Assign 2 Solutions

Thermodynamics of Multicomponent Systems – 3C04 Assignment 2 – Solutions

1. A rigid container is divided into two compartments of equal volume by a partition. One compartment contains 1 mole of ideal gas at 1 atm, and the other contains 1 mole of ideal gas B at 1 atm. Calculate the increase in entropy which occurs when the partition between the two compartments is removed. If the first compartment had contained 2 moles of ideal gas A, what would have been the increase in entropy when the partition was removed? Calculate the corresponding increases in entropy in each of the above two situations if both compartments contained ideal gas A. ANSWER:

1. A) i) Partition Removed: The volume occupied by 1 mole of A doubles -

∴ ∆S A =

R ln 2

The volume occupied by 1 mole of B doubles -

∴ ∆S B =

R ln 2

∴ ∆S A =

2 R ln 2

∴ ∆STOT =

2 R ln 2

ii) First Compartment - 2 Moles of Ideal Gas A: The volume occupied by 2 mole of A doubles -

The volume occupied occupied by 1 mole of B doubles - ∴ ∆S B ∴ ∆STOT =

=

R ln 2

3R ln 2

iii) Partition Removed – All Ideal Gas A: ∴ ∆S TOT = 0 iv) First Compartment - 2 Moles of Ideal Gas A – All Ideal Gas A: The volume occupied by 2 mole of A doubles - ∴ ∆S = 2 R ln 2 Remove the partition - ∆S = 0 The volume occupied by 3 moles of A - ∴ ∆S = 3R ln ( 2 3) ∴ ∆STOT =

2 R ln 2 + 3R ln ( 2 3) = R ln (32 3 2 27 )

*ANSWER FROM THE BACK OF THE BOOK DOESN’T MAKE A LOT OF SENSE TO ME! B) i) Partition Removed: ∆S =

 ( N A + N B )!   N ! N !  A B  

k ln Ω = k ln 

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∆S =

k ( N A

 ∆S = k  ( N A

+

N B ) ln ( N A

+

NB )

+

N B ) ln ( N A

+

N B ) − N A ln N A − N B ln N B 

N B )

− N A

ln N A

+

N A

− N B

ln N B

+

N B 

  N A   N B   N B    ln  +  ln   N N N N N N + + +  A  A  N A + N B   B  B  B   A  X A   X A   X B   X B   ∆S = − R ( N A + N B )   ln  +  ln   X X X X X X X X + + + + A B A B A B A B          ∆S = − k ( N A +

∴ ∆STOT =



−( N A +

N B ) 

N A

2 R ln 2

ii) First Compartment - 2 Moles of Ideal Gas A: ∆S = − R ( N A + ∴ ∆STOT =



  XA   XB  ln  + + + X X X X B   A B   XA + XB  A

N B ) 

X A

  X B    ln     X A + X B  

3R ln 2

iii) Partition Removed – All Ideal Gas A: ∆S = − R ( N A ) ln ( 1) ∴ ∆S TOT =

0

iv) First Compartment - 2 Moles of Ideal Gas A – All Ideal Gas A: 3 2 27) ∴ ∆STOT = 2 R ln 2 + 3R ln ( 2 3) = R ln ( 32 *ANSWER FROM THE BACK OF THE BOOK DOESN’T MAKE A LOT OF SENSE TO ME! o

2. An adiabatic vessel contains 1000g of liquid aluminum at 700 C. Calculate the mass of Cr2O3 at room temperature, which, when added to the liquid aluminum (with which it reacts to form Cr and Al2O3) raises the temperature of the resulting mixture of Al2O3, Cr2O3, and Cr to 1600K.

ANSWER: 2. Reaction --> 2 Al + Cr2O3

=

2Cr + Al2 O3

** All aluminum reacts with Cr2O3 Thermodynamic Data:







2 Al( S ) 2Cr( S ) Al( S )

3 O 2 2( g ) + 3 O2( g ) 2 +

=

o

J

o

J

=

Al2 O3( S )

∆H 298 = −1, 676, 700

=

Cr2 O3( S )

∆H 298 = −1,134, 700 o

Al( l )

∆H 934 = 10,700

J

Al( S )

c p

=

20.67 + 12.38 x10 10 −3 T J / mole.K

Al(l )

c p

=

31.76 J / mole.K

Al2 O3

c p

10 = 106.6 + 17.78 x10

Cr( S )

c p

=

Cr2 O3

c p

10 = 119.37 + 9.30 x10

−3

T

24.43 + 9.87 x10−3 T −3

T

10 − 28.53 x10 − 3.68 x10

5

5

2

T − J / mole.K 2

T − J / mole.K

10 − 15.65 x10

5

2

T − J / mole.K

Using the thermodynamic data above one can solve for the thermodynamic values for the reaction 2 Al( S ) + Cr2 O3 = 2Cr( S ) + Al2O3 as: o

676, 70 700 + 1,134, 70 700 = −541, 00 000 ∆ H 298 = −1, 67 c p and 2 Al( l ) c p

= −5.25 + 3.46 x10 + Cr2 O3 =

2Cr( S )

= −27.43 +

−3

+

T

− 20.24 x10

5

T −2

J

J / mole.K

Al2O3 as:

28.22 x10 10−3 T

Solving for ∆ H 973K for the reaction 2 Al(l )

5

10 − 20.24 x10 + Cr2 O3 =

T

−2

2Cr( S )

J / mole.K +

Al2O3 :

 3.46 x10−3  2 2 5  1  (934 − 298 ) + 20.24 x10  2  934    28.22 x10−3  1  2 2 5  1 −27.43 ( 973 − 934 ) +  −  (973 − 934 ) + 20.24 x10   2  974 934   

∆ H 973 K = −541, 000 − 5.25

∆ H 973 K = −569,116

(934 − 298 ) + 

J

Heat required to raise temperature of 1mol Cr2O3 to 973K:  9.30 x10−3  1   1 2 2 973 − 298 ) +  10 5  ∆ H 973 K = 119.37 (97 −  (973 − 298 ) + 15.65 x10  2  973 298    ∆ H 973 K = 80,920

J

Therefore, total heat available at 973K is: ∆ H 973 K = −569

116 + 80, 92 920 = -488 197 J

−

1   − 2 (10700 ) 298 





∆ H1600 K

 9.87 x10 −3  1 1  2 2 5  x 1 6 0 0 9 7 3 3 . 6 8 10 1 0 = 24.43 (1600 − 973 ) +  − + − ( )    2  1600 973   

∆ H1600 K =

23,130

J

1mol Al2O3: ∆ H1600 K

 17.78 x10 −3  1 1  2 2 5  = 106.6 (1600 − 973 ) +  −  (1600 − 973 ) + 28.53 x10   2  1600 973   

∆ H1600 K = 80,031

J

1mol Cr2O3 : ∆ H1600 K

 9.30 x10−3  2 1600 − 973 ) +  = 119.37 (16  (1600 2  

− 973

2

10 ) + 15.65 x10

5

 1  1600 

−

1   973 

81, 716 J ∆ H1600 K = 81,

Therefore, the heat available at 1600K to increase the temperature of excess Cr 2O3 is: ∆ H1600 K =

488 488,197 ,197--2*23 2*23,1 ,13 30 − 80,03 80,031 1 = 361,90 61,906 6

J

Number of excess moles of Cr2O3 heated: excess moles of Cr2 O3

=

361, 361, 906 81, 81, 716

=

4.43

5.43moles * 151.9904 g/mol = 825.3 g of Cr2O3 per 53.96g of Al2O3 15.3kg of Cr2O3 per kg of Al2O3





3. Calculate the value of ∆G for the reaction: + 3O2 = 3SiO2 quartz + ( )

Si3 N 4

2N 2

At 800K. What percentage error occurs if it is assumed that

∆c p

for the reaction is zero?

ANSWER: Thermodynamic Data: 3Si( S ) + 2 N 2( g ) = Si3 N 4( S ) Si( S )

+ O2( g ) =

Si SiO2( Quartz) o

N 2( g )

∆S 298 = 191.5

O2( g )

∆S 298 =

Si3 N 4( S )

c p

o

=

o

∆H 298 = −744, 800 o

∆H 298 = −910, 900

o

J

∆S 298 = 113

J

∆S 298 =

o

J / mole.K

41.5 J / mole.K

J / mole.K

205.1 J / mole.K 3

70.54 + 98.74 x10 10 − T J / mole.K 3

5

2

SiO2( Quartze) c p

=

43.89 + 1.00 x10 − T

N 2( g )

c p

=

27.87 + 4.27 x10 10 −3 T J / mole.K

O2( g )

c p

=

29.96 + 4.18 x10 −3 T

− 6.02 x10

10 − 1.67 x10

5

T − J / mole.K

T −2 J / mole.K

Using the thermodynamic data above one can solve for the thermodynamic values for the reaction Si3 N 4 + 3O2 = 3SiO2( quartz ) + 2N 2 as: o

910, 90 900 + 744, 80 800 = −1, 98 987, 90 900 ∆ H 298 = −3 * 91 o

∆S 298 =

c p

=

J

2 *191.5 + 3 * 41.5 − 3 * 205.1 − 113 = −220.8

26.99 − 99.74 x10 10 −3 T

10 − 13.05 x10

5

2

T−

J / K

J / mole.K

Therefore at 800K the free energy can be calculated as follows: ∆ H

o 800

 99.74 x10−3  2 8 00 − 298 ) −  = −1, 987, 900 + 26.99 (80  ( 800 2  

o

∆ H 800 = −2,004,587

o 800

∆S

2

− 298

10 ) + 13.05x10

o

J / K

o

004, 587 − ( −250.5 ) ∆G800 = −2, 00 assum sume

c p

=

0

 1   800

J

 13.05 x105  800  −3 = −220.8 + 26.99 ln   − 99.74 x10 ( 800 − 298) +  2  298  

∆S800 = −250.5

5

800 = −1, 804,187 J

 1  2   800

−

1   2982 

−

1   298 





4. One mole of SiCl4 vapor is contained in a 1atm pressure and 350K in a rigid container of fixed volume. The temperature of the container and its contents is cooled to 280K. At what temperature does condensation of the SiCl4 vapor begin and what fraction of the vapor has condensed when the temperature is 280K? ANSWER: Assume that SiCl4 vapor behaves ideally. Therefore, the volume of the container can be calculated as follows: V

=

nRT

=

0.802 0.802057 057 * 350 1

P

=

28.72l

Cooling of the container and its contents changes the gas pressure as: P=

nRT

=

0.802057 0.802057 * T =

28.72

V

2.857 x10−3 T

The saturated vapor pressure of liquid SiCl4 is: ln p ( atm ) =

3620

T  3620

P = exp 

 T

+ 10.96

 

+ 10.96 

 3620  + 10.96   T 

2.857 x10 10−3 T = exp  T

= 328 K

Therefore condensation will begin at 328K. At 280K the saturated vapor pressure is:  3620  + 10.96  = 0.1396atm P = exp   280  The number of moles of SiCl4 in the vapor phase is therefore: n=

PV RT

=

0.1396* 0.1396* 28.72 28.72 0.802 0.80205 057 7 * 280

= 0.175moles

The number of moles in the liquid phase is therefore = 1- 0.175 = 0.825. 82.5% of the vapor has condensed.







o

5. Below the triple point (-56.2 C) the vapor pressure of solid CO2 is given as: ln p ( atm ) = −

3116 T

+ 16.01

The molar latent heat of melting of CO2 is 8330 joules. Calculate the vapor pressure o exerted by liquid CO2 at 25 C and explain why solid CO2 is referred to as “dry ice”. ANSWER: At 216.8K the triple point pressure is:

 3116  + 16.01 = 5.14atm  216.8 

P = exp  −

For vapor in equilibrium with the solid phase: d ln P

=

3116

dT

T2

=

∆ H s →v ( )

RT 2

,907 ∆ H ( s →v ) = 8.314* 3116 = 25,907 ∆ H ( s →l ) = 8,330

J

25, 90 907 − 8, 33 330 = 17, 57 577 J

∴ ∆ H ( l →v ) =

For liquid CO2: ln (5. 5 .14 ) = −

17,577

+

216. R constant = 11.39

constant

At 298K: ln ( P ) = −

J

17,577

8.31 8.314 4 * 298 298 P = 73.3atm

+ 11.39

Assign 2 Solutions

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