5.01  10 11 erg  12

4.803  10

  2.089  10 11 erg  13.0 eV

esu 4.803  10 10 esu  3.90  10 -8 cm

10

(c)

QNa QCl Q Q Q Q  12 Na Na  8 Na Cl r r r 4.803  10 10 esu  4.803  10 10 esu   2.089  10 11 erg  8 4.78  10 -8 cm   1.77  10 11 erg  11.1 eV 6





(d)

QNa QCl Q Q Q Q Q Q  12 Na Na  8 Na Cl  6 Na Na r r r r 10 4.803  10 esu 4.803  10 10 esu 11   1.77  10 erg   6 5.52  10 -8 cm   7.37  10 12 erg  4.60 eV 6





19





4.13 The boiling points of the halides depend upon the strength of intermolecular attractions. The stronger the intermolecular attraction, the higher the boiling point. Dispersion and dipole-dipole interactions are present in all five species listed. The magnitude of the dipole-dipole interactions is similar so the pertinent intermolecular force in these molecules is dispersion. The molecular size increases from left to right. Polarizabilities are greater in larger molecules, which manifests in larger dispersion forces. Therefore, the boiling point increases from left to right.

20

4.14 van der Waals forces hold the Xe atoms together in a molecule of Xe2. The potential energy can be quantified with the Lennard-Jones potential function. The bond length is the r value where the potential is a minimum. From Table 4.2:

 k and

 229 K

  4 .1 A

We start with Equation 4.19: to get   12    6    4        r    r  Differentiation with respect to r yields  12   12 6    6  d  4         0 dr r  r    r  r  where we set this derivative equal to zero to find the minimum. Solving gives 1      2 r 6

or r  6 2  1.12  4.60 A

21

4.15 The data in the following table were taken from Table A.1.1

Species

Tc K 

He CH4 NH3 H2O

5.19 190.6 405.6 647.3

Pc  10 -5 Pa  2.27 46.00 112.77 220.48

The van der Waals a parameter can be calculated using Equation 4.39. 27 RTc 2 a 64 Pc The van der Waals b parameter can be calculated using Equation 4.40. b

RTc 8 Pc

Using the data table and equations listed above, the following table was created; the values of dipole moment and polarizability reported in Table 4.1 are also included.

Species He CH4 NH3 H2O

 J  m3  a  2  mol  0.00346 0.230 0.425 0.554

 m3  b  105    mol  2.38 4.31 3.74 3.05

 [D]

 [cm x 1025]

0 0 1.47 1.85

2.1 26 22.2 14.8

3

The values of a for helium is two orders of magnitude less than the other species since it only has weak dispersion forces (small atom, small The values of a for methane, ammonia, and water are of the same magnitude because the sums of the intermolecular attractions for each molecule are similar. All three molecules have comparable dispersion forces; although slightly weaker in the ammonia and water. However, unlike methane, these two molecules also have dipole-dipole and induction forces. In fact, the strong dipole in water gives it the largest value The values of b are all of the same magnitude, as expected since b scales with size according to the number of atoms in the molecule. Size of Molecules:  Value of b:

CH 4  NH 3  H 2O  He bCH 4  bNH 3  bH 2O  bHe

22

4.16 The van der Waals b parameter can be calculated using Equation 4.40.

b

RTc 8 Pc

Critical point data can be found in Table A.1.1. The following table was made: Tc K 

Species

Pc  10 -5 Pa 

CH4 C6H6 CH3OH

190.6 46.00 562.1 48.94 512.6 80.96  J  Note: Used R  8.314  for calculation of a.  mol  K 

 m3  b  10 5    mol  4.306 11.94 6.58

The equation from page 186 can be rewritten to calculate  .

 3

3b 2N A

The table listed below was created using this equation, and data from Table 4.2 are included along with the percent difference. Species

  1010 m

CH4 C6H6 CH3OH

3.24 4.56 3.74

Table 4.2 Value   1010 m 3.8 5.27 3.6

23

Percent Difference 14.7 13.5 3.8

4.17 The van der Waals a parameter can be calculated using Equation 4.39. The values for the above equation were taken from Table A.1.1, and the following table was made:

Tc K 

Species

 J  m3  a  2  mol  0.2303 1.883 0.9464

Pc  10 -5 Pa 

CH4 C6H6 CH3OH

190.6 46.00 562.1 48.94 512.6 80.96  J  Note: Used R  8.314  for calculation of a.  mol  K  The equation from page 187 can be used to find C6. a

2N a2C6

3 3 3a 3  C6  2N a2 The  values can be found in Table 4.2. The following table can now be created Species

 1010 m

CH4 C6H6 CH3OH

3.8 5.27 3.6



C 6  10 77 J  m 6 1.66 36.3 5.82



To compare the values obtained from Equation 4.13, first calculate the potential energy with Equation 4.13: 

 3  i j 2 r6

 Ii I j   Ii  I j 

   

 II 3  i j  i j 2  Ii  I j

   

Therefore

C6 

We obtain the following values using this equation, and the corresponding percent differences were calculated.

24

Species CH4 C6H6 CH3OH





C 6  10 77 J  m 6 Page 187 Equation 4.13 1.66 1.021 36.3 12.0 5.82 1.36

Percent Difference 63 202 328

The values for the van der Waals a constant have the correct qualitative trends and order of magnitude; however, those values predicted from basic potential theory vary significantly from corresponding states. The basic potential result presented in the text assumes that the species are evenly distributed throughout the volume. It does not take into account the structure given to the fluid through intermolecular forces. In fact, a more careful development includes a radial distribution function, which describes how the molecular density of the fluid varies with r. The radial distribution function depends on pressure and temperature of the fluid.

25

4.18

P

RT a  v  b v2

Redlich-Kwong

P

RT a  1/2 v  b T v(v  b)

Peng-Robinson:

P

RT a(T )  v  b v(v  b)  b(v  b)

van der Waals:

As you may have discovered, these equations are largely empirical with no theoretical justification. They simply represent experimental data better. We can use our knowledge of intermolecular forces, however, to explain why these may work better. If we compare these equations to the van der Waals equation, we note that the first term on the right hand side of all three equation is identical; we accounted for this term as a correction for finite molecular size (or alternatively repulsive interaction due to the Pauli Exclusion Principle). This form represents a hard sphere model. The second term, that which deals with intermolecular attraction, is different in all three models. Both of the later equations include a temperature dependence in this term. We have seen that if attractive forces depend on orientation (dipole-dipole), they fall off with T as a result of the averaging process (recall discussion of Equation 4.11). Another explanation goes as follows: as T increases, the molecules move faster, reducing the effect of intermolecular forces. If we say that the potential energy between two molecules depends on the amount of time that they spend close to each other, then it would be inversely related to velocity (The faster molecules are moving, the less time they spend in the vicinity of other species). In this case, the correction term would go as V-1, where V is the molecular velocity. If we relate molecular velocity to temperature 2 1 2 mV

 32 kT

Therefore the correction term goes as T-1/2, as shown in the Redlich-Kwong equation. The inclusion of a "b" term in the second term may help relax van der Waal's "hard sphere" model with a more realistic potential function, i.e., something closer to a Lennard Jones potential (Figure 4.8) than the Sutherland potential (Figure 4.7) upon which the van der Waals equation is based. It makes sense that this should be included in the force correction since this is taking into account repulsive forces. One example of a more detailed explanation follows: If we look at the Redlich-Kwong equation, it says that if we have 2 species with the same attractive strength (same a -> same magnitude of van der Waals forces), the larger species will have less of an effect on P. The following sketch illustrates how 2 species could have the same van der Waals attractive forces:

26

- +

London = London + Dipole

Species 2 larger and non-polar (larger b, same a)

Species 1 smaller and polar (smaller b, same a)

In the case above, when the two species are the same distance apart, they have the same attractive force; however, the smaller species (1) can get closer before its electron cloud overlaps. Thus it has more opportunity for attractive interactions than the larger species. The Peng-Robinson equation exhibits the most complicated form in an attempt to better fit experimental data.

27

4.19 (a) Work is defined as follows

W    PdV Substitution of the ideal gas law for P yields W  

nRT dV V

  1 L   J  W  2.0 mol 8.314  1000 K ln    mol  K    10 L   W  38.29 kJ (b) Instead of substituting the ideal gas law into the definition of work, the Redlich-Kwong equation is used: v2

W  n 

v1

RT a dv  1 / 2 v  b T v (v  b)

Substituting

 J  R  8.314   mol  K  T  1000 K  n  2 mol  J  K1/2  m 3  a  14.24   2  mol   3 5 m b  2.1110    mol   m3  v1  0.0005    mol   m3  v2  0.005    mol 

28

and evaluating the resulting formula gives W  37.35 kJ

(c) Energy balance:

u  q  w Since the process is reversible

q  Ts and w  u  Ts

To use the steam tables conveniently, we need the initial and final pressures. We can calculate these with the Redlich-Kwong EOS: P1  1.65 MPa P2  15.6 MPa


 kJ  uˆ1  3522.6    kg   kJ  uˆ 2  3462.2    kg 

 kJ  sˆ1  8.101    kg  K   kJ  sˆ2  7.00    kg  K 

Therefore,    kJ    kJ   kJ    kJ  wˆ   3462.2    3522.6     1000 K  7.00   8.101     kg kg kg  K kg  K             kJ  wˆ  1040    kg  W  37.5 kJ 

The answers from the three parts agree very well. Part (a) is not as accurate as Part (b) and Part (c) because water is not an ideal vapor. The value from Part (b) is 1.1 % smaller than the value from Part (c). Clearly, the Redlich-Kwong EOS or steam tables are appropriate for this calculation.

29

4.20 First, we can obtain an expression for the compressibility factor with the van der Waals equation.

1 a Pv v a     b RTv RT v  b RTv 1 v To put this equation in virial form, we can utilize a series expansion: 1  1  x  x 2  x 3  ... 1 x Therefore, 2

1 b b  1        ... b v v 1   v and a   b   2 Pv RT   b    1     ... v RT v This expression can also be expanded in pressure. From Equation 4.58, we know that B'  C'

B RT C  B2

RT 2

where Pv  1  B' P  C ' P 2  .... RT Substituting B and C found above, we get B' 

bRT   a

RT 2

and C ' 

2abRT   a 2

RT 4

Therefore,

30

2   bRT   a  Pv  P   2abRT   a  P 2  ....  1   RT 2   RT RT 4    

31

4.21 Using the virial expansion, the pressure can be written as follows

 1 B C P  RT     ...   v v 2 v3 The virial expansion in pressure is P





RT 1  B' P  C ' P 2  ... v

If we substitute the first expression into the second, we obtain 2   1 B C    1 B C  RT  1 B C    ...    C '  RT       RT   1  B ' RT   2 v  v v 2 v 3  v v v3     v v 2 v3      

If we combine like terms on the right side and set them equal to terms on the left, we find

B  B' RT C  BB' RT  C ' RT 2 Substitute the expression for B’ into the expression for C and solve for C’: C' 

CB

RT 2

and B' 

B RT

32

4.22 At the critical point we have:

Pc 

RTc a  vc  b Tc vc2

(1)

RTc 2a  P     0 2  v Tc vc  b  Tc vc3

(2)

 2  2 RTc 6a  P   2  0 3 Tc vc4  v   vc  b  Tc 

(3)

If we multiply Equation 2 by 2 and Equation 3 by (v-b) and add them together: 0

4a vc3



6avc  b  vc4

this can be solved to give: vc  3b If we plug this back into Equation 2 and solve for a, we get: a

9 vc RTc2 8

Finally, if we plug this back into Equation 1 we can solve for the Berthelot constants in terms of the critical temperature and the critical pressure:

a

27 R 2Tc2 64 Pc

and

b

RTc 8Pc

33

4.23 Before we can find the reduced form, we need to find expressions for a and b in terms of critical point data (Problem 4.22). We find

vc  3b

(1)

27 R 2Tc3 a 64 Pc RTc b 8 Pc

(2) (3)

The Berthelot equation is P

RT a  v  b Tv 2

First, substitute Equation 1 into the first term on the right-hand side and rearrange to get RT a P b  3vr  1 Tv 2 Now, substitute Equation 3: P

8 PcTr a  3v r  1 Tv 2

Substitute Equation 2:

P

8 PcTr  3v r  1

27 R 2Tc3 64 Pc Tv 2

From Equation 3: R 2Tc2  64 Pc2 b 2 Substituting this result, we get P

8PcTr 27 PcTc b 2  3v r  1 Tv 2

34

Finally, substitute Equation 1: P

8PcTr 3PcTc vc2  3v r  1 Tv 2

This can be rewritten as Pr 

8Tr 3  3vr  1 Tr vr2

35

4.24 (a) The virial equation is: z

Pv D B C  1   2  3 ... RT v v v

(1)

We can rewrite this equation: C  Pv   1v  B  ... RT v

z  1v  

(2)

PvT data from the steam tables are given in the steam tables and corresponding values of (z-1)v and 1/v can be calculated. An illustrative plot of (z-1)v vs. 1/v for T = 300 oC is shown below. Plot to determine the second virial coeff at 300 C -115

(z -1)v (cm3 / mol)

-120

-125

(z -1)v -130

2

1.5

1

0.5

0

-135

1/v (mol/l )

We see that at pressures above 1.5 MPa (15 bar) we see that the plot reaches a constant value of around -118 cm3/mol. We may choose to report this value as B. A more careful examination of Equation 2 suggests another possibility. If we plot (z-1)v vs. 1/v, we should get a straight line at low to moderate pressures with an intercept equal to the second virial coefficient, B, and a slope equal to the third virial coefficient, C. At very low pressures, the curve is indeed linear and gives an intercept of -140 cm3/mol, as shown on the next page. The slope of this region would yield the third virial coefficient, C. Can you calculate C?

36

The first value uses much more data while the second method uses limited data in a better range. What value would you be more apt to use? Water exhibits this “odd” general behavior at all temperatures. Plot to determine the second virial coeff at 300 C -115

(z -1)v (cm3 / mol)

-120

y = 177.058x - 140.373 -125 (z -1)v -130

lin

2

1.5

1

0.5

0

-135

1/v (mol/l )

If we do this at other values of T, we can compile a set of second virial coefficients vs. temperature and get an idea of the differences in the two approaches. Temperatures of 200, 250, 300, 350, 400, 500, 600, 700, 800, 900, 1000, and 1200 oC analyzed in this manner. The results are reported in the table and figure below. Also reported were the average value and the value at the lowest pressure used. Note that the average value is indicative of the 1st method above while the value at the lowest pressure used is indicative of the second value used. T 200 250 300 350 400 500 600 700 800 900 1000 1200

B (Level) -214 -156 -118 -89 -72 -49 -34 -24 -17 -13 -9 -3

B (Linear) -226 -173 -140 -102 -103 -83 -73 -68 -66 -51 -50 -51

37

B (AVG) -215 -157 -120 -92 -76 -54 -39 -30 -24 -19 -16 -11

B(1st value) -220 -168 -134 -99 -95 -75 -63 -57 -54 -52 -52 -52

Second Virial Coefficient of Water by 4 Methods 0

B (water) cm 3 /mol

-50

-100 B (average) B (lowest P)

-150

B (Level) -200

B (Linear) CRC 1500

1250

1000

750

500

250

-250

T (K)

For comparison, values of -142.2 cm3/mol and -7160 cm6/mol2 are reported for B and C, respectively, at 260 oC1. Values of B reported in the CRC are also shown on the summary plot. They agree most closely with the first (level) method. (b) There are several alternatives how to solve this problem, each of which comes up with a slightly different result:

Alternative 1: Rewrite the virial equation:   1 BH 2 O P  RT    ... v2  v Take the derivative:   1 2 BH 2 O   P      0  RTc  2   v  v Tc v 3c    c

so  cm3  v BH 2O   c  28   2  mol 

1

J.H. Dymond and E.B. Smith, The Virial Coefficients of Pure Gases and Mixtures, A Critical Compilation, Oxford University Press, Oxford, 1980.

38

Alternative 2: From the virial equation:  cm3  Pv  BH 2O   c c  1vc  43.2    mol   RTc  Alternative 3:   1 BH 2 O C H 2 O P  RT     ... v2 v3  v so   1 2 BH 2O 3C H 2O   P       0  RTc  2  3  v  v Tc v v 4c  c   c

(1)

  2P  6B 12C H 2O      0  RTc  2  H 2O  4 5  v 2   v3 vc vc   Tc   c

(2)

and

Multiply Equation 1 by (4/vc) and add to Equation 2 to get: 

2 v 3c



2 BH 2 O v 4c

0

so  cm3  BH 2O  vc  56    mol 

39

4.25 A trial-and-error is the easiest method for solving this problem. The general method is as follows

1. Guess P sat 2. Calculate values of molar volumes that result at the chosen P sat : vlow , vmid , vhigh 3. Find areas under the curves with the following expressions

 P

v mid

sat



 f v  dv

v low

  f v   P

v high

sat

dv

v mid

[ f v  represents the particular EOS implicit in molar volume being used.] 4. If the values of the expressions are not equal, repeat the process until they are. (a)

1. Guess P sat : P sat  6 bar 

2. Calculate molar volume solutions for: RT a  P sat  1 / 2 v  b T v (v  b )  J  K1/2  m 3    J  41 . 851    8.314  2  363.15 K  mol  mol K      6  105 Pa      m3    3  v  0.0001  363.15 K 1 / 2 v v  0.0001  m     mol   mol     m3   m3   m3   0 . 000551  0 . 00459 vlow  0.000152  v v    high  mid   mol   mol   mol  3. 0.000551 

 a sat  RT dv  1157.19 P       v  b T 1 / 2 v (v  b )     0.000152  0.00459    a  RT    P sat dv  1325.78    v  b T 1 / 2 v (v  b )    0.000551 

40

4. Repeat this process until the areas are equal. This occurs approximately at P sat  6.38 bar 

(b)

1. Guess Psat : Psat  6 bar  2. Calculate molar volume solutions for P sat 

RT a T   v  b v(v  b)  bv  b 

using  m3   J  m3  a  2.066  , b  9  10  6    ,  T   1.188 2  mol   mol   m3   m3   m3   0 . 000551  0 . 00459 vlow  0.000152  v v    high  mid   mol   mol   mol  3.  sat  RT a T   P  v  b  v(v  b)  bv  b  dv  1568.48  0.000152  0.000551



 RT a T  sat   v  b  v(v  b)  bv  b   P dv  1100.19  0.000551  0.00459



4. Repeat this process until the areas are equal. This occurs approximately at P sat  4.945 bar 

The value calculated with the Redlich-Kwong EOS is 11.9% greater than the measured value of 5.7 bar. The Peng-Robinson EOS results in a value that is 13.25% less than the measured value.

41

4.26 First, calculate a , b , and  T  . From Table A.1.1:

Pc  48.84 Pa 

Tc  305.4 K 

w  0.099

Now we can calculate the required parameters.

a

   J  305.4 K  0.45724 8.314    mol  K     48.84  105 Pa 

2

 J  m3   0.6036   2  mol 

  J  0.07780 8.314  305.4 K   mol  K     4.045  10  5 b 5 48.84  10 Pa 



  T   1  0.37464  1.542260.099  0.269920.0992   T   1.12

 m3     mol 



 243.15 K   1    3 05.4 K  

2

Now, we can find the three solutions to the Peng-Robinson EOS: P sat 

RT a T   v  b v(v  b)  bv  b 

Using the values calculated above and P sat  10.6  105 Pa   J  R  8.314   mol  K  we get  m3  4 v  7.199  10  5   , v  2.167 10  mol 

 m3   3 3 m   , v 1 . 578 10      mol   mol 

The molar volume of saturated ethane liquid is the smallest value from the list above, while the molar volume of saturated ethane vapor is the largest value. Therefore,

42

 g  36.0694  3  MW ethane  mol   1 m   0.501  g    liq   3 liq 3 3   3   100  cm  v  cm  5 m 7.199  10    mol   g  36.0694  3  MW ethane  mol   1 m   0.0229  g    vap   3  3   100 3 cm 3  v vap  cm  3 m 1.578 10    mol  Both of the densities calculated with the Peng-Robinson EOS are larger than the reported values. The liquid density is 7.05% larger, and the vapor density is 18.7% greater.

43

4.27 (a) We will use the Redlich-Kwong EOS in order to obtain an accurate estimate. First, calculate the a and b parameters:

a

 J  m 3  K1 / 2  0.42748R 2Tc2.5  1.56   Pc mol 2  

 m3     mol  (Note: Critical data for nitrogen obtained in Table A.1.1.) 0.08664 RTc  2.69  10  5 b Pc

Now use the EOS to find the molar volume: P

RT a  1 / 2 v  b T vv  b 

Assume the temperature of the gas is 22 ºC, substitute values, and find the molar volume with a solver function:  m3  v  1.97 10  4    mol  Now calculate the total volume of gas in the 30,000 units: Vtotal  30000 units 43 liter/unit   1.29  10 6 liters  1290 m 3 Therefore, the number of moles is: n

1290 m 3  6.55  10 6 mol 3 m  1.97  10  4    mol 

and the mass





m  6.55  10 6 mol 0.02801 kg/mol  1.83  105 kg Now, calculate the value of the gas:





Value  $6.1 / kg  1.83  105 kg  $1,116,300

44

If we use the ideal gas law, we find:   J  295 K   8.314  RT   mol  K    1.98  10  4  v P 12400000 Pa

 m3     mol 

Following the steps above, Value  $1,113,200 The value calculated using the ideal gas law is $3100 less than the value calculated using the Redlich-Kwong EOS. (b) First, calculate the a and b parameters:

 J  m 3  K1 / 2  0.42748R 2Tc2.5  1.74  a  Pc mol 2    m3     mol  (Note: Critical data for oxygen obtained in Table A.1.1.) b

0.08664 RTc  2.21  10  5 Pc

Assume the temperature of the gas is 22 ºC, substitute values, and find the molar volume with a solver function:  m3  v  1.53  10 4    mol  Following the steps presented in Part (a), we find that

Value  $2,428,000 With the ideal gas law, we find  m3  v  1.64  10 4    mol  which provides

Value  $2,265,000

45

The value found with the ideal gas law is $163,000 less than the value found using the RedlichKwong EOS.

46

4.28 First, rewrite the Redlich-Kwong equation in cubic form.

v3 

RT 2  a RT ab  v   b  b 2 v  0 P P  PT 1 / 2  PT 0.5

or at the critical point v3 

 RTc 2  a RT ab v   c b  b 2 v  0 0 . 5 0.5   Pc P P T P T c c c  c c 

Now expand v  vc 3  0 v 3  3v 2 vc  3vvc2  vc3  0 Setting the coefficients equal we get the following expressions. 3vc 

RTc Pc

3vc2 

a

vc3 

(1)

PcTc0.5 ab



RTc b  b2 Pc

(2) (3)

Pc Tc0.5

Using Equation 2, find an expression for a:   RT a   3vc2  C  b 2  Pc Tc0.5 PC  

Substitute the above expression into Equation 3 to get b 3  3vc b 2  3vc2b  vc3  0 The one real root to this equation is





b  21 / 3  1 vc

(4)

Substitute Equation 4 into Equation 1:

47

b

0.08664 RTc Pc

(Equation 4.48)

Now, substitute Equation 1 and Equation 4.48 into the expression for a to get a

0.42748RTc2.5 Pc

(Equation 4.47)

Pv To verify Equation 4.50, substitute Equation 1 into z c  c c RTc

 RT Pc  c 3Pc zc   RTc

  1 3

(Equation 4.50)

The Redlich-Kwong equation is P

RT a  v  b T 0.5vv  b 

We can use Equation 4 to get 0.2599 RT a b P  vr  0.2599 T 0.5vvc vr  0.2599 

Now, substitute Equation 4.48 P

3Tr Pc a  0 . 5 v r  0.2599 T vvc v r  0.2599 

Replacing a with Equation 4.47 yields P

3Tr Pc 0.42748R 2Tc2.5  v r  0.2599 PC T 0.5 vvc v r  0.2599

From Equation 4.48 R 2Tc2 

Pc2 b 2

0.086642

48

which upon substitution yields P

3Tr Pc Pc b 2 0.42748  v r  0.2599 0.08664 2 Tr0.5 vvc v r  0.2599

Now, use Equation 4: P

3Tr Pc Pc 0.42748  0.2599 2  v r  0.2599 Tr0.5 v r v r  0.2599 0.086642

Therefore, Pr 

3Tr 1  0 . 5 vr  0.2599 0.2599Tr vr vr  0.2599

49

(Equation 4.49)

4.29 Since we are concerned with liquid water, we can base all our calculations on saturated liquid water since the molar volume of liquids are weakly dependent on pressure. Therefore, our results are also applicable to sub-cooled water (the pressure is greater than the saturation pressure). To find the thermal expansion coefficient, we will use the following approximation

 vˆ sat T  5 º C   vˆ sat T  º C   1  vˆ sat T  5 º C   vˆ sat T  º C        10 K vˆ sat T   T  5 º C   T  5 º C   vˆ sat T    This approximation is valid even though the saturation pressure changes because molar volume is weakly dependent on pressure. From the steam tables



1

 m3  vˆ sat 15 º C   0.001001    kg   m3  ˆv sat 20 º C   0.001002    kg 

 m3  vˆ sat 95 º C   0.001040    kg   m3  ˆv sat 100 º C   0.001044    kg 

 m3  vˆ sat 25 º C   0.001003    kg 

 m3  vˆ sat 105 º C   0.001047    kg 

Using these values, we can calculate the thermal expansion coefficients   m3   m3     0 . 001003 0 . 001001     1  kg kg  m3             20 º C    0.001002       10 K  kg            m3   m3     0.001040   1  0.001047   kg  kg    m3         100 º C    0.001044       10 K  kg         

   100 º C   6.71  10 K 

 20 º C   2.0  10  4 K -1 4

-1

The accuracy of these values may be limited since they are based on the small differences between liquid volumes. To calculate the isothermal compressibility, a similar approximation will be used. It is

50



 dvˆ sat  vˆ sat T   dP 1

   T

The derivatives are determined from the following graph. Specific Volume of Liquid Water vs. Pressure

v = -5E-07*P + 0.001044 R2 = 0.9949

0.001040

3

Specific Volume [m /kg]

0.001050

0.001030

T = 20ºC T = 100 ºC

0.001020

Linear (T = 100 ºC)

Linear (T = 20ºC)

0.001010

v = -5E-07*P + 0.001002 R2 = 0.9979

0.001000 0.000990 0

5

10

15

20

Pressure [MPa]

It is clear that  dvˆ sat   dP 

  dvˆ sat      5  10  7   dP  T  20 0 º C  T 100 0 º C

 m3     kg  MPa 

Therefore,   m3    20 º C    0.001002       kg   

1

  5  10  7  

 m3       kg MPa   

 1  1  4.99  10 10     Pa   MPa 

 20 º C   4.99  10  4 

  m3    100 º C    0.001044      kg   

1

  5 10  7  

 m3      kg MPa     1  1  4.79 10 10    100 º C   4.79 10  4    MPa   Pa 

51

25

4.30 (a) Substitute critical data into the Rackett equation:

vcalc 

8.314190.6 0.29056  0.087750.0081 1111 / 190   5 2/7

46  10  cm 3  vcalc  38    mol  Now, calculate the error.

 v calc  vexp Error    vexp  Error  0.8%

    100 %    

(b)

 cm 3  vcalc  54.2    mol 

Error  1.1%

 cm 3  vcalc  161.2    mol 

Error  10.8%

 cm 3  vcalc  20.5    mol 

Error  13.5%

 cm 3  vcalc  68.5    mol 

Error  19.7%

(c)

(d)

(e)

1-hexanol had the largest percent error, while ethane and methane had the smallest percent errors. The alkenes have low percentile errors due to their nonpolar nature, which the alcohols and acids had large percentile errors due to being polar substances and exhibiting hydrogen bonds.

52

4.31 (a) Since we are given the temperature and pressure, we can make use of the generalized compressibility charts. Using data from Table A.1.1, the required quantities can be found.

343.15 K  T   1.12 305.4 K  Tc 30 bar  P   0.616 Pr  Pc 48.74 bar  w  0.099

Tr 

By double interpolation of the charts z 0   0.8376 z 1  0.0168

and

z  z 0   wz 1  0.8376  0.099 0.0168  0.8393

Therefore,

V

znRT  P

 30 kg   J   8.314  343.15 K   mol  K    0.03007 kg/mol 

0.8393

 

30 10

5

Pa 

V  0.796 m 3

(b) The Redlich-Kwong EOS should give reasonably accurate results. Room temperature was assumed to be 25 ºC. The molar volume is required for the calculations, so

v V 

  

V 0.1 m 3   7.518  10  5 n   40 kg      0 . 03007 kg/mol  

 m3     mol 

Using data from Table A.1.1 and Equations 4.47 and 4.48,  J  m 3  K1 / 2  a  9.88   2  mol 

53

 3 5 m   b 4.51 10    mol  Substituting these values into the Redlich-Kwong EOS and evaluating gives P  191.3 bar 

54

4.32 (a) For propane

MW  propane  0.0441  kg   mol 

n 

50 kg   1130 mol  kg  0.0441   mol 

Now calculate the volume:

V 

nRT  P

1130 mol 8.314  

 

J  50  273.15 K   mol  K  

35  105 Pa 

V  0.870 m 3

(b) The Redlich-Kwong EOS is

P

RT a  v  b T 1 / 2 vv  b 

a

 J  m 3  K1 / 2  0.42748R 2Tc2.5  18.33   Pc mol 2  

where

 m3     mol  (Note: Critical data for propane obtained in Table A.1.1.) b

0.08664 RTc  6.28  10  5 Pc

Now that these values are known, there is only one unknown in the Redlich-Kwong EOS: v . Using a numerical technique, e.g., the solver function on a graphing calculator  m3  v  0.000109    mol 

 

V  0.124 m 3

55

(c) The Peng-Robinson EOS is

P

RT a T   v  b vv  b   bv  b 

where

 

  1   1  Tr Tr 

2

50  273.15 K  0.873 T  370 K Tc

  0.37464  1.542260.152  0.269920.11522  0.605

  1.081 a

 J  m3  0.45724RTc 2  1.02   Pc  mol 

0.07780 RTc b  5.64  10  5 Pc

 m3     mol 

Now every variable in the Peng-Robinson EOS is known, except v.  m3  v  0.0000942    mol 

 

V  0.107 m 3

(d) We must calculate the reduce temperature and pressure to use the compressibility charts:

323.15 K T   0.873 370 K Tc 35 bar P   0.825 Pr  Pc 42.44 bar

Tr 

By double-interpolation on the compressibility charts (Appendix C), z 0   0.1349 z 1  0.052

Therefore,

56

Pv  z  z 0   wz 0   0.1349  0.152 0.052  0.127 RT  m3  0.127 RT  0.00009749  v  P  mol  V  0.111 m 3 (e) From ThermoSolver Using the Peng-Robinson equation,

 m3  v  0.00009429    mol 

 

V  0.107 m 3

Using the generalized compressibility charts:  m3  v  0.0000971    mol  Therefore,

 

V  0.110 m 3

57

4.33 Note: Multiple possibilities exist for which substance to use in the vial. The solution using one possibility is illustrated below. (a) The substance must have a critical temperature above room temperature but below the temperature of one’s hand. From the Appendix A.1.2, we that for carbon dioxide

Tc  304.2 K  31.1 º C Clearly, CO2 is a suitable substance, and it is safe to use. (b) The vial must be able to withstand the pressure of the substance at its critical point. Therefore, the vial must withstand 73.76 bar (the critical pressure of carbon dioxide). (c) Since the substance passes through its critical point, the molar volume at that state is constrained by the critical temperature and pressure. To estimate the molar volume, we can use the PengRobinson EOS. Calculate the necessary parameters for the EOS: 2

a  0.45724

  J   8.314   304.2 K 2   mol  K    73.76  10 5 Pa 

 J  m3   0.40   2  mol 

  J   8.314  304.2 K   mol  K    b  0.07780  3.4  10  4 5 73.76  10 Pa  α 1

 m3     mol 

Substitute the above parameters and solve for the molar volume:  cm 3   m3  vc  1.18  10  4  118      mol   mol  Now, we can calculate the required amount of CO2. n

 

100 cm 3 V   0.847 mol vc  cm 3  118    mol 

58

(d) If the vial contains less substance than needed, the molar volume will be greater. Therefore, the substance will not pass through the critical point. As the substance is heated, it will go through a transition from a saturated liquid and vapor mixture to superheated vapor. Then, it will become a supercritical fluid once the critical temperature is exceeded. P

critical point

liquid

part c

part d not enough CO2

vapor

liquid - vapor

v

59

4.34 Methane: The following quantities are required to calculate the molar volume with the Peng-Robinson equation.

 

 T   1   1  Tr  T   0.9626

2

a

 J  m3  0.45724RTc 2  0.25   Pc  mol 

b

0.07780 RTc  2.68  10  5 Pc

T  Tr Tc  209.66 K 

 m3     mol 

P  Pr Pc  55.2  10 5 Pa  The molar volume is the only unknown in the following equation. P

RT a T   v  b vv  b   bv  b 

v  1.83  10

4

 m3     mol 

Therefore,





  3  4 m  55.2  10 Pa  1.83  10     mol     z  0.58   J   8.314  209.66 K   mol  K    5

From the compressibility charts, z 0   0.5984 z 1  0.0897

Therefore, z  z 0   wz 1  0.5984  0.0080.0897 z  0.599

60

Methanol:

Following the procedure outlined above, the Peng-Robinson equation produces  m3  v  3.14 10  4    mol   z  0.651 Using the compressibility charts: z 0   0.5984 z 1  0.0897

From Table A.1.1, w  0.559

Therefore, z  z 0   wz 1  0.5984  0.559 0.0897 z  0.649 Summary:

Methane z  0.58 (Peng-Robinson) z  0.599 (Compressibility charts) The value from the Peng-Robinson EOS is 3.2% smaller than the value from the charts. Methanol z  0.651 (Peng-Robinson) z  0.649 (Compressibility charts) The value from the Peng-Robinson EOS is 0.31% smaller than the value from the charts.

61

4.35 From Table A.1.1:

Tc  405.6 K Pc  112.77 bar w  0.25 Calculate reduced temperature and pressure:

92  273.15 K  0.9 T  405.6 K Tc 306.5 bar P   2.718 Pr  Pc 112.77 bar

Tr 

From interpolation of the compressibility charts: z 0   0.4133

z 1  0.1351

Therefore, Pv  z  z 0   wz 1  0.38 RT and v

0.38RT  3.76  10  5 P

 m3     mol 

Since the temperature of the ammonia in this system is below the critical temperature, we know the ammonia is not a supercritical fluid. The pressure is greater than the critical pressure, so the ammonia is a liquid.

62

4.36 Let the subscript “ace” represent acetylene and “but” represent n-butane. First, convert the given quantities of acetylene and n-butane into moles. nace 

nbut 

30 kg  26.038  10

-3

50 kg  58.123 10

-3

 kg   mol   kg   mol 

 1152.2 mol  860.2 mol

Therefore, y ace  0.573 ybut  0.427 We can also calculate a and b parameters for pure species using the following equations 0.42748 R 2Tc2.5 a Pc b

0.08664 RTc Pc

Substituting critical data from Table A.1.1 gives  J  K1/2  m3  aace  8.03   2  mol   J  K1/2  m 3  abut  29.07   2   mol

 m3  bace  3.62 10  5    mol   m3  bbut  8.08110  5    mol 

Now, we can use mixing rules to calculate the parameters for the mixture. amix  y12 a1  2 y1 y 2 a12  y 22 a2  J  K1/2  m3  a12  8.03  29.07 1  0.092  13.87   2  mol   J  K1/2  m 3   amix  14.72   2  mol 

63

bmix  y1b1  y 2b2  5.53  10

5

 m3     mol 

Substitution of the mixture parameters into the Redlich-Kwong EOS results in an equation with one unknown.  m3  v  0.00111    mol    m3    V  nace  nbut v  1152.2 mol  860.2 mol 0.00111     mol    

 

V  2.23 m 3

64

4.37 First, we need to calculate the a and b parameters for species (1) and (2). From Table A.1.1 and A.1.2:

CO2 (1):

Toluene (2):

Tc  304.2 K 

Pc  73.76 bar 

a1 

 J  m 3  K1/2  0.42748R 2Tc2.5  6.466   Pc mol  

b1 

0.08664 RTc  2.97  10  5 Pc

 m3     mol 

Tc  591.7 K 

Pc  41.14 bar 

 J  m 3  K1/2  0.42748R 2Tc2.5 a2   61.17   Pc mol   b1 

0.08664 RTc  1.036  10  4 Pc

 m3     mol 

We can use the mixing rules to calculate bmix. bmix  y1b1  y 2b2  2 mol  5 bmix    2.97  10  5 mol    3 5 m bmix  7.40  10    mol 

 m 3    3 mol  4     1.036  10   mol 5 mol     

 m3      mol   

Now we can substitute the known values into the Redlich-Kwong EOS. Note that the molar volume can be calculated as follows v

 

 m3  0.01 m 3 V   0.002   5 mol ntot  mol 

We can then solve for amix.  J  m 3  K1/2  a mix  31.0   mol  

65

From the mixing rules, we know  J  m3  K1/2  2 2 amix  31.0    y1 a1  2 y1 y2 a12  y2 a2 mol   Therefore,  J  m 3  K1/2  a12  16.55   mol   Equation 4.79 states a12  a1a 2 1  k12 

Substitution of the values provides: k12  0.17

66

4.38 (a) We can match the species by looking at the b parameter alone. The b parameter is directly related to the size of the molecule. Because sizeC 2 H 6  size H 2 O  size H 2 , bC 2 H 6  bH 2 O  bH 2 . Therefore,

 J  m3  a   mol  0.564

 m3  b   mol 

Species

6.38  10 5

C2H6 (3)

0.025

2.66  10 5

H2 (1)

0.561

3.05 10 5

H2O (2)

These values are also consistent with what we would expect for the magnitude of van der Waals interactions given by a. (b) The van der Waals parameters for the mixture can be calculated according to the mixing rules. From Equations 4.81 and 4.82 a mix  y12 a1  y1 y 2 a12  y1 y3 a13  y 2 y1 a21  y 22 a 2  y 2 y3 a 23  y3 y1 a31  y3 y 2 a32  y32 a3 a mix  y12 a1  2 y1 y 2 a12  2 y1 y3 a13  y 22 a2  2 y 2 y3 a 23  y32 a3 bmix  y1b1  y 2b2  y3b3

Calculate mole fractions: n1 5 mol   0.5 n1  n2  n3 10 mol y 2  0 .4 y3  0.1 y1 

Since binary interaction parameters are not available, we must use Equation 4.78 to calculate a12 , a13 , a23 .  J  m3  a12  a1a2  0.118    mol   J  m3  a13  0.119    mol 

67

 J  m3  a23  0.562    mol  Now we can find the numerical values for the van der Waals parameters.  J  m3  amix  0.206    mol   m3  bmix  3.19  105    mol  (c) For the mixture, the van der Waals equation is P

a RT  mix v  bmix v2

The molar volume is calculated as follows v

 

 m3  0.0125 m 3 V   0.00125   10 mol n1  n2  n3  mol 

Therefore,  J  m3    J  0 . 206    8.314   300 K  mol  mol K       P  2   3   3  m3    0.00125  m   3.19  10  5  m     0.00125       mol     mol    mol   P  1.92 MPa

68

4.39 (a) To calculate the molar volume of the mixture, we will use the virial expansion in pressure since it is more accurate at moderate pressures. But first, we must calculate the second virial coefficient for the mixture.

Bmix  y12 B1  2 y1 y 2 B12  y 22 B2     cm 3    cm 3    cm 3   2         0 . 75 110 2 0 . 25 0 . 75 153 Bmix  0.252   625                  mol    mol    mol     

 cm 3  Bmix  158.3    mol  Therefore, RT  Bmix  RT P   Bmix 1  P  RT  P  cm 3   m3  v  2445.6    0.002446    mol   mol 

v

(b) We can estimate k12 using an EOS where the a parameter is the only unknown. b can be calculated as follows

bmix  y1b1  y 2b2 Using the Redlich-Kwong EOS,  bmix  0.25 5.83 10  5  

  m3   5     0.75 2.97 10   mol    

 m3    3 5 m     3.69  10     mol    mol 

Substitute the known properties into the Redlich-Kwong EOS and solve for a.  J  K1/2  m 3  amix  8.693   2  mol  Now we can calculate k12 as follows a mix  y12 a1  2 y1 y 2 a12  y 22 a2

69

 J  K1/2  m 3  a1  28.03   2  mol   J  K1/2  m 3  a2  6.466   2  mol   J  K1/2  m 3   a12  8.81   2  mol 

(Calculated using Equation 4.47)

From Equation 4.79 a12  a1a 2 1  k12  8.81 k12  1   0.35 6.466  28.03

70

4.40 During this process the gas in Tank A undergoes an expansion from 7 bar to 3.28 bar. During this isentropic process, the gas cools. If attractive forces are present, they will manifest themselves more in state 1 at the higher pressure than in state 2. (While T has some effect to counter this trend, we expect it will be secondary to the effect of pressure). Thus, in the nonideal case, additional energy must be supplied to “pull” the molecules apart. Since the tank is insulated, the only place that this can come from is the kinetic energy of the molecules. Thus, they slow down even more than in the ideal gas case, and the final temperature is lower. If we had an equation of state that appropriately described the non-ideal behavior, we could apply the concepts of the thermodynamic web that we have learned in this chapter to solve for the final temperature.

71

4.41 Select the “Equation of State Solver” from the main menu. Enter the pressure and temperature and solve for the molar volume using the Lee-Kesler EOS. The program provides

 m3  v  1.07  10  4    mol  To calculate the volume occupied by 10 kilograms of butane, we need to know the number of moles present in 10 kg. n

10 kg   172.1 mol 0.05812 kg/mol

Therefore,  V  172.1 mol1.07  10  4  

 

 m3   3     0.018 m   mol  

The answers provided in Example 4.9 agree well with this answer. The answer found using the Redlich-Kwong EOS is 16.7% larger than the answer from ThermoSolver, but the answer from the compressibility charts is only 5.56% larger.

72

4.42 Using ThermoSolver should be straightforward; thus, only the answers are provided. (a)

Tc  305.4 K 

Pc  48.84 bar 

 kJ  h f ,298  84.68    mol 

(b)

Tsat  299.42 K 

(c) The percent difference is calculated as follows

 v LK  v PR  z LK  z PR % Difference   or  100 % v LK z LK   (i). Quantity v z

Lee Kessler EOS  m3  8.34  10  5    mol  0.1383

Peng Robinson EOS  m3  8.70  10  5    mol  0.1444

% Difference

Lee Kessler EOS  m3  3.68  10  4    mol  0.5859

Peng Robinson EOS  m3  3.58  10  4    mol  0.5711

% Difference

4.3 4.4

(i). Quantity v z

73

2.7 2.5



5.1 (a) Following the example given by Equation 5.5a in the text  u   u  du    dT    dP  P T  T  P

(b)  u   u  du    dT    ds  T  s  s T (c)  u   u  du    dh    ds  s  h  h  s

2

5.2. The internal energy can be written as follows  u   u  du    dT    dv  T  v  v T Substituting Equations 5.38 and 5.40

  P    u   u    P    cv and    T   v T   T  v  T  v  into the above expression yields

  P   du  cv dT  T    P  dv   T  v  From the ideal gas law, we have

R  P      T  v v Therefore,

 RT  du  cv dT    P  dv  v  which upon noting that P 

RT for an ideal gas, becomes v

du  cv dT

3

5.3 The heat capacity at constant pressure can be defined mathematically as follows

 h   u   Pv    u   v  cP         P  T  T  P   P  T  v  T  P For an ideal gas:

R  v      T  P P Therefore,

 u  cP    R  T  v One mathematical definition of du is

 u   u  du    dT    dP  T  P  P T  u  We can now rewrite   :  T  v  u   u   T   u   P             cv  T  v  T  P  T  v  P T  T  v For an ideal gas:

 u    0  P  T so

 u  cv     T  P Substituting this result into our expression for c P gives

c P  cv  R

4

5.4 In terms of P, v, and T, the cyclic equation is

 P   T   v  1         T  v  v  P  P T For the ideal gas law: Pv  RT

so the derivatives become:

R  P      T  v v P  T      v  P R  RT  v  v     2  P  P T P Therefore,

 P   T   v   R  P   v              1  T  v  v  P  P T  v  R  P  The ideal gas law follows the cyclic rule.

5

5.5 For a pure species two independent, intensive properties constrains the state of the system. If we specify these variables, all other properties are fixed. Thus, if we hold T and P constant h cannot change, i.e.,

 h  0    v T , P

6

5.6 Expansion of the enthalpy term in the numerator results in  Ts  vP   h      T  T  s  s  P   h      v  T  s  T  s

Using a Maxwell relation  s   h    v    v  P  T  s  s   T   h        v  T  P  v  P  T  s We can show that c  s    P   T  P T

(use thermodynamic web)

1  RT 2a 2ab   T          v  P R  v  b v 2 v 3 

(differentiate van der Waals EOS)

Therefore,  v vc  RT 2a 2ab  2a  b    h    c P     1      P   2 3  T  s RT  v  b v v   v  b vRT  v 

7

5.7  h    :  P T

 s   Ts  vP   h    T   v    P  P T T  P T 



R  v   s  2    1  B' P  C ' P     P  T  P  P T







RT  h  1  B ' P  C ' P 2  v  v  v    P P   T  h    0  P T

 h    :  P  s  h   Ts  vP   v    P s  P  s  RT  h  1  B' P  C ' P 2    P  P  s





 h   :   T  P  h    cP   T  P

(Definition of cP)

 h   :   T  s  P   Ts  vP   h     v    T  T  s s  T  s  c  T  c P  s   P   P    P     P     T  v  P T R 1  B' P  C ' P 2  T  P  s T  T  s



8



 

Pv 1 1  B' P  C ' P 2  h   c  c   P P RT 1  B' P  C ' P 2  T  s 1  B' P  C ' P 2  h    cP   T  s





9

 

5.8 (a) A sketch of the process is provided below

well insulated

m

s = 0

T1 P1

T2 P2

The diagram shows an infinitesimal amount of mass being placed on top of the piston of a piston-cylinder assembly. The increase in mass causes the gas in the piston to be compressed. Because the mass increases infinitesimally and the piston is well insulated, the compression is reversible and adiabatic. For a reversible, adiabatic process the change in entropy is zero. Therefore, the compression changes the internal energy of the gas at constant entropy as the pressure increases. (b) To determine the sign of the relation, consider an energy balance on the piston. Neglecting potential and kinetic energy changes, we obtain

U  Q  W Since the process is adiabatic, the energy balance reduces to U  W

As the pressure increases on the piston, the piston compresses. Positive work is done on the system; hence, the change in internal energy is positive. We have justified the statement

 u    0  P  s

10

5.9 (a) By definition: 1  v   v  T  P

  and

1  v   v  P  T

  

Dividing, we get:  v      v   P   T  P         v   T  P  v  T    P  T

where derivative inversion was used. Applying the cyclic rule:  v   P   T  1         T  P  v  T  P  v

Hence,

  P      T  v (b) If we write T = T(v,P), we get:

 T   T  dT    dv    dP  P  v  v  P

(1)

From Equations 5.33 and 5.36 ds 

cv c  P   v  dT    dv  P dT    dP T T  T  v  T  P

We can solve for dT to get:

11

dT 

T  P  T  v    dP   dv  c P  cv  T  v c P  cv  T  P

(2)

For Equations 1 and 2 to be equal, each term on the left hand side must be equal. Hence,

T  T      v  P c P  cv

 P     T  v

or

  v   P   v  c P  cv  T      T     T  P  T  v  T  P where the result from part a was used. Applying the definition of the thermal expansion coefficient:

Tv  P   v  c P  cv  T        T  v  T  P

2

12

5.10 We need data for acetone, benzene, and copper. A table of values for the molar volume, thermal expansion coefficient and isothermal compressibility are taken from Table 4.4:

 

 m3  v  10 6    mol  73.33 86.89 7.11

Species Acetone Benzene Copper

 

  103 K -1

  1010 Pa -1

1.49 1.24 0.0486

12.7 9.4 0.091

We can calculate the difference in heat capacity use the result from Problem 5.9b: c P  cv 

vT 2



or

  73.33  10  6  c p  cv  

Species

Acetone Benzene Copper

 



 m3   3 -1 2   293 K  1.49  10 K   mol   12.7  10

10

Pa  -1

 J  c p  cv   mol  K  37.6 41.6 0.5

 J  cp   mol  K  125.6 135.6 22.6

 J   37.6   mol  K 

% difference 30% 31% 2%

We can compare values to that of the heat capacity given in Appendix A2.2. While we often assume that cP and cv are equal for condensed phases, this may not be the case.

13

5.11 We know from Equations 4.71 and 4.72

1  v   v  T  P

 

and

1  v   v  P T

  

Maxwell relation:  P   s       v T  T  v

Employing the cyclic rule gives  P   v   P          v T  T  P  T  v

which can be rewritten as 1  v    v  T  P  P   s        v T  T  v  1  P    v  v T Therefore,

  s      v T  Maxwell Relation:  s   v        P T  T  P From Equation 4.71:  v     v  T  P Therefore,  s      v  P T

14

5.12 (a) An isochor on a Mollier diagram can be represented mathematically as

 h     s  v This can be rewritten:  h   Ts  vP   P      T   s  s  v  v  s  v Employing the appropriate Maxwell relation and cyclic rule results in  h   T   s     T  v     s  v  v  T  s  v We know T  T   s   P  and          s  v c v  v  T  T  v For an ideal gas: R  s   P        v  T  T  v v Therefore,  T R R  h   T 1     T  v cv v cv   s  v 

(b) In Part (a), we found

T  h    T  v cv  s  v

 P     T  v

For a van der Waals gas:

15

R  P      T  v v  b Therefore, RT  h    T  cv  s  v

 v    v b

16

5.13 (a) The cyclic rule can be employed to give

 T   T   s          s  P  P T  P  s Substitution of Equations 5.19 and 5.31 yields T  T      P  s c P

 v     T  P

For an ideal gas: R  v      T  P P Therefore, RT 1 v  T      P cP cP  P  s (b) Separation of variables provides

T R P  T cP P Integration provides T ln 2  T1

 P   ln 2   P1

R

 cP  

which can be rewritten as R  P2  cP  

T2  T1  P1 

The ideal gas law is now employed

17

R

P2 v 2  P2  cP   P1v1  P1   R 1 c P2 P

  v

2



 R 1 c P1 P

  v 1

where 1

R c P  R cv 1    cP cP cP k

If we raise both sides of the equation by a power of k, we find P2 v 2k  P1v1k

 Pv k  const. (c) In Part (a), we found

T  v   T        P  s c P  T  P Using the derivative inversion rule, we find for the van der Waals equation Rv 3 v  b   v      T  P RTv 3  2av  b 2

Therefore, 1 RTv 3 v  b   T      P  s c P RTv 3  2av  b 2

18

5.14 The development of Equation 5.48 is analogous to the development of Equation E5.3D. We want to know how the heat capacity changes with pressure, so consider

 c P   P

  T

which can be rewritten as  c P   P

   h      h             T  P  T  P  T  T  P T  P

 h  Consider the   term:  P T  h   Ts  vP   s   v    T    v  T   v    P  P T  T  P  P T  T  c  Substitution of this expression back into the equation for  P  results in  P T     v    c P       T     v   P T  T   T  P  P   2v  T  v   c P      v     T     2   T  T  P  P T  T  P  T  P   2v   c P     T  2   T   P T  P

Therefore, c Preal

Preal

c Pideal

Pideal

 dc P 



   2v     dP  T  2    T   P 

and ideal  c Preal  c P

Preal    2v T    2 Pideal   T



    dP   P 

19

5.15 In order to solve this problem we need to relate the change in entropy from 10 to 12 bar to the change in molar volume (for which we have complete data). First, we can rewrite the change in entropy as 12 bar

s  s 2  s1 

 s    dP  P T 10 bar



Applying a Maxwell relation, we can relate the above equation to the change in molar volume: 12 bar

s 2  s1 

12 bar

 s   v    P T dP  s1     T  P dP 10 bar 10 bar

As 10 bar:  v   v  4      5.60  10  T  P  T  P

 m3     kg  K 

At 12 bar:  v   v  4      4.80 10  T  P  T  P

 m3     kg  K 

 v  To integrate the above entropy equation, we need an expression that relates   to pressure.  T  P Thus, we will fit a line to the data. We obtain   v  10      4.0  10  T  P 

 m3    3  4 m    P  9.6  10     kg  K  Pa    kg  K 

Now integrate the equation to find the entropy:

 4.0 10

1.210 6 Pa

s 2  s1 

1.010 6 Pa

10

P  9.6 10 4 dP 5.4960  0.104  kgkJ K   5.392  kgkJ K 

20

5.16 A schematic of the process follows:

Propane in v1 = 600 cm3/mol T1= 350 oC

Turbine

ws Propane P = 1 atm 2 out

We also know the ideal gas heat capacity from Table A.2.1:

cP  1.213  28.785 10  3 T  8.824 10  6 T 2 R Since this process is isentropic (s=0), we can construct a path such that the sum of s is zero. (a) T, v as independent variables Choosing T and v as the independent variables, (and changing T under ideal gas conditions), we get:

 s=0

s1

step 1

 s2

volume

v2 ,T2

Ideal step 2 Gas

v1 ,T 1

Temperature

or in mathematical terms:  s   s  ds    dT    dv  0  T  v  v T However, From Equation 5.33:

21

c  P  ds  v dT  dv T v T To get s1 P

a RT  2 vb v

so

R  P   T v v  b

and v  dv  2 R dv  R ln v2  b       v1  b  v1 T v v1 v  b

v 2 P

s1   ds  

or, using the ideal gas law, we can put s1 in terms of T2: RT2  b  P  s1  R ln  2 v1  b    

For step 2 T2

T2

T1

623.15 K

c s 2   v dT  R T



0.213  28.785  10  3 T  8.824  10  6 T 2 dT T

Now add both steps s  s1  s 2  0  RT2   P  b T2  8.824  10  6 2 3   0.213 ln    ln  2  28 . 785  10 T  623 . 15 K  T2  623.15 K 2  2 2  v1  b   623.15     



Substitute T1  623.15 K



v1  600 cm 3 /mol P2  1 atm



 cm 3  atm  R  82.06    mol  K 

22



and solve for T2: T2  448.3 K 

(b) T, P as independent variables

Choosing T and P as the independent variables, (and changing T under ideal gas conditions), we get:

P1,T1

step 1

Pressure

s=0

step 2

s1

Ideal Gas

s2

P2,T2

Temperature

Mathematically, the entropy is defined as follows  s   s  ds    dT    dP  0  P T  T  P

Using the appropriate relationships, the expression can be rewritten as c  v  ds  P dT    dP  0 T  T  P

For the van der Waals equation  R   v  b    

 v      T  P  RT 2a     2 v 3   v  b  Therefore,

23

s 

T2



T1

cP dT  T

P2



 R   v  b    

2a  P1      v  b 2 v 3 

dP  0

RT

We can’t integrate the second term of the expression as it is, so we need to rewrite dP in terms of the other variables. For the van der Waals equation at constant temperature:  2a RT  dP     dv 3 v  b 2   v Substituting this into the entropy expression, we get 1.213  28.785 10  3 T  8.824  10  6 T 2 s   dT  T T2

v2

T1

v1

Upon substituting T1  623.15 K v1  600 cm 3 v2 

RT2 (gas acts ideally at 1 atm) P2

 cm 3  b  91    mol   cm 3  atm  R  82.06    mol  K 

we obtain one equation for one unknown. Solving, we get T2  448.3 K

24

R

 v  b dv  0

5.17 (a) Attractive forces dominate. If we examine the expression for z, we see that at any absolute temperature and pressure, z  1. The intermolecular attractions cause the molar volume to deviate negatively from ideality and are stronger than the repulsive interactions. (b) Energy balance: h2  h1  q

Alternative 1: path through ideal gas state Because the gas is not ideal under these conditions, we have to create a hypothetical path that connects the initial and final states through three steps. One hypothetical path is shown below: P [bar] P,T1

P,T2

q = h

50

step 3

step 1

h1

h 3

h2 step 2

0

ideal gas 300

500

Choosing T and P as the independent properties:  h   h  dh    dT    dP  T  P  P T or using Equation 5.46   v   dh  c P dT   T    v  dP   T  P  The given EOS can be rewritten as 1  v  R  aT 1 / 2  P 

25

T [K]

Taking the derivative gives: R  v   0.5     0.5aRT  T  P P

so





dh  cP dT  0.5aRT 0.5 dP For step 1

 0.5aRT1

0

0.5

h1 

50 bar

dP  0.5aRT10.5 P  252  molJ 

For step 2

 3.58  3.02 10

500 K

h2  R

3

300 K



 J  T  0.875T  0.5 dT  7961    mol 

For step 3:

 0.5aRT2

50 bar

h3 

0.5

0

dP  0.5aRT20.5P  323  molJ 

Finally summing up the three terms, we get,  J  q  h1  h2  h3  7888   mol  Alternative 2: real heat capacity For a real gas h  c Preal

From Equation 5.48:   2v T   T 2 ideak  P P real

c Preal  c Pideal 

  dP  P

For the given EOS

26

 1 v  R  aT 1 / 2  P  Therefore,   2v     0.25aRT 1.5  T 2   P

and   2v T   T 2 ideak  P P real



 

P 50 bar   dP   0.25aRT 0.5 dP  0.875 K 1/2 RT 0.5   P P ideak 0 bar real

We can combine this result with the expression for c Preal and find the enthalpy change.

 3.58  3.02 10

500 K

h  R

3



T  0.875T  0.5 dT

300 K

 J  q  h  7888   mol  The answers is equivalent to that calculated in alternative 1

27

5.18 (a) Calculate the temperature of the gas using the van der Waals equation. The van der Waals equation is given by:

P

RT a  2 v b v

First, we need to find the molar volume and pressure of state 1.

  

 m3  V Al 0.1 m 2 0.4 m    0.00016  v1  1   n n 250 mol  mol  P1 

mg  Patm  A

10000 kg 9.81  m2    0.1 m 2

 

s  

 1.01325 10 5 Pa   1.08 10 6 Pa 

Substituting these equations into the van der Waals equation above gives  J  m3    J  0 .5    8.314  T1  mol   mol  K    6 1.08  10 Pa    2 3   m3  3    5 m m 0.00016    0.00016    4  10    mol  mol    mol     T1  297.5 K Since the process is isothermal, the following path can be used to calculate internal energy: v2 ,T2 = T 1

v

u s

v1 ,T1

T

Thus, we can write the change in internal energy as:

28

 u   u   u  du    dT    dv    dv  v T  v T  T  v Using Equation 5.40 v2

u 

  P 



 T  T  v  P  dv

v1

For the van der Waals EOS: P

RT a  2 v b v

so R  P      T  v v  b Therefore, v2

u 

a

 v 2 dv

v1

We can assume the gas in state 2 is an ideal gas since the final pressure is atmospheric. Therefore, we calculate v 2 , v2 

 m3  RT2  0.0244   P2  mol 

and  J  m3  0 . 5   0.0244  mol   J  dv  3104.5  u    mol  v2 0.00016 or U  776.1 kJ 

29

(b) From the definition of entropy: suniv  s sys  s surr

First, let’s solve for s sys using the thermodynamic web.  s   s  ds sys    dT    dv  T  v  v T Since the process is isothermal,  s  ds sys    dv  v T v2

 P   s sys     dv T  v  v 1

Again, for the van der Waals equation, R  P      T  v v  b Substitution of this expression into the equation for entropy yields v2

s sys 

R

 v  b dv

v1

 J  8.314   mol  K  dv  44.17  J  s sys    mol  K   3 0.00016 v  4  10  5 m    mol  J S sys  11042.5   K 0.0244

The change in entropy of the surroundings will be calculated as follows s surr 

Qsurr Tsurr

where 30

Qsurr  Q

(Q is the heat transfer for the system)

Application of the first law provides Q  U  W

We know the change in internal energy from part a, so let’s calculate W using v2

W  n  Pdv v1

Since the external pressure is constant, m  250 mol1.01325  10 5 Pa  0.0244   

W

W  614030 J 

  m3     0.00016    mol mol      



3

Now calculate heat transfer. Q  776100 J    614030 J   1.39  10 6 J 

Therefore,

S surr 

 1.39  10 6 J  J  4672   297.5 K  K 

and the entropy change of the universe is: J J J S univ  11042.5    4672    6370.5   K  K  K 

31

5.19 First, calculate the initial and final pressure of the system.

Pi  10  10

5

P f  10  10

 20000 kg 9.81 m/s 2  Pa    4.92  10 6 Pa  2

  Pa   30000 kg9.81 m/s   6.89  10 0.05 m  0.05 m

5

2

6

2

Pa 

To find the final temperature, we can perform an energy balance. Since the system is wellinsulated, all of the work done by adding the third block is converted into internal energy. The energy balance is u  w

To find the work, we need the initial and final molar volumes, which we can obtain from the given EOS:



vi  8.37 10 4 m 3 /mol vf 



8.314T f





 25  6.89  10 6 1   Tf   



 3.2  10 -5 m 3 /mol



Now, calculate the work       8.314T f w   Pf v f  vi   6.89 10 6 Pa   3.2  10 -5  8.37  10 4   25    6  6 . 89 10 1     Tf        We also need to find an expression for the change in internal energy with only one variable: Tf. To find the change in internal energy, we can create a hypothetical path shown below:











step 2 v

ideal gas

step 3

step 1



vi ,Ti u = -P (v -v ) f f i vf ,Tf

500

Tf

32

T

For step 1, we calculate the change in internal energy as follows v  RT / Plow



u1 

vi

u1 

 u    dv   v T

v  RT / Plow



vi

  P     P dv T    T  v 

v  RT / Plow 

2  2  aRTi  dv  aRTi ln RTi / Plow  b    T  a 2  v  b  T  a 2  vi  b  i  i 



vi

Similarly, for step 3: vf

v

f   P    u  dv T P u3           v T    T  v dv v  RT / Plow v  RT / Plow

 aRT 2  dv   aRT f 2 vf b f     ln   T  a 2  v  b T  a 2  RT f / Plow  b    f v  RT / Plow  f  vf

u3 









Insert the expression for the final molar volume into the equation for u3 :

 8.314T f u3  ln T f  a 2  6.89  10 6 1  25 / T f RT f / Plow  b



aRT f 2











   

Since the pressure is low (molar volume is big) during the second step, we can use the ideal heat capacity to calculate the change in internal energy. u 2 

Tf

Tf

Ti  500 K

Ti  500 K

 cv dT 



 20  R  0.05T dT





u 2  11.686 T f  500  0.025 T f2  500 2



If we set the sum of the three steps in the internal energy calculation equal to the work and choose an arbitrary value for Plow, 100 Pa for example, we obtain one equation with one unknown:

33





 RT / P  b    11.686 T f  500  0.025 T f2  500 2  ln i low 2 Ti  a   vi  b 



aRTi 2



  8.314T f  ln  2  6 Tf  a  6.89  10 1  25 / T f RT f / Plow  b        8.314T f  6.89  10 6 Pa   3.2  10 -5  8.37  10  4     25  6   6.89  10 1      Tf   



aRT f 2



















Solving for Tf we get T f  536.2 K

The piston-cylinder assembly is well-insulated, so suniv  s sys

Since the gas in the cylinder is not ideal, we must construct a hypothetical path, such as one shown below, to calculate the change in entropy during this process.

P Pf ,Tf

6.9

step 1

step 3

Pi ,Ti

4.9

ideal gas Plow

step 2 536

500

For steps 1 and 3 Plow

Plow

 s  s1     dP    P T Pi Pi

  v     dP     T  P 

34

T

Pf

P

f   v    s  s3     dP       dP P T T    P  Plow Plow 

We can differentiate the given EOS as required: Plow   RTi 2a  Ti    RTi 2a  Ti   Plow dP  ln s1    2 2     a T P a T      Pi i i  Pi 



s3 



  RT 2a  T f f   a  T 2 P f Plow  Pf





  

 dP   RT f 2a  T f  ln Pf   a  T f 2  Plow  

For step 2 Tf

Tf

Tf

c  s   20  s 2     dT   P dT     0.05  dT T  T  P T  Ti Ti Ti Tf    0.05 T f  Ti s2  20 ln  T  i 





Sum all of the steps to obtain the change in entropy for the entire process suniv  s sys  s1  s 2  s3





RT f 2a  T f Tf   Pf   RTi 2a  Ti   Plow    0.05 T f  Ti      20 ln ln ln T  P  2 a  Ti 2  a T i low  Pi      f Arbitrarily choose Plow (try 100 Pa), substitute numerical values, and evaluate:



suniv 

 J  suniv  0.388   mol  K    J  J S univ  2 mol 0.388    0.766     mol  K   K  

35







5.20 A schematic of the process is given by: well insulated

V=1L

V=2L

V=1L

T = 500 K

T=?

Vacuum

nCO =1 mole

nCO=1 mole

State i

State f

(a) The following equation was developed in Chapter 5:

  2P    T  T 2  dv v  v ideal   v

cvreal  cvideal 

For the van der Waals EOS  2P   0  T 2    Therefore, cvreal  cvideal From Appendix A.2:  3100   J    R  22.0  cvreal  R 3.376  5.57  10  4 500   mol  K  500 2   (b) As the diaphragm ruptures, the total internal energy of the system remains constant. Because the volume available to the molecules increases, the average distance between molecules also increases. Due to the increase in intermolecular distances, the potential energies increase. Since the total internal energy does not change, the kinetic energy must compensate by decreasing. Therefore, the temperature, which is a manifestation of molecular kinetic energy, decreases.

36

(c) Because the heat capacity is ideal under these circumstances we can create a two-step hypothetical path to connect the initial and final states. One hypothetical path is shown below: v

vf ,Tf step 2

u 2 s2

step 1

vi ,Ti

u1, s1

Tf

500

For the first section of the path, we have Tf

Tf

u1 



cvreal dT

  cvideal dT Ti

Ti



Tf



 3100  4 2.376  5.57 10 T  2  dT T  500 K i  25773.4 u1  2.32 10  3 T f2  19.75T f   10507.4 Tf

u1  R







For the second step, we can use the following equation vf

u 2 

 u 

  v T dv

vi

If we apply Equation 5.40, we can rewrite the above equation as vf

u 2 

  P    T P       T  v  dv vi

For the van der Waals EOS, P 

RT a  2 , v b v

37

T

R  P      T  v v  b Therefore, v f  0.002

v f  0.002

a  RT   J  u1    v  b  P dv   v 2 dv  73.7  mol  v i  0.001 v i  0.001 Now set the sum of the two internal energies equal to zero and solve for Tf:





u1  u 2  2.32  10  3 T f2  19.75T f 

25773.4  10507.4  73.7  0 Tf

T f  497 K

(d) Since the system is well-insulated suniv  s sys

To solve for the change in entropy use the following development:  s   s  ds sys    dT    dv  v T  T  v Using the thermodynamic web, the following relationships can be proven cv  s      T  v T  P   s       v T  T  v For the van der Waals EOS R  P      T  v v  b  Now we can combine everything and calculate the change in entropy

38

497 K

s sys 



500 K

cv dT  T

0.002

R

 v  b  dv 0.001





0.002 497 K  2.376 dv 3100  dT  s sys  R     5.57  10  4    5 500 K  T T3  0.001 v  3.95  10  J  suniv  s sys  5.80    mol  K 



39

  



5.21 A schematic of the process is given by: well insulated

V = 0.1 m3

V = 0.1 m3

T = 300 K

V = 0.2 m3 T=?

Vacuum

nA =400 moles

nA=400 moles

State f

State i

Energy balance: u  0

Because the gas is not ideal under these conditions, we have to create a hypothetical path that connects the initial and final states through three steps. One hypothetical path is shown below: step 2 v [m 3/mol]

ideal gas step 1

step 3

0.2/400

vf ,Tf u

0.1/400

vi ,Ti

Tf

300

For the first section of the path, we have v 

u1 

 u    dv  v T vi



If we apply Equation 5.40, we can rewrite the above equation as v 

u1 

  P     P  dv T   T  v  vi 



40

T [K]

For the van der Waals EOS

R a  P      2  T  v v  b T v 2 Therefore, v 

u1 

v 

 RT  a 2a  J  P     4  v  b Tv 2  dv   4 T v 2 dv  1120  mol  i vi  2.510 vi  2.510

Similarly for step 3: v f  510 4

u3 



v

 RT  a  v  b  2  P  dv  Tv  

v f  510 4



v 

2a

Tf v

2

dv 

 168000  J   mol  K  Tf

For step 2, the molar volume is infinite, so we can use the ideal heat capacity given in the problem statement to calculate the change in internal energy:

u 2 



3 R T f  300 K 2



If we set sum of the changes in internal energy for each step, we obtain one equation for one unknown:





 168000  J  3  J  u1  u 2  u3  1120    8.314   T f  300 K  0   Tf  mol  K    mol  2 

Solve for Tf: T f  261.6 K

41

5.22 A schematic of the process is shown below: Ethane 3 MPa; 500K

T surr = 293 K Initially: vacuum

(a) Consider the tank as the system. Since kinetic and potential energy effects are negligible, the open system, unsteady-state energy balance (Equation 2.47) is  dU      nin hin   n out hout  Q  W s dt   sys in out

The process is adiabatic and no shaft work is done. Furthermore, there is one inlet stream and no outlet stream. The energy balance reduces to  dU    nin hin   dt  sys

Integration must now be performed U2

t

U1

0

 dU   n

h dt

in in

t

n2 u 2  n1u1  hin  n in dt  nin hin  (n2  n1 )hin 0

Since the tank is initially a vacuum, n1=0, and the relation reduces to: u 2  hin

42

As is typical for problems involving the thermodynamic web, this problem can be solved in several possible ways. To illustrate we present two alternatives below: Alternative 1: path through ideal gas state Substituting the definition of enthalpy:

u 2  uin  Pin vin

or u2 at 3 MPa, T   uin at 3 MPa, 500 K   Pin vin

(1)

From the equation of state:





  J  Pin vin  RT 1  B' P    8.314  552 K  1  2.8  10 8 3  10 6 Pa  3,800  mol K    



 J   mol  (2)

The change in internal energy can be found from the following path: P

step 1

Plow

u1

u3

step 3

3 MPa

u2 step 2 ideal gas T2

500 K

T

For steps 1 and 3, we need to determine how the internal energy changes with pressure at constant temperature: From the fundamental property relation and the appropriate Maxwell relation:

 u   s   v   v   v     T    P   T    P   P T  P T  P T  T  P  P T From the equation of state

RT  u  1  B'P   P  RT2    B' RT    P  P T  P 

43

So for step 1:  u  u1     dP    B ' RTdP B ' RT Pin  349 [J/mol] P  T Pin  Pin 0

0

(3)

and for step 3:  u  u 3     dP    B ' RTdP   B ' RT P2  0.7T P  T 0 0 P2

P2

(4)

For step 2 T T  h   Pv    u  u 2     dT   c P  R dT    dT    T  P  T  P  T  P 500  500  500 T

or u 2  R

 0.131  19.225 10

T2

3



T  5.56110 6 T 2 dT

T1 500 K

Substituting Equations 2, 3, 4, and 5 into 1 and solving for T gives: T2  552 K

Alternative 2: real heat capacity Starting with:

u 2  hin

The above equation is equivalent to h2  P2 v2  hin  h2  hin  P2 v2

To calculate the enthalpy difference, we can use the real heat capacity ideal c Preal  c P 

P   2  v T  2   Pideal   T



   dP   P 

For the truncated viral equation,

44

(5)

  2v    0  T 2   P

Therefore, ideal c Preal  c P

Now, we can calculate the change in enthalpy and equate it to the flow work term. T2

 cP

ideal

dT  P2 v2

T1  500 K

 1.131  19.225 10

T2

R

3



T  5.561 10 6 T 2 dT  P2 v2  RT2 1  B' P2 

T1 500 K

Integrate and solve for T2: T2  552 K

(b) In order to solve the problem, we will need to find the final pressure. To do so, first we need to calculate the molar volume. Using the information from Part (a) and the truncated virial equation to do this

v

RT 1  B' P   P

  J  552 K   8.314   mol  K    6

3  10 Pa

1  2.8 10 3 10 Pa  8

6

 m3  v  0.0014    mol 

This quantity will not change as the tank cools, so now we can calculate the final pressure.   m3   P2  0.0014    mol      1  2.8 10 8 P2   J  293 K   8.314   mol  K   





Solve for P2 : 45

P2  1.66  10 6 Pa

The entropy change of the universe can be expressed as follows: S univ  S sys  S surr

To solve for the change in entropy of the system start with the following relationship:  s   s  ds sys    dT    dP  T  P  P T

Alternative 1: path through ideal gas state Using the proper relationships, the above equation can be rewritten as ds sys 

cP  v  dT    dP T  T  P

We can then use the following solution path: P

s3

1.66 MPa step 1

s 1 s 2

Plow

step 2 ideal gas 500 K

Choosing a value of 1 Pa for Plow, for step 1: R  v  ' s1     dP    1  B P dP T  P P 1.66 MPa  1.66 MPa 1 Pa

1 Pa

For step 3, 3 MPa

s1 



1 Pa

 v    dP   T  P

3 MPa



1 Pa

step 3

3 MPa







R 1  B ' P dP P

46

T2

T

For step 2: 293 K cP 1.131  dT R   19.225  10 3  5.561 10 6 T  dT    T T  552 K 552 K  293 K

s 2 

Adding together steps 1, 2 and 3:  J  s sys  46.9   mol  K  ______________________________________________________________________________ Alternative 2: real heat capacity Using the proper relationships, the above equation can be rewritten as c real  v  ds sys  P dT    dP T  T  P

For the truncated virial equation  v  1     R  B '   T  P P 

Now, substitute the proper values into the expression for entropy and integrate:

293 K

s sys  R



552 K

1.6610 6 Pa

1.131 3 6   T  19.225  10  5.561  10 T  dT  R

1    B' dP P  310 ^ Pa



 J  s sys  46.9   mol  K  ______________________________________________________________________________

In order to calculate the change in entropy of the surroundings, first perform an energy balance. u  q

Rewrite the above equation as follows h  Pv   q

47

Since the real heat capacity is equal to ideal heat capacity and the molar volume does not change, we obtain the following equation Tf

 cP

ideal





dT  v P f  Pi  q

Ti T f  293K

R



1.131  19.225 10

Ti  552 K

3





 J  q  15845   mol 

Therefore,  J  q surr  15845   mol 

and  J  s surr  54.08   mol  K 

Before combining the two entropies to obtain the entropy change of the universe, find the number of moles in the tank. n



  m3   6 6 T  5.561  10  6 T 2 dT   0.0014    1.66  10 Pa - 3  10 Pa  q   mol    

 

0.05 m 3  75.7 mol  m3  0.0014    mol 

Now, calculate the entropy change of the universe.   J   J  S univ  75.7 mol 54.08   46.9     mol  K   mol  K    J S univ  544   K

48

5.23 First, focus on the numerator of the second term of the expression given in the problem statement. We can rewrite the numerator as follows:





uTr , v r  uTideal  uTr , v r  uTideal  uTideal  uTideal r ,vr r ,vr   r ,vr r ,vr  



For an ideal gas, we know uTideal  uTideal 0 r ,vr r ,vr  

Therefore, uTr , v r  uTideal  uTr , v r  uTideal ,vr ,vr  r r

Substitute this relationship into the expression given in the problem statement: uTdep ,v r

r

RTc

uTr , v r  uTideal uTr , v r  uTideal ,vr ,vr  r r   RTc RTc

Now, we need to find an expression for uTr , vr  uTideal . Note that the temperature is constant. r ,vr   Equation 5.41 reduces to the following at constant temperature:   P   duT  T    P  dv   T  v 

The pressure can be written as P

zRT v

and substituted into the expression for the differential internal energy   RT  z   RT 2  z   RT  zRT       duT  T  dv       dv  v  v  v T   v    v  T  v  v 

Applying the Principle of Corresponding States

49

 T 2  z   r  RTc  vr  Tr 

duTr

    dvr  v r 

If we integrate the above expression, we obtain vr  2 uTr , v r  uTideal T  z  r, vv       r   RTc  RTc  v r  Tr  v v  v   r v

duTr

  dv r  

Therefore, uTdep ,v r

RTc

r

vr  2 uTr , v r  uTideal uTr , v r  uTideal T  z  v , ,vr  r r r       r  RTc RTc  v r  Tr  v v   r

50

  dv r  

5.24 We write enthalpy in terms of the independent variables T and v:

 h   h  dh    dT    dv  T v  v T

using the fundamental property relation: dh  Tds  vdP

At constant temperature, we get:   P   P   dhT  T    v   dv  v T    T  v

For the Redlich-Kwong EOS 1 R a  P      3 / 2  T  v v  b 2 T vv  b   RT a a  P        v T v  b 2 T 1 / 2 v 2 v  b  T 1 / 2 vv  b 2

Therefore,  RT  3 RTv a a    dhT    dv  v  b v  b 2 2 T 1 / 2 vv  b  T 1 / 2 v  b 2 

To find the enthalpy departure function, we can integrate as follows v

h

dep



 dhT 

v 

 RT  RTv a a 3       v  b v  b 2 2 T 1 / 2 vv  b  T 1 / 2 v  b 2  dv  v   v

Since temperature is constant, we obtain h dep 

3a RTb a  v    ln  v  b 2bT 1 / 2  v  b  T 1 / 2 v  b 

To calculate the entropy departure we need to be careful. From Equation 5.64, we have:





gas gas ideal gas gas sT , P  sTideal  sT , P  sTideal  sTideal ,P , P  0  sT , P , P 0

51



However, since we have a P explicit equation of state, we want to put this equation in terms of v. Let’s look at converting each state. The first two states are straight -forward sT , P  sT , v

and gas ideal gas sTideal , P  0  sT , v  

For the third state, however, we must realize that the ideal gas volume v’ at the T and P of the system is different from the volume of the system, v. In order to see this we can compare the equation of state for an ideal gas at T and P P

RT v'

to a real gas at T and P P

RT a  vb T vv  b 

The volume calculated by the ideal gas equation, v’, is clearly different from the volume, v, calculated by the Redlich-Kwong equation. Hence: gas gas  ideal gas gas  sTideal  s ideal' gas  sTideal s '  sTideal  ,P ,v , v , T ,v T v  

Thus,







gas gas ideal gas gas  ideal gas gas   sT , v  sTideal  sTideal  sTideal sT , P  sTideal  ,P , v    sT , v , v     sT , v ' ,v 

Using a Maxwell relation:  ds   P       dv T  T  v

Therefore,  P  dsT    dv  T  v

52

For the Redlich-Kwong EOS 1 R a  P      3 / 2  T  v v  b 2 T vv  b  so

s



ideal gas T , v  sT , v   

v  R

 a 1    dv 3/ 2  2 v b    T v v b   v  



For an ideal gas R  P      T  v v

so





gas gas  sTideal sTideal ,v ,v  

v

R  v  dv v 



Finally: v' dv v' RT ideal gas ideal gas s '  sT , v R  R ln  R ln T ,v v v Pv v



Integrating and adding together the three terms gives: s dep  R ln

v  b   v

RT  v  ln  R ln  Pv 2bT 3 / 2  v  b  a

53

5.25 Calculate the reduced temperature and pressure:

Tc  647.3 K 

Pc  220.48 bar 

(Table A.1.2)

w  0.344 300 bar   1.36 220.48 bar  673.15 K Tr   1.04 647.3 K Pr 

By double interpolation of data from Tables C.3 and C.4

 h dep  Tr , Pr  RTc 

   

( 0)

 2.921

 h dep  Tr , Pr  RTc 

   

(1)

 s dep  Tr , Pr  R 

   

(1)

 h dep Tr , Pr  w  RTc 

   

 1.459

From Tables C.5 and C.6:  s dep  Tr , Pr  R 

   

( 0)

 2.292

 1.405

Now we can calculate the departure functions   dep  hTr , Pr h dep  RTc      RTc  

   

( 0)

(1) 

   

  J   J  h dep   8.314  647.3  2.921  0.344 1.459  18421    mol  K    mol     dep  sTr , Pr s dep  R     R  

   

( 0)

 s dep Tr , Pr  w  R 

   

(1) 

   

  J   J  s dep   8.314   2.292  0.344 1.405  23.07    mol  K    mol  K  

54

To use the steam tables for calculating the departure functions, we can use the following relationships. h dep  hT , P  hTideal ,P s dep  sT , P  sTideal ,P

From the steam tables  kJ   kJ  hT , P  2151.0   and sT , P  4.4728    kg   kg  K 

We need to calculate the ideal enthalpies and entropies using the steam tables’ reference state. vap 0 .01º C   hTideal , P  h

673.15 K c ideal dT p 273.16 K



 kJ  from the steam tables and heat capacity data from Table A.2.2. We can get h vap  45.1   mol  Using this information, we obtain  kJ    kJ   hTideal   0.008314   , P  45.1   mol mol  K   

 

673.15 K 

 3.47  1.45  10  273.16 K 



 kJ  hTideal , P  59.14   mol 

Now, calculate the ideal entropy. vap 0 .01º C  sTideal , P  s

673.15 K c ideal p



273.16 K

T

P  dT  R ln 2   P1 

From the steam tables:  kJ  s vap 0.01 º C   0.165   mol  K 

Substitute values into the entropy expression:

55

3

0.121  10 5  T  dT T2 

673.15 K  5 kJ    3.47 30    3 0.121  10  dT      1 . 45 10 ln        mol  K   273.16 K  T  0.000613  T3 

  sTideal , P  0.165   0.008314  

 J  sTideal , P  107  mol  K   

Now, calculate the departure functions:  kJ   kJ   kJ  h dep  2151.0   0.0180148 kg/mol  59.14   20.4    mol   mol   kg   kJ  0.0180148 kg/mol  0.107  kJ   0.0264  kJ  s dep  4.4728    mol  K   mol  K   kg  K 

Table of Results Generalized Percent Difference Steam Tables Tables (Based on steam tables)  kJ  h dep   mol   kJ  s dep   mol  K 

-18.62

-20.4

9.9

-0.0231

-0.0264

12.5

56

5.26 State 1 is at 300 K and 30 bar. State 2 is at 400 K and 50 bar. The reduced temperature and pressures are

30 bar   0.616 48.74 bar  300 K T1, r   0.982 305.4 K P1, r 

50 bar   1.026 48.74 bar  400 K T2,r   1.31 305.4 K P2,r 

and

  0.099 By double interpolation of data in Tables C.3 and C.4 ( 0)

 h dep  T1, r , P1, r  RT c 

   

 h dep  T2 , r , P2 , r  RT c 

   

(1)

 0.825

 h dep  T1, r , P1, r  RT c 

   

 0.711

 h dep  T2 , r , P2 , r  RT c 

   

( 0)

 0.799

(1)

 0.196

Therefore,  h dep   T1, r , P1, r   RT   0.825  0.099 0.799   0.904 c     h dep   T2 ,r , P2 ,r   0.711  0.099 0.196  0.730  RTc   

The ideal enthalpy change from 300 K to 400 K can be calculated using ideal cP data from Table A.2.1. 400 K

hTideal R T 1

2

1.131  19.225 10

3

T  5.56110  6T 2 dT  717.39 R

300 K

The total entropy change is

57

dep

dep

h  hT , P  hTideal  hT , P T2 1, r 1, r 1 2,r 2,r h  R  0.904TC  717.39  0.730TC    J  h   8.314   0.904305.4 K   717.39 K  0.730305.4 K   mol  K     J  h  6406.2    mol 

Using the data in Table C.5 and C.6  s dep  T1, r , P1, r  R   s dep  T2 , r , P2 , r  R 

   0.601  0.099 0.756   0.676       0.394  0.099 0.224   0.416 

Substituting heat capacity data into Equation 3.62, we get 400 K 1.131  19.225  10 3 T  5.561  10 6 T 2  50 bar  dT  ln s ideal  R    T  30 bar  300 K s ideal  1.542 R

Therefore, s  sT , P  s ideal  sT , P  R0.676  1.542  0.416  1, r 1, r 2,r 2,r dep

dep

 J  s  14.98    mol  K 

58

5.27 The turbine is isentropic. Therefore, we know the following dep

dep

s  sT , P  s ideal  sT , P  0 1, r 1, r 2,r 2,r

Using the van der Waals EOS, we can find P1,r, which leaves one unknown in the above equation: T2.

P1 

   3  82.06  cm  atm  623.15 K     mol  K      3   3  600  cm   91  cm      mol    mol  



 atm  cm 3  91  10 5   2   mol   3   600  cm      mol   

P1  75.19 atm  76.19 bar 

2

Calculate reduced temperature and pressures using data from Table A.1.1 76.19 bar   1.8 42.44 bar  623.15 K T1, r   1.68 370.0 K P1, r 

P2,r 

1.013 bar   0.024 42.44 bar 

Also,

  0.152 From Tables C.5 and C.6:  s dep  T1, r , P1, r  R 

   0.327  0.152 0.102   0.343  

Substituting heat capacity data into Equation 3.62, we get s

ideal

 T2 1.213  28.785  10  3 T  8.824  10  6 T 2   1 atm   R dT  ln  75.19 atm  T    623.15 K 

Therefore,

59

  s dep T2 1.213  28.785  10 3 T  8.824  10 6 T 2   T2 , r , P2 , r R 0.343  dT  4 . 32    T R  623.15 K   We can solve this using a guess-and-check method T2  600 K : T2,r  1.62

 J  s  33.84    mol  K  T2  450 K : T2,r  1.22

 J  s  0.77    mol  K  T2  446.6 K : T2,r  1.21

 J  s  0   mol  K  Therefore, T2  446.6 K

60

     s  

5.28 A reversible process requires the minimum amount of work. Since the process is reversible and adiabatic s  0

which can be rewritten as dep

dep

s  sT , P  s ideal  sT , P  0 1, r 1, r 2,r 2,r Calculate reduced temperature and pressures using data from Table A.1.1 1 bar   0.0217 46.0 bar  300 K T1, r   1.57 190.6 K P1, r 

P2,r 

10 bar   0.217 46.0 bar 

From Tables C.5 and C.6:  s dep  T1, r , P1, r  R 

    0.00457  0.008 0.0028  0.0046 

Substituting heat capacity data into Equation 3.62, we get s

ideal

 T2 1.702  9.081 103T  2.164  106 T 2   10 bar   R  dT  ln  T 300 K  1 bar   

Therefore,   s dep T2 3 6 2     T T 1 . 213 28 . 785 10 8 . 824 10   T2 , r , P2 , r s  R 0.0046   dT  2.303   T R  300 K   We can solve using a guess-and-check method T2  400 K : T2,r  2.10  J  s  4.98    mol  K  T2  385 K :

T2,r  2.02

61

    

 J  s  1.42    mol  K  T2  379 K :

T2,r  1.99  J  s  0.018    mol  K 

Therefore, T2  379 K

An energy balance reveals that h2  h1  h  ws We can calculate the enthalpy using departure functions. From Tables C.3 and C.4:  h dep   T1, r , P1, r   RT   0.0965  0.008 0.011  0.0966 c    h dep   T2 , r , P2 , r   RT   0.0614  0.00890.015  0.0613 c  

Ideal heat capacity data can be used to determine the ideal change in enthalpy 379 K  h ideal  R   1.702  9.081  10  3 T  2.164  10  6 T 2 dT  300 K  Therefore, 379 K    J   3 6 2      190 . 6 K 0 . 0966 0 . 0613 1 . 702 9 . 081 10 T 2 . 164 10 T dT h   8.314            mol  K      300 K

 J  w s  h  3034.2    mol  and   mol    J    3034 . 2 W S  1 / 30      mol    101.1 W  s     

62

5.29 Equation 4.71 states

1  v     v  T  P  v   v     T  P This can be substituted into Equation 5.75 to give

 JT 

vT  1 cP

63

5.30 For an ideal gas

R  v      T  P P Therefore,  RT   v  P   v  v   0  JT   cP cP This result could also be reasoned from a physical argument.

64

5.31 The van der Waals equation is given by:

P

RT a  v  b v2

(1)

The thermal expansion coefficient is given by: 1  v   1   T       v  T  P  v   v  P

 

(2)

Solving Equation 1 for T: a  v  b   T   P  2   v  R   Differentiating by applying the chain rule,  a  1  v  b  2a Pv 3  av  2ab  T        P  2     v  P  v  R  R  v3 Rv 3 Substitution into Equation 2 gives



Rv 2 Pv 3  av  2ab

Substituting Equation 1 for P gives b in terms of R, T, v, a , and b:



Rv 2 v  b  2 RTv 3  2av  b 

The isothermal compressibility is given by: 1  v   1   P        v  P  T  v   v  T

  

From the van der Waals equation: RT 2a  RTv 3  2av  b   P       2 v  b 2 v 3 v 3 v  b   v T

2

so

65

(3)



v 2 v  b  2 RTv 3  2av  b  2

For the Joule-Thomson coefficient, we can use Equation 5.75:

 JT  c Pideal

  v   T    v    T  P  Preal 2   v    T  2   dP Pideal    T  P 

Substituting the van der Waals equation into Equation 3 gives RTv 3  2av  b  T 2av  b   T       3 v  b Rv v  b  Rv 3  v  P 2

(4)

Thus, the second derivative becomes:   2T  1  T  2a 6av  b  T  2        3 2 Rv 4 v  b  v  b  v  P Rv  v  P

or simplifying using Equation 4,   2T  2a v  3b   2   Rv 4  v  P

(5)

Substituting Equations 5 and 4 into Equation 5.75 gives:  bRTv 3  2avv  b  2 RTv 3  2av  b   Preal  RTv 4  dP c Pideal    2av  3b   Pideal  2

 JT

At a given temperature the integral in pressure can be rewritten in terms of volume using the van der Waals equation to give:

66

 bRTv 3  2avv  b  2 RTv 3  2av  b  vreal  RTv  RTv 3  2av  b 2 dv    2av  3b   v  b 2 videal  2

 JT  c Pideal

67

5.32 We can solve this problem by using the form of the Joule-Thomson coefficient given in Equation 5.75. The following approximation can be made

 vˆ   vˆ       T  P  T  P At 300 ºC, vˆ350 º C,1MPa   vˆ250 º C,1MPa   vˆ     350  250 º C  T  P  m3   m3  0.28247     0.23268   vˆ   kg   kg     350  250 º C  T  P  m3   m3   vˆ     0.0005     0.0005   T  P  kg  º C   kg  K 

A similar process was followed to find cP.  ˆ  hˆ     h  cˆ P       dT  P  T  P

At 300 ºC,  hˆ  hˆ350 º C,1MPa   hˆ250 º C,1MPa      T  350  250 º C  P

 kJ   kJ  3157.7    2942.6    kg   kg   350  250 º C

 hˆ     T   P  hˆ   ˆ        h   2.15  kJ   2.15  kJ  cˆ P       kg  º C   kg  K   dT  P  T  P Now,  JT can be found.

68

3  3      vˆ   573.15 K  0.0005  m    0.25794  m       T       T   vˆ    kg kg K          P   JT   cˆ P  kJ  2.15    kg  K 

 m3  K    kJ 

 JT  0.0133 

69

5.33 At the inversion line, the Joule-Thomson coefficient is zero. From Equation 5.75:

 JT  c Pideal

   v  T    v     T  P 0 Preal 2   v    T  2   dP   T  P  Pideal 

This is true when the numerator is zero, i.e.,   v   T    v   0   T  P  For the van der Waals equation, we have P

RT a  2 v b v

Solving for T: a  v  b   T   P  2   v  R   so a  1  v  b  2a Pv 3  av  2ab  T       P  2    v  R  R  v3 Rv 3  v  P  Substituting for P: RTv 3  2av  b   T     v  b Rv 3  v  P

2

Hence,  bRTv 3  2avv  b   v  T   v  0  2 RTv 3  2av  b   T  P

2

Solving for T: 2avv  b  bRv 3

2

T

(1)

70

Substituting this value of T back into the van der Waals equation gives

P

2avv  b  a a2v  3b   2  bv 3 v bv 2

(2)

We can solve Equations 1 and 2 by picking a value of v and solving for T and P. For N2, the critical temperature and pressure are given by Tc = 126.2 [K] and Pc = 33.84 [bar], respectively. Thus, we can find the van der Waals constants a and b: 2

 RT   Jm 3  a  27  c  = 0.137  2 64 Pc  mol 

b

 m3  RTc = 3.88  10 -5   8 Pc  mol 

Using these values in Equations (1) and (2), we get the following plot: Joule-Thomson inversion line 1000

800

600

400

200

0 0

100

200 T [K]

71

300

400

5.34 We can solve this problem using departure functions, so first find the reduced temperatures and pressures.

50 bar   0.99 50.36 bar  273.15 K T1, r   0.967 282.4 K P1, r 

P2,r 

10 bar   0.2 50.36 bar 

Since the ethylene is in two-phase equilibrium when it leaves the throttling device, the temperature is constrained. From the vapor-liquid dome in Figure 5.5: T2,r  0.76 T2  214.6

The process is isenthalpic, so the following expression holds dep

dep

h  hT , P  hTideal  hT , P  0 T2 1, r 1, r 1 2,r 2,r Therefore, dep

dep

hT , P  hT , P  hTideal T2 2,r 2,r 1, r 1, r 1 From Table A.2.1:

 1.424  14.394 10

214.6 K

hTideal 1  T2

R

3



T  4.392  10  6 T 2 dT

273.15 K

From Tables C.3 and C.4   0.085 :  h dep  T1,r , P1,r  RT c 

   3.678  0.085 3.51  3.976  

Now we can solve for the enthalpy departure at state 2.

72

hTdep, P 2,r

2,r

RTc hTdep, P 2,r

2,r

RTc

214.6 K   1 3 6 2  1 . 424 14 . 394 10 T 4 . 392 10 T dT   3.976       282.4 K   273.15 K  

 3.01

We can calculate the quality of the water using the following relation hTdep, P 2,r

2,r

RTc

 1  x 

hTdep,, liq P 2,r

2,r

RTc

x

hTdep,, vap P 2,r

2,r

RTc

where x represents the quality. From Figures 5.5 and 5.6: hTdep,, liq P 2,r

2,r

RTc hTdep,, vap P 2,r

2,r

RTc

 4.6  0.085 5.5  5.068  0.4  0.0859 0.75  0.464

Thus, x  0.447

55.3% of the inlet stream is liquefied.

73

5.35 Density is calculated from molar volume as follows:



MW v

2 Substitute the above into the expression for Vsound :

  P 2 Vsound     MW     v

    1  P  MW   1 / v   s    s

The following can be shown using differentials: v 1     v v2

Therefore,  P  v 2  P  2 Vsound        MW  v  s    s

74

5.36 From Problem 5.35: 2 Vsound

 P  v 2  P         MW  v  s    s

The thermodynamic web gives:  P   s   T   P   T   s   P   s   P                        v  s  s  v  v  P  v  s  T  s  s  v  v T  s T  T  P T  P       v  s  cv

c  s   P   c P          P  cv  v T  s T  T 

 P   T       T  v  v  P

If we treat air as an ideal gas consisting of diatomic molecules only R  P      T  v v

cP 7  cv 5

P  T      v  P R

Therefore,  P   7  P         v  s  5  v  and Vsound 

v 2  7  P      MW  5  v 

7  RT    5  MW 

Vsound  343 m/s

The lightening bolt is 1360 m away.

75

5.37 From Problem 5.35: 2 Vsound

 P  v 2  P         MW  v  s    s

The thermodynamic web gives:  P   s   T   P   T   s   P   s   P                        v  s  s  v  v  P  v  s  T  s  s  v  v T  s T  T  P so  T  P   T   c P   P             v  s  cv  T  v  v  P  T 

For liquids c P  cv water at 20 ºC, so  P   P   T          v  s  T  v  v  P However, the cyclic rule gives:  P   T   v  1         T v  v  P  P T So  P   P       v  s  v T From the steam tables, for saturated water at 20 oC:  m3  ˆ  . 001002 P = 2.34 kPa and v    kg  For subcooled water at 20 oC: P = 5 MPa and vˆ  . 0009995

m3    kg   76

So  kg kPa  5000  2.34  P   P   P           vˆ  s  vˆ T  vˆ  .0009995  .001002  m 3  and  P  Vsound  vˆ 2    1414 m/s  vˆ  s

77

5.38 (a) The fundamental property relation for internal energy is

dU  Qrev  Wrev Substituting the proper relationships for work and heat, we obtain dU  TdS  Fdz

The fundamental property relation for the Helmholtz energy is dA  dU  d TS   dU  TdS  SdT

Substitute the expression for the internal energy differential: dA  Fdz  SdT

(b) First, relate the entropy differential to temperature and length.  S   S  dS    dT    dz  T  z  Z T

Now we need to find expressions for the partial derivatives.  u   TS  Fz   S  nc z  n     T  T  T  z  z  T  z Therefore, nc  S  a    z  n  b   T  T  z T  The following statement is true mathematically (order of differentiation does not matter):  Z

 A     A           T  z  T T  Z T  z

Furthermore,

78

 Z

 A    S         Z T  T  z T

 T

 A    F         Z T  z  T  z

 S   F        k z  z0   Z T  T  z

Substituting the expressions for the partial derivatives into the expression for the entropy differential, we obtain a  dS  n  b dT  k z  z 0 dz T  (c) First, start with an expression for the internal energy differential:

 U   U  dU    dT    dz  T  z  Z T From information given in the problem statement:  U     na  bT   T  z Using the expression for internal energy developed in Part (a) and information from Part (b)  U   S     T    F  T  k  z  z 0   kT  z  z 0   0  Z T  z T Therefore, dU  na  bT dT  0dz  na  bT dT

(d) We showed in Part (c) that

 U  FU    0  z T

79

 S  Using the expression for   developed in Part B, we obtain  z T  S  FS  T    kT  z  z0   z T (e) First, perform an energy balance for the adiabatic process. dU  W

Substitute expressions for internal energy and work.

na  bT dT  Fdz  kT z  z 0 dz Rearrangement gives dT kT  z  z 0   dz na  bT  The right-hand side of the above equation is always positive, so the temperature increases as the rubber is stretched.

80

5.39 The second law states that for a process to be possible,

suniv  0 To see if this condition is satisfied, we must add the entropy change of the system to the entropy change of the surroundings. For this isothermal process, the entropy change can be written ds 

cv  P   P  dT    dv    dv T  T  v  T  v

Applying the van der Waals equation: ds 

R dv v b

Integrating s sys  R ln

v2  J   11.5  v1  mol K 

For the entropy change of the surroundings, we use the value of heat given in Example 5.2:  J  q  q surr  600   mol  Hence the entropy change of the surroundings is: s surr 

q surr  600  J    1.6  373 Tsurr  mol K 

and  J  suniv  s sys  ssurr  9.9   mol K  Since the entropy change of the universe is positive we say this process is possible and that it is irreversible. Under these conditions propane exhibits attractive intermolecular forces (dispersion). The closer they are together, on average, the lower the energy. That we need to put work into this system says that the work needed to separate the propane molecules is greater than the work we get out during the irreversible expansion.

81

5.40 A schematic of the process is given by: Gas A in Pi = 100 bar

Turbine

ws

Ti = 600 K

Gas A out

Pf = 20 bar Tf = 445 K

The energy balance for this process is provided below: h  wS Because the gas is not ideal under these conditions, we have to create a hypothetical path that connects the initial and final states through three steps. One hypothetical path is shown below: Pi ,Ti

100 ideal gas P [bar]

step 1

h Pf ,Tf

Plow

step 3

20 step 2

445

600

For the first section of the path, we have P 0

h1 

 h    dP  P T P1



If we apply Equation 5.45 we can rewrite the above equation as P 0

h1 

   v    v  dP  Ti   T  P  Pi 



For the given EOS:

82

T [K]

R aP  v      2  T  P P Ti

Therefore, P 0

v 0

 2aP    aP J   dP  2467    P v dP b h1        5   Ti T  mol    Pi 10010 Pi 10010 5  i Similarly for step 3 Pf  2010 5



h3 

P 0

 2aP   J    b dP  250    Tf   mol   

For step 2, the pressure is zero, so we can use the ideal heat capacity given in the problem statement to calculate the enthalpy change. h2 

Tf

445 K

Ti

600 K

 J 

 c P dT   30  0.02T dT  6270  mol 

Now sum each part to find the total change in enthalpy:  J  h  h1  h2  h3  8487   mol   J  ws  8487   mol  In other words, for every mole of gas that flows through the turbine, 8487 joules of work are produced.

83



6.1 (a) The Clausius-Clapeyron equation:

 P sat  h vap 1 1  ln i   i   R T 373 [K]   101 kPa

vap

dPisat hi dT 2 sat  RT Pi

or

so

Pi

sat

 h vap 1 1   101 kPaexp i     R T 373 [K]

(b) and (c) Using

 kJ  h vap  40.626   mol  we obtain the following table T [K]

Eqn 6.24 [kPa]

Steam Tables [kPa]

% Difference

273.156 278.15 283.15 288.15 293.15 298.15 303.15 308.15 313.15 318.15 323.15 328.15 333.15 338.15 343.15 348.15 353.15 358.15 363.15 368.15 373.15

0.84 1.16 1.58 2.13 2.84 3.76 4.93 6.40 8.24 10.54 13.36 16.82 21.04 26.13 32.26 39.58 48.28 58.56 70.67 84.84 101.35

0.6113 0.87 1.23 1.71 2.34 3.17 4.25 5.63 7.38 9.59 12.35 15.76 19.94 25.03 31.19 38.58 47.39 57.84 70.14 84.55 101.35

37.30% 33.02% 28.30% 24.51% 21.51% 18.62% 15.94% 13.68% 11.71% 9.86% 8.19% 6.75% 5.50% 4.40% 3.42% 2.58% 1.87% 1.25% 0.75% 0.34% 0.00%

2

The logarithmic trend is well-represented. However, at lower temperatures the Clausius kJ  Clapeyron equation is up to 37% off. The actual heat of vaporization changes from 2501.3    kg   kJ  at 0.01 oC to 2257.0   at 100 oC, a difference of around 10%.  kg  100.00

10.00

Eqn 6.23 Tables

1.00

0.10 273

283

293

303

313

323

333

343

353

363

373

Temperature [K]

(d) For 100 ºC to 200 ºC, we obtain the following table: T [K]

Eqn 6.24 [kPa]

Steam Tables [kPa]

% Difference

373.15 378.15 383.15 388.15 393.15 398.15 403.15 408.15 413.15 418.15 423.15 428.15 433.15 438.15 443.15 448.15 453.15 458.15 463.15

101.35 120.51 142.64 168.11 197.30 230.63 268.55 311.54 360.11 414.82 476.24 545.00 621.74 707.16 801.99 906.98 1022.93 1150.68 1291.10

101.35 120.82 143.28 169.06 198.53 232.1 270.1 313 361.3 415.5 475.9 543.1 617.8 700.5 791.7 892 1002.2 1122.7 1254.4

0.00% 0.26% 0.44% 0.56% 0.62% 0.63% 0.57% 0.47% 0.33% 0.16% 0.07% 0.35% 0.64% 0.95% 1.30% 1.68% 2.07% 2.49% 2.93%

3

468.15 473.15

1445.10 1613.62

1397.8 1553.8

3.38% 3.85%

10000.00

Eqn 6.23

1000.00

Tables

100.00 373

383

393

403

413

423

433

443

453

463

473

Temperature [K]

Over this range the Clausius-Clapeyron equation represents the data well and is no more than 4 % off. The actual heat of vaporization changes from 2257.0 kJ/kg  at 100 oC to 1940.7 kJ/kg  at 200 oC, a difference of around 15%. (e) The heat of vaporization can be corrected for temperature as follows hvap T  

Tb



T

l cP dT

 hvap Tb    c Pv dT Tb

T

We can acquire heat capacity data from Appendix A.2, but to simplify the analysis, we will use an average heat capacity for the vapor. hvap T   75.4373.15  T   40626  34.13T  373.15

hvap T   56026  41.27T

Substitute this expression into the Clausius-Clapeyron equation dPisat Pisat



56026  41.27T  dT RT 2

Integrate:

4

1  1  1  T   Pisat  101.35 kPa  exp   56026    41.27 ln    T 373.15   373.15   R 

Now plot the data as before from 0.01 ºC to 200 ºC. T [K]

Eqn 6.24 [kPa]

Steam Tables [kPa]

% Difference

273.16 278.15 283.15 288.15 293.15 298.15 303.15 308.15 313.15 318.15 323.15 328.15 333.15 338.15 343.15 348.15 353.15 358.15 363.15 368.15 373.15 378.15 383.15 388.15 393.15 398.15 403.15 408.15 413.15 418.15 423.15 428.15 433.15 438.15 443.15 448.15 453.15 458.15 463.15

0.64 0.91 1.28 1.78 2.43 3.29 4.39 5.81 7.60 9.86 12.66 16.12 20.35 25.49 31.68 39.10 47.91 58.32 70.54 84.80 101.35 120.46 142.40 167.48 196.00 228.29 264.70 305.56 351.26 402.16 458.64 521.10 589.92 665.51 748.27 838.59 936.89 1043.55 1158.98

0.6113 0.87 1.23 1.71 2.34 3.17 4.25 5.63 7.38 9.59 12.35 15.76 19.94 25.03 31.19 38.58 47.39 57.84 70.14 84.55 101.35 120.82 143.28 169.06 198.53 232.1 270.1 313 361.3 415.5 475.9 543.1 617.8 700.5 791.7 892 1002.2 1122.7 1254.4

4.95 4.96 4.24 3.88 3.86 3.65 3.35 3.17 3.03 2.78 2.51 2.28 2.06 1.84 1.58 1.34 1.09 0.82 0.57 0.29 0.00 0.30 0.61 0.94 1.28 1.64 2.00 2.38 2.78 3.21 3.63 4.05 4.51 4.99 5.49 5.99 6.52 7.05 7.61

5

468.15 473.15

1283.56 1417.67

1397.8 1553.8

8.17 8.76

Water Saturation Pressure as a Function of Temperature 10000.00

Equation

Steam Tables

Psat (kPa)

1000.00

100.00

10.00

1.00 250

300

350

0.10

400

450

500

T (K)

The agreement between the two values at lower temperatures improves significantly at lower temperatures, but actually worsens at higher temperatures. The agreement could potentially be improved by not averaging the heat capacity.

6

6.2 We can find the required pressure by applying the Clapeyron equation: dP hl  h s  dT vl  v s T





We can find the molar volume of water ice from any number of reference books. At 0 ºC and 1 bar:  kg    1000    m3  l

l

 v  1.80  10

5

 m3     mol 

 m3   v s  1.97  10  5    mol 

 kg    m3 

 s  917 

Also at 0 ºC and 1 bar,

 J  h l  h s  6010   mol 









If we assume that h l  h s and v l  v s are independent of temperature and pressure, we can separate variables in the Clapeyron equation and integrate. P2  P1 

T  ln 2  v v  T1 

hl  h s



l

s



P2  1  10 5 Pa   so

1.80 10

6010 J/mol 5

 1.97  10

P2  66.1 bar 

7

5

 268.15 K   ln 273 . 15 K m / mol  



3



6.3 (a) At 1 bar, the gas will act as an ideal gas.

  J  300 K   8.314   m3  RT   mol  K    0.0249  v   P 1  10 5 Pa   mol  The number of moles of vapor are found as follows (neglect molar volume of liquid) V n   v v

 

0.001 m 3  m3  0.0249    mol 

n v  0.0402 mol

(b) At 21 bar, the gas will not behave ideally. Since we are assuming that the molar volume of liquid is negligible and the heat of vaporization is independent of temperature, the Clapeyron equation becomes

dP h vap  v dT vT The molar volume using pressure expansion of the virial equation is vv 





RT 1  B ' P  1   RT   B' P P  

Substituting this expression into the Clapeyron equation yields

dP  dT

h vap 1  R  B ' T 2 P 

Separation of variables yields P2  21 105 Par



P1 1 105 Pa

h vap 1 '   B dP  R P 

T2

dT 2 T1 300 K T



and integration results in

8

P ln 2  P1

  h vap   B ' P2  P1   R 

1 1     T2 T1 

We can substitute values for given quantities and constants to solve for T2. T2  523.3 K

(c) Using the virial equation, 3   1 1    J  7  m         v v  RT   B '    8.314  523 . 3 K 1 10     5   P    mol  K    J   21  10 Pa

 m3  v v  0.00164    mol 

We can assume the volume occupied by the liquid is negligible. Therefore, nv 

V  vv

 

0.001 m 3  m3  0.00164    mol 

n v  0.61 mol

9

6.4 We can use the following computational path to solve for pressure at which graphite and diamond are in equilibrium at 25 oC. graphite

P = 1 [atm] g  2866

g1 

diamond

 J  mol

P

 v graphdP

g 3 

1

1

 v diam dP P

P graphite

g 2  0

diamond

Summing together the three steps we get:  J  g 1[atm]  2866   g1  g 2  g 3  mol  To find the change in Gibbs energy with pressure, we apply the fundamental property relation, Equation 5.9. At constant temperature: 0 dgi  v idP  sidT If the solid is assumed incompressible, we can integrate to get

g i   vi dP  vi P Thus the sum of Gibbs energy becomes P

1





 J  g 1[atm]  2866     v graph dP  0   v diam dP  v graph  v diam P  1  mol  1 P

Solving





 1   cm 3    g   1  m 3    J    1 5   2866      3  P  1.01  10 Pa   12        6    mol    2.26 3.51   g    mol   10  cm   

or

P = 1,514 [MPa] = 15,143 [bar]

10

6.5 From the Clausius-Clapeyron equation: s hTfus h lAl  h Al dP   s s dT v lAl  v Al T v lAl  v Al T



 



(I)

where h Tfus is the enthalpy of fusion at temperature T. We can get the molar volumes from the densities:  kg  0.027  MW  mol   1.17  10  5 v lAl    kg  ρl 2,300    m3 

 m3     mol 

 kg  0.027  MW  mol   1.00  10  5 s v Al    kg  ρs 2,700    m3 

 m3     mol 

and

so  m3  s v lAl  v Al  1.7  10 6    mol  We can use the following path to calculate for h Tfus . solid

liquid

T h Tfus

933

 cP dT

T

T

933

 cPl dT

s

T = 933 [K] solid

h fus  10,711

 J  mol 

liquid

11

hTfus 

T 933.45  J  s l c P dT  10,711  cP dT    mol  T 933.45





 J   J  l Using c Ps  20.608  0.0138T   and c P  31.748   , we get:  mol K   mol K  h Tfus  5,819.9  11.68T  0.0069T 2

Back into Equation (I) gives:  5,819.9  11.68T  0.0069T 2  dP    dT 1.7  10  6 T  





Integrating: 100 [bar]

 5,819.9  11.68T  0.0069T 2   dT  dP    6  1 . 7 10 T   1 933.45 [K] T





or

100  1105 1.7 10  6   5819.9 ln 933T.45   11.68T  993.45  0.00345T 2  993.452  solving for T gives T = 934.91 [K]

12

6.6 We can assume that silver acts as an ideal gas at 1500 K. We can also assume the molar volume of the vapor is much greater than the molar volume of liquid. Therefore, we can use Equation 6.22

dP sat P sat



h vap dT RT 2

This can be rearranged to show dP sat h vap P sat  dT RT 2 This relation assumes 1. vv>>vl 2. Silver acts as an ideal gas We can differentiate the expression for pressure in the problem statement to obtain  14260 0.458    14260     0.458 ln T   12.23   exp 2 dT T   T  T  sat dP  14260 0.458  sat   P 2 dT T   T

dP sat

Therefore, vap  14260 0.458  h    2 T  RT 2  T

Therefore, h vap  R14260  0.458T 

At 1500 K,   J  14260  0.4581500 K  h vap   8.314   mol  K     kJ  h vap  112.8   mol 

13

6.7 For a single component system:

  Gi  g i  g From the fundamental property relation given by Equation 5.9:

dg   sdT  vdP We can identify a phase transition from the vertical line of the g vs. T plot, as indicated below. Since this transition is vertical, i.e., the temperature is constant, the pressure must also be constant. Thus, we can differentiate the Gibbs energy with respect to temperature at constant pressure to get:  g     s  T  P

Hence the slope of a plot of g (or ) vs. T at any temperature must be the negative of the value of entropy on the plot for s vs. T. The resulting curve is sketched below. 

s

phase transition; vertical line indicates P = const



T 

  

Thick line denotes lowest value of 

Slope = the negative value of s at the same T on the curve above

 T*

T

14

6.8 The ferrite phase has stronger bonds. At room temperature, iron is in the ferrite phase. The heating to 912 ºC has the effect of increasing the entropy contribution to the Gibbs energy. At a high enough temperature, the austenite phase becomes stable, so that its entropy must be greater than the ferrite phase. If the entropy of the austenite phase is greater, the enthalpy of the ferrite phase must be greater or else the austenite phase would be stable over the entire temperature range. Hence, the ferrite phase has stronger bonds.

15

6.9 Since the pressures are low, we can assume ideal gas behavior. We can also assume that the molar volume of the vapor is much greater than the molar volume of liquid and the heat of vaporization is independent of temperature. Therefore, we can rearrange Equation 6.24 to obtain

h vap

h vap

  J    760 torr    8.314   ln   mol  K    400 torr    1 1       353.25 K 333.75 K   kJ   32.3   mol 

This value is 7.7% smaller than the reported value.

16

6.10 (a) The freezing point occurs where there is a discontinuity in the g vs. T plot, as indicated below. The liquid is at a temperature higher than the freezing point and the solid at lower temperature. These are demarked below. The melting temperature is 250 K, which occurs at a value g = 3,000 [J/mol]

Pure species "a" 6000

Gibb's Energy, g (J/mol)

1 unit up New Freezing point

4,000 solid Freezing point

2,000 liquid

100

200

1.2 units up

300

Temperature, T (K) (b) At constant pressure, the entropy can be found from Equation 5.14. For the solid we have: g 1,000  g   J    10  s      T 10  T  P  mol K 

And for the liquid, we get: g 2,000  g   J    40  s     T 50  T  P  mol K 

(c) As we change pressure, we can see how the Gibbs energy changes at any given temperature by Equation 5.14:

17

 g    v  P  T

Assuming the molar volumes of the liquid and vapor stay constant over the temperature range around the melting point, we see that the Gibbs energy of the liquid increases by 1.2 times the Gibbs energy of the solid, since the molar volume of the liquid is 20% larger. The Gibbs energy of the new freezing point at higher pressure is schematically drawn on the plot above. For convenience, we choose the solid to increase by 1 unit on the plot. Thus, the liquid increases by 1.2 units. As the sketch shows, the freezing point, where the two lines intersect, will shift to higher temperature.

18

6.11 For a single component system the fundamental property relation, Equation 5.9, gives:

dg   sdT  vdP We can identify a phase transition from the vertical line of the g vs. P plot, as indicated below. Since this transition is vertical, i.e., the pressure is constant, the temperature must also be constant. Thus, we can differentiate the Gibbs energy with respect to pressure at constant temperature to get:  g    v  P  T

Hence the slope of a plot of g vs. P must have a slope that matches the plot for v vs. T. Since the molar volume of phase  is about twice the value of phase , its slope should be twice as big. The resulting curve is sketched below. v

phase transition; vertical line indicates T = const



v v

 P

g

 Slope of top line is

Straight line since v is constant



 Thick line denotes lowest value of g

 P*

P

19

about twice as big as the slope of bottom line

6.12 The saturation pressure can be found using the Clausius-Clapeyron equation with the assumption that the heat of vaporization is independent of temperature. First, we need to use the given data for the 63.5 ºC and 78.4 ºC to find the heat of vaporization.

 P sat  R ln 2  P sat  1

    J    760 torr   ln     8.314   mol  K    400 torr    vap  h  1 1   1 1  351.55 K  336.65 K      T2 T1   kJ  h vap  42.39   mol  Now we can calculate the vapor pressure at 100 ºC.   h vap sat sat P3  P2 exp  R 

   kJ    42.39     1  1  1 1 mol            760 torr  exp  kJ   373.15 K 351.55 K    T3 T2    0.008314    mol  K   

P3sat  1760 torr  2.32 atm

In comparison, ThermoSolver gives a value of 2.23 atm, using the Antoine equation.

20

6.13 We can show using the Chain Rule that

  gi    T  T  

  Tg i  g i T         T2    1  g i   g i          T T  T  P T 2    P  P

Using fundamental property relations, Equation 5.14 states  g i   T

    si P

Therefore,

  gi    T  T  

      Tsi  g i   Tsi  hi  Tsi    hi  T2 T2 T2  P

21

6.14 Let T1  922 K, T2  1,300 K g 2  h2  T2 s 2 dh  c P dT 1,300 K

h2  h1 

 c P dT

922 K

 J  h2  h1  c P T2  T1   39,116   mol  c ds  P dT T T   J  s2  s1  c P ln 2   85.10   mol K   T1   J  g 2  h2  T2 s2  71,500   mol  Alternative solution using the result from Problem 6.13:

d g T   dT  g    T 2

    hT   h1  c P T  T1  We must leave h as a function of T  T2 T2 

 h  c T  T   g  d  T      1 PT 2 1  dT 922 K g 1300 K

   T 1

1 1 T  g g  J        h1  c PT1     c P ln 2   55.01   T  2  T 1  mol K   T2 T1   T1 

 J  g 2  71,500   mol 

22

6.15 A possible hypothetical solution path is presented below: monoclinic

orthorhombic T = 298 [K]

g 1

g 3

T = 368.3 [K] monoclinic

g = 0 2

orthorhombic

From the diagram, we see that the Gibbs energy for steps one and three can be calculated as follows:  g m g1    T 298 K  368 K

  dT P

and  g o    T  dT 368 K  P respectively. We can apply Equation 5.14 from the thermodynamic web 298 K

g 3 

 g     s  T  P At 368 K, sulfur undergoes a phase transition, so g 3m68o K 0 mo Using these above relationships, the expression for g 298 K becomes mo g 298 K 

368 K m

s

dT  0 

298 K

298 K o

s

368 K

dT  

368 K

298 K

298 K

368 K

 13.8  0.066T dT   11  0.071T dT

 J  m o g 298 K  79.5   mol  Therefore, the transition from the monoclinic to orthorhombic state occurs spontaneously. The orthorhombic state is more stable. 23

6.16 At the phase transition, the following is true g g      T  Sr ( s )  T  Sr (l )

Using the thermodynamic web, the following can be shown (see Problem 6.13)

h  g / T      2  T  P T The enthalpies can be written as follows T

h l T   49179 

 35.146 dT  35.146T  3540

h s T   20285 

1500 K T

 37.656 dT  37.656T  16305.4

900 K

g can also be calculated at 900 K for solid Sr and 1500 K for liquid Sr. T  gl  T 

  J    83.85    mol  K   ref

 gs   T 

  J    68.68    mol  K   ref

g We can find   at any temperature using the differential equation as follows T  g

h  g / T    dT    2 dT T  P T

 d  T    

Substituting our expressions, we get g l /T

T

 35.146T  3540  g dT  d  T      2 T   83.85 1500 K

24

 gl  T 

   35.146 ln(T )  3540  176.04  T 

g s /T

T

 37.656T  16305.4  g dT  d  T      2 T   68.68 900 K  gs   T   gl Set  T 

   37.646 ln(T )  16305.4  205.5  T 

  gl   T  

  and solve for T:  

T melt  1059.8 K

The enthalpy of melting is defined as



 

h fus  h s T melt  h l T melt



Using the expressions developed above  kJ   kJ   kJ   7.41   33.71  h fus  26.30     mol   mol   mol 

25

6.17 At the phase transition, the temperature and Gibb’s energy of both phases must be equal. Mathematically, this is equivalent to g g      T  SiO2 ( s )  T  SiO 2 (l )

Using the thermodynamic web, the following can be shown (see Problem 6.13) h  g / T      2  T  P T

Using the definition of enthalpy, we can write the following hT 

T

href

Tref

 dh   c P dT

 hT   href 

T

 c P dT

Tref

The enthalpies can be written as follows h T   738440  l

h s T   856840 

T

 85.772 dT

2500 K T

 53.466  0.02706T  1.27  10

5 2

1100 K

g can also be calculated at 1100 K for solid SiO2 and 2500 K for liquid SiO2. T  gl  T 

  J    487.3    mol  K   ref

 gs   T 

  J    903.5     mol  K   ref

26



T  2.19  10  9 T 3 dT

We can substitute our expressions for

g and hT  into the above differential equation and T

separate variables to obtain T       dT 738440 85 . 772 l  g /T T   g 2500 K   d    dT  T    2 T   487.3 2500 K    





T   5 2 9 3          856840 5 3.466 0 . 02706 1 . 27 10 2 . 19 10 T T T dT s  g /T T   g 1100 K   d    dT  T    2 T   903.5 1100 K    

Integration provides g l  9.5268  10 9  85.772 ln(T )  1052.4T   487.5 T T g s  1.82  10 10 T 4  2.12  10 6 T 3  0.01353T 2  927190.9  53.466T ln(T )  1230T   903.5 T T

 gl If we plot  T 

  gs    T  

  vs. T, we obtain the following:  

27

There are three solutions, but only the solution between 1100 K and 2500 K is physically meaningful. If we magnify the plot near the middle solution, we find T  1983 K

The enthalpy of fusion is defined as



 

h fus  h s T melt  h l T melt



Using the expressions developed above  kJ   kJ   kJ   9.72  h fus  792.5    782.78     mol   mol   mol 

28

6.18 From the Clausius-Clapeyron equation sat dPCS



2

dT

h vap



T vv  vl



Assuming: v v  v l

we get sat dPCS



2

dT

h vap

(I)

Tv v

The saturation pressure is given by: sat ln PCS  62.7839  2

4.7063  10 3  6.7794 ln T  8.0194  10  3 T T

(II)

sat 5  4.48 10 Pa. Taking the derivative of Equation II At T = 373 K, PCS 2

sat d ln PCS

2

dT



sat dPCS

1

2

sat PCS

dT



4.7063  10 3 T2



6.7794  8.0194  10  3 (III) T

2

Plugging Equation III into Equation I,  sat h vap  4.7063  10 3 6.7794    8.0194  10  3  PCS  2 T T2 Tv v    kJ  vap Solving for vv using hCS gives:  24.050  2  mol   h vap  4.7063  10 3 6.7794 v    8.0194  10 3   sat T TPCS T2   v

2

z

Pv B  0.878  1  RT v 29

1

 m3   6.08  10 3    mol 

or  m3   cm3  B  7.4  10  4    740    mol   mol 

This value is about 50% higher than the reported value.

Alternative solution: Following similar development as Problem 6.3: sat

dPCS dT

2

h vap    1 RT 2   B ' sat  PCS 2

  h vap 1 sat  sat  B' dPCS  2 dT 2 P RT  CS2 

We must be careful about the limits of integration. We need to pick a value of T close so enthalpy of vaporization is not too different, but far enough away to avoid round off error. If we sat 5 choose T = 378 K, Equation I gives PCS  5.04 10 Pa. Integrating: 2

4.4810 5 Pa

 373 h vap sat  sat  B dPCS    2 dT 2 P RT 5   378 5.04 10 Pa CS2

1

'

4.48  105  ' h vap  1 1  5 5  B ln 4.48  10  5.04  10    5  R 373 378  5.04  10 





Solving for B’ gives: B'  2.55  107 Pa

or m 3  B  B ' RT   7.9  10 4   mol

30

6.19 Calculate vA, vB, v, VA, VB, and V from the ideal gas law:

RT  0.05 m 3 / mol P RT vB   0.05 m 3 / mol P RT v   0.05 m 3 / mol P vA 

V A  n A v A  0.1 m 3

V B  n B v B  0.15 m 3 V  ntot v  0.25 m 3

We calculate the partial molar volumes as follows  V     n A  n B  RT   RT  0.05 m 3 / mol  V A    P  P  n A T , P, n B n A   V V B    n B

    n A  n B  RT   RT  0.05 m 3 / mol   P  P T , P, n A n B 

To find the remaining quantities, we can apply Equations 6.44 and 6.46 Vmix  n A V A  v A   n B V B  v B  Vmix  20.05  0.05  30.05  0.05  0

vmix  x A V A  v A   x B V B  v B  vmix  0

31

6.20 (a) For a pure species property

va  v y a  1 Substitution yields

 cm 3  va  1001  800  2.510   100    mol  (b) From Equation 6.29

 V Va    na

   n b ,T , P

We can find V by multiplying the given expression for molar volume by the total number of moles.  V  na  nb 100 ya  80 yb  2.5 

y a yb  n n   100na  80nb  2.5 a b y a yb  na  nb

Differentiating with respect to na we get,

Va 

 na

 n n  nb na nb 100na  80nb  2.5 a b   100  2.5  2.5 na  nb  n na  nb na  nb 2  b

so

Va  100  2.5 yb 1  ya   100  2.5 yb2 To find the molar volume at infinite dilution, we can use the following relation Va  lim Va ya 0

 cm 3  Va  102.5    mol 

32

(c) Since species A contributes more to a mixture than to a pure species,

v mix  0 Note: The Gibbs-Duhem equation says that species B also contributes more.

33

6.21 Calculate mole fractions: n1 1 mol   0.2 ntot 5 mol

y1 

y3  0.4

y 2  0.4

Calculate v. Obtain an expression for v:

v

RT P

 B  2 A 1  P  RT  y1  y 2   RT     

Substitute values:

v

   3  82.06  cm  atm  500 K    mol  K    50 atm

1  50   9.0 10 2

5



0.2  0.4  3.0  10  5 

 cm 3  v  919.0    mol  Calculate V.

V  ntot v   cm 3   V  5 mol 919.0    mol    

 

V  4595 cm 3

Calculate v1. The value of v1 can be found by substituting y1=1 into the expression for v1.

v1 

   3  82.06  cm  atm  500 K    mol  K    50 atm 

1  50   9.0 10 2

 cm 3  v1  698    mol 

34

5



1  0  3.0  10  5 

Calculate v2.

v2 

   3  82.06  cm  atm  500 K     mol  K    50 atm 

1  50   9.0 10 2

5



0  1  3.0  10  5 

 cm 3  v2  1067    mol 

Calculate v3.

v3 

   3  82.06  cm  atm  500 K    mol  K    50 atm

1  50  3.0 10  2

5

 cm 3  v3  882    mol  Calculate V1 . From Equation 6.29:

  nv    V    V1      n1  n 2 , n 3 ,T , P  n1  n 2 , n 3 ,T , P We can substitute the expression for V into this derivative and use the fact that ntot  n1  n2  n3 to obtain V1 

  RT  B    2 A               n n n P n n n n n 1 2 3 1 2 1 2 3   RT    n1  P  RT   n 2 , n 3 ,T , P

Differentiating we get V1 

RT P

 B  2 A 1  P  RT  RT     

Substitute values:

35

V1 

   3  82.06  cm  atm  500 K    mol  K    50 atm

1  50   9 10 2

 cm 3  V1  697.5    mol 

36

5

 3  10  5



6.22 (a) By definition:

 H H a    na

  T , P , n b , n c

h  5,000 xa  3,000 xb  2,200 xc  500 xa xb xc J/mol n  na  nb  nc H  nh   5,000na  3,000nb  2,200nc  500

na nbnc

na  nb  nc 2

 H  nbn c 2na nbnc     5,000  500   na T ,P,n ,n na  n b  nc 2 na  nb  n c 3    b c H  Ha     5,000  500xb x c 1 2xa  J/mol na T ,P,n ,n b c (b) xa  xb  xc 

1 3

Ha  5,018.5 J/mol (c) xa  1 , xb  xc  0

Ha  5,000 J/mol (d) xb  1 , xa  xc  0

Hb  hb  3,000 J/mol

37

6.23 Let the subscript “1” designate CO2, and “2” designate propane. To calculate the partial molar volumes, the following formulas will be used:

V1  v  y 2

dv dy 2

v  y1V1  y 2V2 Expressions can’t be obtained for the molar volume with the van der Waals EOS; therefore, the problem will be solved graphically. First, obtain an expression for the pressure that contains the mole fractions of CO2 and propane: a mix  y12 a1  2 y1 y 2 a1a 2  y 22 a 2 bmix  y1b1  y 2 b2 y12 a1  2 y1 y 2 a1a2  y 22 a2 RT P  v   y1b1  y 2 b2  v2 Solve for a and b using data from the appendices.  J  m3  a1  0.366    mol 2   J  m3  a 2  0.941   2  mol 

 mol  b1  4.29  10  5    m3   mol  b2  9.06  10  5    m3 

Now we can create a spreadsheet with the following headings: y1

y2

amix

bmix

v

The last column contains the molar volumes obtained by solving the van der Waals equation with the spreadsheet’s solver function. After the table is completed, we create the following graph.

38

v vs. y 2 1.48E-03 1.46E-03

v (m 3 mol -1)

1.44E-03 1.42E-03 1.40E-03 1.38E-03 1.36E-03 y = -0.00008x 2 - 0.00008x + 0.00147 R2 = 0.99946

1.34E-03 1.32E-03 1.30E-03 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y2 (Mole Fraction)

From the line of best fit, we find v  8  10  5 y 22  8  10  5 y 2  0.00147

Therefore, dv  1.6  10  4 y 2  8  10  5 dy 2 and



V1  v  y 2  1.6  10 4 y 2  8  10 5



We can find the partial molar volume of propane from the following relationship v  y1V1  y 2V2 V2 

v  y1V1 y2

Tabulate the values of the partial molar volumes in the spreadsheet and create the following graph

39

Partial Molar Volumes as a Function of Carbon Dioxide Mole Fraction

Partial Molar Volume (m

3

mol -1)

0.0016

V1 V2

0.00155 0.0015 0.00145 0.0014 0.00135 0.0013 0.00125 0

0.1

0.2

0.3

0.4 0.5 0.6 y1 (CO 2 mole fraction)

40

0.7

0.8

0.9

1

6.24 (a)

ga  40

kJ mol

Ga  g  x b

dg dx b

dg  40  60  RT  ln xa  1  ln x b  1  5x a  5xb dx b Ga  40xa  60x b  RT xa ln x a  x b ln xb   5xa x b

 x b40  60  RT  ln xa  1  ln x b  1  5x a  5xb 

Ga  40xa  xb   RT xa  xb  ln x a  5xb2

Ga   40  RT ln x a  5 xb2 Ga   40 

kJ kJ 8.314(300) ln 0.2  5(0.64)   40.8 mol mol 1000

Ga   Gmix  ng  x a ga  xb gb  g  xa ga  x bgb   RT x a ln xa  xb ln x b   5 x a xb Gmix  nT RT x a ln xa  xb ln x b   5xa x b   2.2 kJ

(b)

gmix  hmix  Tsmix

Assume the entropy of mixing is ideal: hmix  5x a xb  0

so h  hmix  hsensibleheat  0

so

hsensibleheat  0 and T goes down

41

6.25 To find V1 and V2 , we can read values directly from the graphs. Calculate mole fractions

x1 

1  0.2 5

At x1  0.2 ,  cm 3  V1  46.5    mol   cm 3  V2  69.8    mol  The following relationships are employed to calculate the molar volumes of pure species

v1  lim V1 x1 1

v 2  lim V2  lim V2 x 2 1

x1  0

From the graph

 cm 3  v1  50    mol   cm 3  v2  70    mol  Therefore,

   280 cm 

V1  50 cm 3 V2

3

To calculate the total volume, we can use

V  n1V1  n2V2 Substituting the values, we find

 

V  146.5  469.8  325.7 cm3

42

Therefore

v

V 325.7  ntot 1 4

 cm 3  v  65.14    mol 

Using Equation 6.43 we can calculate the change in volume.

Vmix  V  V1  V2   325.7  50  280

 

Vmix  4.3 cm 3

43

6.26 (a) Expression for hmix : hmix   X i H i  X i hi X Cd H Cd  X Sn H Sn    X Cd hCd  X Sn hSn 

Multiply both sides by the total number of moles

H mix  nCd H Cd  n Sn H Sn   nCd hCd  n Sn hSn  Therefore,

H mix Cd

 H mix    nCd

   H Cd  hCd  n Sn ,T , P

(b) We can show by repeating Part (a) for Sn that

H mix Sn  H Sn  hSn Equation 6.65:

dhmix dX Sn H mix Sn  hmix  X Cd dhmix dX Cd

H mix Cd

 hmix  X Sn

Since, hmix  13000 X Cd X Sn

We get,

dhmix d 13000 X Cd X Sn   13000 X Cd  X Sn   dX Sn dX Sn dhmix d 13000 X Cd X Sn   13000 X Sn  X Cd   dX Cd dX Cd Therefore, for 3 moles of cadmium and 2 moles of tin at 500 ºC:

44

2 H Cd  hCd  H mix Cd  13000 X Cd X Sn  X Sn13000 X Cd  X Sn   13000 X Sn

 J  H Cd  hCd  H mix Cd  2080   mol  and

 J  2 H Sn  hSn  H mix Sn  13000 X Cd  4680   mol  (c) Equation 6.37: nCd d H mix Cd  n Sn d H mix Sn  0

Differentiate with respect to XCd:

nCd

d H mix Cd dxCd

 n Sn

d H mix Sn dxCd

0

where

d H mix Cd dX Cd

d H mix Sn dX Sn

 26000 X Sn  26000  26000 X Cd  26000 X Cd

Therefore,

nCd


 nSn


 xCd ntot  26000  26000 xCd   1  xCd ntot 26000 xCd

Inspection of the above expression reveals that

nCd


 n Sn


0

(d) A graphical solution can be found using the tangent-slope method discussed on pages 285-287:

A plot of a line tangent to the enthalpy of mixing curve at XCd = 0.6, is given below:

45

Heat of mixin g in cadmium (Cd )-Tin (Sn) s ystem 6000 5400

  J   mol

4800 4200 3600

data

 hmix

3000 2400 1800 1200 600 0 0

0 .2

fit to:   J   hmix  13, 000 xCd x Sn    mol

0 .4

0 .6

0 .8

1

xCd

The intercepts give the respective partial molar quantities as follows:

 J  H Cd  hCd  2050   mol   J  H Sn  hSn  4800   mol  The values using the graphical method are reasonably close to the analytical method.

46

6.27 The following can be shown with the Gibbs-Duhem equation

0  x1V1  x 2V2 Differentiation with respect to x1: 0  x1

dV1 dV  x2 2 dx1 dx1

If the partial molar volume of species 1 is constant, the Gibbs-Duhem equation simplifies to 0

dV2 dx1

Therefore, the partial molar volume of species 2 is also constant. Note that in this case, since the partial molar volume of species 1 is constant:

V1  v1 and similarly for species 2: `

V2  v2

Hence, the molar volume can be written:

v  x1V1  x2V2  x1v1  x2v2 This is known as Amagat’s law.

47

6.28 (a) Let species 1 represent HCl and species 2 represent H2O. An expression for the enthalpy of the solution is

h  x1h1  x 2 h2  hmix which can be written ~ h  x1h1  x2 h2  x1hs

Therefore, ~ H  n1h1  n2 h2  n1hs

To use the heat of solution data in Table 6.1, we need to determine the values of n1 and n2 consistent with the convention used in the table. As seen in Example 6.6,

x1 

n1 1  n1  n2  1  n

For this problem x1  0.2

Therefore,

n1  1 n2  n  4 Now we can find expressions for the partial molar enthalpies.

 H   H H 2 O  H 2    n  2  n1 , T , P ~ dhs H H 2 O  H 2  h2  n1 dn2

~ dhs  H H 2 O  hH 2 O  H 2  h2  n1 dn2 Using the data in Table 6.1 for n  4 ,

48

~ J   hs  61,204   mol solute  J J      64049    56852  ~ ~ ~  dhs hs n  5  hs n  3  mol solute   mol solute    5 mol  3 mol 5 mol  3 mol dn2 ~ dhs J    3598.5  dn2  mol  mol solute  Therefore,

 J    J  H H 2 O  hH 2 O  1 mol solute  3598.5    3599    mol  mol solute    mol   We can calculate H HCl  hHCl using Equation 6.46

H HCl  hHCl 



hmix  x H 2 O H H 2 O  hH 2 O x HCl

  xHCl h~s  xH O H H O  hH O  2

2

2

x HCl

    J   J     0.8  3599   0.2  61204    mol    mol    J      H HCl  hHCl   46808  0.2  mol  (b) For n1  2 and n2  80 ,  n  n2  82   1  n   1  1  40 2  n1 

The new values for the number of moles consistent with Table 6.1

n1  1 mol

n2  40 mol Using the data in the table for n  40 , ~ ~ ~ dhs hs n  50   hs n  30  50 mol  30 mol dn2 Interpolating the data in Table 6.1

49

~ J   hs n  30  72428   mol solute  Therefore, J J      73729    72428  ~  dhs J   mol solute   mol solute   65.05  and   50 mol H 2O  30 mol H 2O dn2  mol  mol solute     J J   H H 2 O  hH 2 O  1 mol solute  65.05    65.05     mol  mol solute     mol H 2 O 

50

6.29 First perform an energy balance on the mixing process.

hmix  q We can calculate hmix using data from Table 6.1. Referring to Equation E6.7A, we find ~ hmix  x HCl hs

Calculate x HCl :

x HCl

wHCl 0.30 MW HCl 36.46    0.175 0 . 30 0.70 w wHCl H 2O   MW HCl MW H 2O 36.46 18.0148

Heats of data are tabulated for a solution containing one mole of the solute for various amounts of water. Thus, we need to calculate how many moles of water must be added to HCl to obtain the above mole fraction. 1 mol HCl ; where n is the number of moles of H2O 1 mol HCl  n n  4.71 mol H 2 O 

x HCl 

By interpolation of data from Table 6.1, we get ~  J  hs  63224   mol 

(for n  4.71 )

Therefore,

  J   J  hmix  0.175  63224    11064    mol    mol   and

 J  q  hmix  11064   mol 

51

6.30 To calculate the enthalpy of mixing from Table 6.1, we must use the following expression

~ hmix  x H 2 SO 4 hs The mole fraction of sulfuric acid is

x H 2 SO4 

1 1 n

where n is the number of moles of water. Equation 6.47 states



hmix  74.4 x H 2 SO 4 x H 2 O 1  0.561x H 2 SO

4



For n  1 , x H 2 SO 4  0.5 and x H 2 O  0.5 Table 6.1:

~  J  hs  31087   mol    J   J    15543.5   hmix  0.5  31087    mol    mol  

Equation 6.47: hmix  74.40.50.51  0.5610.5

 J  hmix  13383   mol  The following table was made n [mol H2O]

x H 2 SO 4

1 2 3 4 5 10 20 50 100

0.5 0.333333 0.25 0.2 0.166667 0.090909 0.047619 0.019608 0.009901

hmix [kJ/mol]

hmix [kJ/mol]

(Table 6.1)

(Eq. 6.47)

-15543.5 -14978.7 -13001.8 -11414 -10174.2 -6367.27 -3548.43 -1497.22 -762.238

-13382.7 -13441.6 -11993.5 -10568.4 -9367.17 -5835.17 -3284.01 -1414.49 -725.289

52

% Difference 14.94 10.82 8.07 7.69 8.26 8.72 7.74 5.68 4.97

As you can see, the percent difference between the two methods decreases as the mole fraction of sulfuric acid decreases. Although Equation 6.47 fit data at 21 ºC, while the Table 6.1 tabulates data taken at 25 ºC, we do not expect the temperature dependence to account for all the observed difference. The table and equation come from different experimental data sets, and also represent measurement uncertainty. Nevertheless, the agreement is reasonable.

53

6.31 To calculate the enthalpy of mixing from Table 6.1, we must use the following expression ~ hmix  x HCl hs

The mole fraction of HCl is

x HCl 

1 1 n

where n is the number of moles of water. The following table was made using these two equations. n [mol H2O]

x1

~  J  hs   mol HCl 

 J  hmix   mol 

1 2 3 4 5 10 20 50

0.5 0.333 0.25 0.2 0.167 0.091 0.048 0.020

-26225 -48819 -56852 -61204 -64049 -69488 -71777 -73729

-13112.5 -16273 -14213 -12240.8 -10674.8 -6317.09 -3417.95 -1445.67

100

0.0099

-73848

-731.168

54

6.32 A schematic for the process is given below. The inlet streams are labeled “1” and “2” and the exit stream “3”. q

50 wt% NaOH

10 wt% NaOH

50 wt% H2O

90 wt% H2O

Stream 1

Stream 3

H 2O

Stream 2

The energy balance for this process reduces to

Q  H 3  H 2  H1 We first convert from weight percentage to mole fraction. For stream 1,

x NaOH ,1

wNaOH 0.50 MW NaOH 40    0.311 0.50 0.50 w H 2O wNaOH   MW NaOH MW H 2O 40 18.0148

and for stream 3,

x NaOH ,3

wNaOH 0.10 MW NaOH 40    0.048 0.10 0.90 wH 2 O wNaOH   MW NaOH MW H 2O 40 18.0148

We now calculate the moles of water per mole of NaOH so that we can use Table 6.1:

x NaOH 

1 1  nH 2 O

Therefore, for every mole of NaOH n H 2 O,1  2.21

55

n H 2 O,3  19.8

Since enthalpy is a state function, we can choose any hypothetical path to calculate the change in enthalpy. One such path is shown below. The box in our original schematic is depicted with dashed lines below. We pick a basis of 1 mole NaOH. In step A, the inlet stream is separated into its pure components. In step B, 17.6 additional moles of water are added to the pure water stream. Finally the H2O and NaOH streams are remixed

1 mol NaOH

Step C

Step A

1 mol NaOH

2.2 mol H2O

1 mol NaOH 19.8 mol H2O

Step B

hmix=0 2.2 mol H2O

19.8 mol H2O

17.6 mol H2O

The enthalpy change is found by adding each step

H 3  H 2  H 1  H A  H B  H C Since H B represents the mixing of water with water, H B  0 . The enthalpies of mixing for steps A and C can be related to enthalpy of solution data from Table 6.1: ~ J   hs ,1  23906   mol NaOH  ~ J   hs ,3  42858   mol NaOH  Note: The enthalpy of solution for Stream 1 is calculated by extrapolation. Generally, extrapolation should be avoided, but it is necessary to complete this problem, and we are not extrapolating very far. For step A, we need the negative value of the heat of solution of stream 1. Thus for a basis of 1 mole NaOH: H A   23906  J 

56

while for step C:

H C  42858 J  Now adding the enthalpies of each step per 1 mole of NaOH: H  23906  0  42858  18952 J 

To get the total heat that must be removed per mole of product solution, we divide by the number of moles of product per mol of NaOH: q

H  J   910  n NaOH  mol 

57

6.33 The partial molar property can be written as follows:

 n K  K1   T   n1 T , P ,n2 ,n3 Applying the chain rule to the above relationship:

 n K1  k  T  n1

  k     nT   T , P ,n2 ,n3  n1  T , P ,n2 ,n3

 k   K1  k  nT   n1  T , P ,n2 ,n3

(1)

 k   . At constant T and P, we can write, Now focus on nT   n  1 T , P ,n2 ,n3  k   k   k     dx3 dx2   dx1   dk    x1 T , P, x2 , x3  x2 T , P, x1 , x3  x3 T , P, x1 , x2 Therefore,

 k   k   x1   k   x 2               n1  T , P ,n2 ,n3  x1  T , P , x2 , x3  n1  T , P ,n2 ,n3  x2 T , P , x1 , x3  n1  T , P ,n2 ,n3  k   x3        x3  T , P , x1 , x2  n1  T , P ,n2 ,n3 but, x1 

n1 n1  n2  n3

so  x1  n1 1 1      1  x1  2 nT  n1 T , P ,n2 ,n3 n1  n2  n3 n1  n2  n3  Similarly,

58

(3)

(2)

 x 2  n2 x     2 2 nT n1  n2  n3   n1  T , P ,n2 ,n3

(4)

 x3  n3 x     3 2 nT n1  n2  n3   n1 T , P ,n2 ,n3

(5)

and

Substituting Equations 2, 3, 4, and 5 into the expression 1 for K1 and simplifying:    k     1  x1    k   x2    k   x3  K1  k    x1 T , P, x 2 , x 3  x 2 T , P, x1 , x 3  x3 T , P, x1 , x 2 Utilize the fact that x1  x 2  x3  1    k    k   k    (6)   x3  k        K1  k  x 2    x1 T , P, x , x  x3 T , P, x , x   x1 T , P, x , x  x 2 T , P, x , x  2 3 1 3  2 3  1 2   When we hold species 3 constant:  k   k    dk   dx1   dx 2  x 2 T , P, x1 , x3  x1 T , P, x2 , x3  k   k   dx1   k   dx2               x 2  T , P , x3  x1  T , P , x2 , x3  dx2   x2  T , P , x1 , x3  dx2  Thus,  k   k   k           x2 T ,P , x3  x1 T ,P , x2 , x3  x2 T ,P , x1 , x3

(7)

When we hold species 2 constant, a similar analysis shows:  k   k   k           x1 T , P , x2 , x3  x3 T , P , x1 , x2  x3 T , P , x2 Substituting Equations 7 and 8 into Equation 6 gives

59

(8)

 k   k    K1  k  x 2   x3   x2  T , P , x3  x3  T , P , x2 Furthermore, the above analysis can be extended to m components. In general,  k   K i  k   xm  m i  xm T , P , x ji ,m

60

6.34 The expression can be found by employing the Gibbs-Duhem equation:

0  n1V1  n2V2 Differentiate with respect to x1 and then divide by the total number of moles: 0  x1

dV dV dV dV1  x 2 2  x1 1  1  x1  2 dx1 dx1 dx1 dx1

Differentiate the expression given in the problem statement. dV1  5.28  5.28 x1 dx1 Substitution of this result into the Gibbs-Duhem equation and rearrangement yields dV2 5.28 x12  5.28 x1   5.28 x1 dx1 x1  1

Integrate:  cm 3  V2  2.64 x12  C    mol  To determine C, we can use the density information given in the problem statement. V2  x 2  1  V2  x1  0  C  v 2 where C  v2 

1  2     MW2 



 cm 3   109.58   mol   0.768 g/cm 3       84.16 g/mol    1





Therefore,  cm 3  V2  2.64 x12  109.58    mol 

61

6.35 An expression for the enthalpy of the solution is

h  x1h1  x 2 h2  hmix which is equivalent to ~ h  x1h1  x 2 h2  x1hs

Multiplication by the total number of moles yields ~ H  n1h1  n2 h2  n1hs

To use the heat of solution data in Table 6.1, we need to determine the values of n1 and n2 consistent with the convention used in the table. As seen in Example 6.6, x1 

n1 1  n1  n2  1  n

For this problem x1  0.33

Therefore, n1  1 n2  n  2 Now we can find expressions for the partial molar enthalpies.  H   H H 2 O  H 2    n2  n1 ,T , P ~ dhs H H 2 O  H 2  h2  n1 dn2

~ dhs  H H 2 O  hH 2 O  H 2  h2  n1 dn2 Using the data in Table 6.1 for n  2 ,

62

J J      2,787    812  ~ ~ ~  dhs hs n  3  hs n  1  mol solute   mol solute    3 mol  1 mol 3 mol H 2O  1 mol H 2O dn2 ~ dhs J    987.5  dn2  mol  mol solute  Therefore,  J    J  H H 2 O  hH 2 O  1 mol solute  987.5    987.5    mol  mol solute    mol   Calculate the partial molar enthalpy:  J  H H 2O  hH 2 O  987.5   mol  From the saturated steam tables at 25 ºC:  kJ  hˆH 2 O  104.87    kg   kJ   hH 2 O  1.89   mol  Now we can find the partial molar enthalpy       kJ J J H H 2 O  1.89    0.988    0.90    mol H 2 O   mol H 2 O   mol H 2 O 

63

6.36 (a) Calculate the mole fraction of sulfuric acid

x1 

w1 MW 1 w1 w2  MW 1 MW 2

0.20 98.078   0.044 0.20 0.80  98.078 18.0148

Calculate n to use in Table 6.1: 1  0.044 1 n n  21.7 mol H 2 O x1 

Interpolating from Table 6.1 ~  J  hs  74621   mol  Now calculate the heat transfer ~   J   q  hmix  x1hs  0.044  74621   mol     J  q  3283   mol  (b) Calculate the mole fraction of pure sulfuric acid. Consider a mixture of 20 kg of 18 M sulfuric acid and 80 kg of water. Find the mass of sulfuric acid present.

VH 2 SO4 

mH 2 SO4



 H 2 SO4

20 kg  10.9 L 1.84 kg/L

nH 2 SO4  VM  10.9 L 18 mol/L  196.2 mol





m H 2 SO 4  n H 2 SO4 MW H 2 SO 4  196.2 mol0.098 kg/mol  19.2 kg

Since both the initial (i) and final (f) states contain mixtures, to get the enthalpy of mixing, we need to calculate the relative differences follows:

64

~ ~ q  hmix  x1, f hs, f  x1,i hs ,i calculate the mole fraction in the final state w1 MW 1

0.192 98.078   0.042 x1, f  w1 w2 0.192 0.808   MW 1 MW 2 98.078 18.0148

Calculate n to use in Table 6.1: 1  0.042 1 n n  22.8 mol H 2 O x1, f 

Interpolating from Table 6.1 ~  J  hs , f  74689   mol  For the initial 18 M sulfuric acid: w1 MW 1

0.192 98.078   0.81 x1,i  w1 w2 0.192 0.008   MW 1 MW 2 98.078 18.0148

Calculate n to use in Table 6.1: 1  0.81 1 n n  0.23 mol H 2 O x1,i 

We must extrapolate from Table 6.1. To do this we wish to extend the trend at low water concentration. A plot of the data in Table 6.1 is useful. A semi-log plot follows:

65

0

3200 [J/mol] enthalp of solution [J/mol solute]

-10000 -20000 -30000 -40000 -50000 -60000 -70000 -80000 -90000 0.1

1

10

100

n

~  J  hs ,i  3200   mol  Now calculate the heat transfer ~ ~    J   J  q  hmix  x1, f hs , f  x1,i hs,i  0.042  74689    0.81  3200     mol    mol      J  q  710   mol 

(c) Calculate the mole fraction of sodium hydroxide

x1 

w1 MW 1 w1 w2  MW 1 MW 2

0.20 40   0.101 0.20 0.80  40 18.0148

Calculate the n value to use in Table 6.1: 1  0.101 1 n n  8.9 mol H 2 O x1 

Interpolating from Table 6.1 ~  J  hs  41458   mol 

66

Now calculate the heat transfer ~   J   q  hmix  x1hs  0.101  41458   mol     J  q  4187   mol  (d) Calculate the mole fraction of ammonia

x1 

w1 MW 1 w1 w2  MW 1 MW 2

0.20 17.03   0.209 0.20 0.80  17.03 18.0148

Calculate the n value to use in Table 6.1: 1  0.209 1 n n  3.78 mol H 2 O x1 

Interpolating from Table 6.1 ~  J  hs  33153   mol  Now calculate the heat transfer ~   J   q  hmix  x1hs  0.209  33153   mol     J  q  6929   mol 

67

6.37 Let species 1 designate ethanol and species 2 designate water. We need to obtain an expression for the molar volume, so first, convert the given the mass fractions and densities to mole fractions and molar volumes.

x1 

Mole fractions:

Specific molar volumes:

w1 MW 1 w1 w2  MW 1 MW 2

v  vˆMW mixture where MW mixture  x1 MW 1  x 2 MW 2

Using this set of equations, the following table was made. Mole Frac. EtOH 0.000 0.042 0.089 0.144 0.207 0.281 0.370 0.477 0.610 0.779 1.000

Mole Frac. H2O v (ml/mol) 1.000 18.05 0.958 19.54 0.911 21.18 0.856 23.11 0.793 25.47 0.719 28.34 0.630 31.85 0.523 36.19 0.390 41.65 0.221 48.73 0.000 58.36

The following graph plots the data. The trendline relates v to x1. 60.00

v (ml/mol)

50.00 40.00 2 v = 4.5491x1 + 35.918x1 + 17.957

30.00

R2 = 1

20.00 10.00 0.00 0.000

0.200

0.400

0.600

Mole Fraction EtOH (x1)

68

0.800

1.000

Now, we can calculate V1 .  V    nv       V1    n1  n 2 , n 3 ,T , P  n1  n 2 , n 3 ,T , P We can substitute the trendline for V into this derivative and use the fact that ntot  n1  n2 to obtain  n12   V1  4.5491  35.918n1  17.957n1  n2   n1  n2  n1  

n 2 ,T , P

Differentiating we get  2n n  n   n 2  2 1  53.875  ml  V1  4.5491 1 1   mol    n1  n2 2    ml  V1  4.5491 2 x1  x12  53.875   mol 





Calculate V2 : V2 

 n2

2    4.5491 n1  35.918n1  17.957n1  n2    n1  n2   n1 ,T , P 

Differentiating we get   n2   ml  1 V2  4.5491   17.957    mol   n1  n2 2   ml  V2  4.5491x12  17.957   mol 

Plotting V1 and V2 vs. x1 we obtain

69

60 59

20 19

58 57 56 55

18 17 16 15

54 53 52

14 13 12 0

0.2

0.4

0.6

0.8

H2O Partial Molar Volume (ml/mol)

EtOH Partial Molar Volume (ml/mol)

Partial Molar Volumes vs. EtOH Mole Fractions

EtOH H2O

1

Mole Fraction EtOH (x1)

(b) v mix can be calculated using Equation 6.46





v mix  x H 2 O V H 2 O  v H 2 O  x EtOH V EtOH  v EtOH  From the data table in Part (a),  ml  v EtOH  v x1 1  58.36   mol   ml  v H 2 O  v x1  0  18.05   mol  Using the expressions for partial molar volumes





 ml  V1  4.5491 20.5  0.52  53.875  57.29   mol   ml  V2  4.54910.52  17.957  16.82    mol  Therefore,    ml    ml      0.5 57.29  58.36  v mix  0.516.82  18.05    mol    mol      ml  vmix  1.15    mol 

70

6.38 We can use the density data given in the problem statement to determine the pure species properties. For pure ethanol x1  1 :

46 g/mol

 cm 3    58.55  v1   1 0.7857 g/cm 3  mol  MW1





 

  cm 3   3 V1  n1v1  3 mol 58.55     175.7 cm    mol   

For pure formamide  x1  0  : 45 g/mol

 cm 3   39.77   2 1.1314 g/cm 3  mol    cm 3   3 V2  n2 v 2  1 mol 39.77     39.77 cm   mol     v2 

MW2







 

To calculate V and v, interpolate in the data table to obtain the density of the mixture when x1  0.75 :



  0.8550 g/cm 3



Therefore, v

where

MW



MW  0.7546 g/mol  0.2545 g/mol  45.75 g/mol

Substitute numerical values: v

45.75 g/mol



0.8550 g/cm

3





 53.51 cm 3 /mol

  

  cm 3   3 V  nv  4 mol 53.51     214.0 cm    mol   

Now, we can calculate the volume change of mixing by using Equation 6.43:

71

 

 

 

Vmix  V  V1  V2  214 cm 3  175.7 cm 3  39.77 cm 3

 

Vmix  1.47 cm 3

The intensive volume change of mixing:

 



Vmix  1.47 cm 3 v mix    0.368 cm 3 / mol n 4



We can use Equation 6.65 to determine the partial molar volume of formamide: V2  v  x1

dv v  v  x1 dx1 x1

From the provided data table 45.8 45.7  v  0.8401 0.8701  19.50 cm 3 / mol x1 0.8009  0.6986





Therefore,





 





V2  53.51 cm 3 / mol  0.75 19.5 cm 3 / mol  38.89 cm 3 / mol



Now calculate the partial molar volume of ethanol: V  n1V1  n2V2

 









214.0 cm 3  1 mol 38.89 cm 3 /mol V1   58.37 cm 3 / mol 3 mol

72



6.39 Using the definition of G, the Gibbs energy of mixing of an ideal gas can be rewritten in terms of the enthalpy of mixing and the entropy of mixing: ideal gas ideal gas ideal gas g mix  hmix  Tsmix

Since an ideal gas exerts no intermolecular interactions, ideal gas hmix 0

From Equation 6.48: ideal gas smix   R xa ln xa  xb ln xb 

so ideal gas g mix  RT  xa ln xa  xb ln xb 

To find the partial molar Gibbs energy of mixing of species a, we apply Equation 6.29:

Gmix a    nng mix  

a

 T , P , nb

Applying the expression above  nb  na   RT na ln na  nb ln nb  na  nb lnna  nb  ng mix  RT  na ln  nb ln na  nb na  nb   where the mathematical relation of logarithms was used. Thus,

 n  nb   RT ln x n  RT ln na  a  0  ln na  nb   a a na  nb  na a   T , P , nb  at infinite dilution xa goes to zero, and the ln term blows up,

Gmix a    nng mix 

Gmix a  a  g a   As chemical engineers, we are often interested in the limiting case of infinite dilution. We see that even for ideal gas mixtures the chemical potential in this limit is not mathematically well.behaved, In Chapter 7, we will develop a different function, the fugacity, which behaves better.

73

6.40 At equilibrium l H

2

O

v  H 2O

or G Hl O  G Hv O 2 2

Since the water in the liquid phase is pure GHl 2O  g Hl 2O  hHl 2O  Ts Hl 2O

The enthalpy and entropy of liquids are not sensitive to pressure changes. We can use data from the saturated steam tables at 25 ºC to determine the Gibbs energy. l gˆ H  104.87 kJ/kg   298.15 K 0.3673 kJ/kg  K  2O l gˆ H  4.640 kJ/kg  2O







l l gH  MWH 2 O gˆ H  0.0180148 kg/mol 4.640 kJ/kg   0.0836 kJ/mol 2O 2O Therefore, v H

2O

l  gH  83.6 J/mol 2O

74

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