perfect gas expansion

PERFECTGAS EXPANSION REPORT December 4, 2012

ABSTRACT/SUMMARY:

This experiment had been done on 30 th November 2012 in the thermodynamics’ laboratory. The aim of this experiment is to determine the properties of measurement/PVT. The equipment that had been used is called Perfect Gas Expansion and by using this kind of equipment, all 7 experiments were conducted successfully. For the first experiment, we conducted to show the Boyle’s Law and to determine the ratio of volume. In this experiment, the experiment is done for three times from pressurized chamber to vacuum chamber, from atmospheric atmospheric chamber to pressurized chamber and increase increase the gas of both chamber and let it merge by opening the valve no 2,V02. Next, the second experiment is to determine the Gay-Lussac Lawand it also done repeatedly for three times to get the average value of the temperature at pressurize and depressurize vessels. After getting the average value, the graph of pressure vs temperature is plotted. In the third experiment, we need to determine the ratio of heat capacity. Only the pressurised chamber and compressive pump are used during this experiment. The last experiment is to demonstrate the isentropic expansion process. In this experiment, the pressure and the temperature of pressurised chamber is taken before and after the expansion occur. Based on all the experiments that was conducted, all the data which are about the reading before and after the setting are recorded into the data as below.


INTRODUCTION:

Experiment of measurement properties or PVT deals with ideal gas. An ideal gas is a gas that obeys the relationship PV=RT. In this definition P and T are the absolute pressure and absolute temperature respectively and R is the particular gas constant. The particular gas constant depends on the molecular weight of the gas. The perfect gas expansion which allow students familiarize with several fundamental thermodynamic processes can be manipulate by monitored the digital indicator on the control panel. Therefore, this apparatus should not harm students. However, students should take care about their safety during the experiment. The most important thing that student should do is open the valve slowly when releasing the gas inside the vessel to atmosphere because there are high pressure gas inside the vessel that being released by the valve that can be harm to students. The equipment that used is such like below:


Gas particles in a box collide with its walls and transfer momentum to them during each collision. The gas pressure is equal to the momentum delivered to a unit area of a wall, during a unit time. Ideal gas particles do not collide with each other but only with the walls. A single particle moves arbitrarily along some direction until it strikes a wall. It then bounces back, changes direction and speed and moves towards another wall. The gas expansion equations are derived directly from the law of conservation of linear momentum and the law of conservation of energy.

AIMS: For each experiment, they have a different aims and objectives which listed as below: EXPERIMENT 1: - To determine the relationship between pressure and volume of an ideal gas - To compare the experiment result with theoretical result.

EXPERIMENT 2: - To determine the relationship between pressure pressure and temperature temperature of an ideal gas.

EXPERIMENT3: - To demonstrate the isentropic expansion process.

EXPERIMENT 4: - To study the respond of the pressurize vessel following stepwise depressurization.

EXPERIMENT 5: - To study the response of the pressurized vessel following a b rief depressurization.


EXPERIMENT 6: - To determine the ratio and compares it to the theoretical value.

EXPERIMENT 7: - To determine the ratio of heat capacity.

THEORY: Boyle’s law experiment and determination of volume ratio

Boyle's Law states that the product of the pressure and volume for a gas is a constant for a fixed amount of gas at a fixed temperature. Written in mathematical terms, this law is

P x V = constant A common use for this law is to predict on how a change in pressure will alter the volume of gas or vice versa. Therefore, for initial values of p 1 and V1, which change to final values of p2 and V2, the following equation applies

P1 x V1 = P2 x V2 (for fixed amount of gas at constant temperature) The graph shows how the pressure and volume vary according to Boyles Law at two difference temperatures. Then it can be conclude that, the pressure and volume gas is indirectly related which is if the pressure of the chamber is increase then the volume of the gas inside the chamber also decrease. Besides, it also involves the kinetic energy. If we decrease the volume of a gas, thus means that the same number of gas particles are now going to come in contact with each


other and with the sides of the container much more often. The pressure is also measure the frequency of collision of gas particle with each other and with the side of the container they are in. Thus if the volume decrease, the pressure will naturally increase. The opposite is true if the volume of the gas is increased, the gas particles collide less frequently and the pressure will decrease.

.


At lower temperatures the volume and pressure values are lower. Any volume or pressure units can be used as long as both P's and both V's have the same units. The particle theory and simple arithmetical values is used to explain Boyles Law. 

When the volume of gas is compress into half, the collision of the gas will increase and thus the pressure will increase double compare to the origin value.



But if the volume of the gas is doubled or increase in the factor of two, the collision drop and decrease thus the pressure will decrease into half compare to the origin.

Gay-Lussac Law theory Compare to the Boyle’s Law, the expression of Gay -Lussac’s Law is used for each of

the two relationship named after the French chemist Joseph Louis Gay-Lussac (1778-1850) and which concern the properties of gases, though it is more usually applied to his law of combining volumes. One law relates to volumes before and after chemical reaction while the other concerns the pressure and temperature relationship for a sample of gas. According to Gay-Lussac’s law, for a given amount of gas held at constant volume, the pressure is proportional to the absolute temperature. Mathematically,

Where, kG is the appropriate proportionality constant. Besides, Gay-Lussac’ law also tells us that it may be dangerous to heat a gas in a closed container. The increased pressure might cause the container to explode. Therefore, for initial values of p1 and T1, which change to final values of p 2 and T2, the following equation applies


In all calculations, the absolute or Kelvin scale of temperature must be used for T (K = oC + 273).

The graph shows how the pressure and temperature vary according to Gay-Lussac Law. Based on Gay-Lussac it stated that the pressure exerted on a container’s sides by an ideal is proportional to the absolute temperature of the gas. This follows from the kinetic theory which stated that by increasing the temperature of the gas, the molecules ‘speed

increase meaning an increased amount of collisions with the container walls.


Determination of ratio of heat capacity theory For a perfect gas, Cp = Cv + R

Where, Cp = molar heat capacity at constant pressure, and Cv = molar heat capacity at constant volume.

For a real gas a relationship may be defined between the heat capacity, which is dependent on the equation of state, although it is more complex than that for a perfect gas. The heat capacity ratio may then be determined experimentally experimentally using a two step process. 1. An adiabatic reversible expansion from the initial pressure Ps to an intermediate pressure Pi {Ps, Vs, Ts}

{Pi, Vi, Ti}

2. A return of the temperature temperature to its original value T s at constant volume Vi {Pi, Vi, Ti}

{Pf, Vi, Ts}

For a reversible adiabatic expression dq = 0 From the First Law of Thermodynamics, dU = dq + dW Therefore during the expansion process dU = dW

or

dU = -pdV


At constant volume the heat capacity relates the change in temperature to the change in internal energy dU = CvdT Substituting in to equation x, CvdT = -pdV Substituting in the ideal gas law l aw and then integrating gives

 T i   V     R ln i   T s   V s 

C v ln 

Now, for an ideal gas T i T s



P iV i P sV s

Therefore,

 P i

C v  ln

 P s

 ln

V i 

V

   R ln i V s  V s

Rearranging and substituting in from equation x,

ln

P i P s



C p C v

ln

V i V s

During the return of the temperature to the starting value,


V i



P s

P s



V s

P f

Thus,

ln

P i

C p C v

ln

P s P f

Rearranging gives the relationship in its required required form: C p C v



 ln P i ln P s  ln P f ln P s

Isentropic expansion process theory In thermodynamics, an isentropic process or can be called isoentropic process is a process takes place from initiation to completion without an increase or decrease in the entropy of the system. The entropy of the system remains in constant. Entropy is a type of energy (like heat, work, and enthalpy) and is by definition energy which is lost in a process which is characterized by:

ΔS = 0

or

S1 = S2

If a process is both reversible and adiabatic, then it is an isentropic process. An isentropic process is an idealization of an actual process, and serves as a limiting case for an actual process. For adiabatic, there is no transfer of heat energy.


APPARATUS AND EQUIPMENT:

Perfect gas expansion apparatus, model TH 11


PROCEDURES: General start-up 1. The equipments are connected to single phase power supply and the unit is switch on. 2. Then, open all valves and the pressure reading panel. This is to make sure that the chambers are under atmospheric pressure. 3. After that, close all the valves. 4. Next, connect the pipe from compressive port of the pump to pressure chamber or connect the pipe from vacuum port of the pump to vacuum chamber. The connect must not does at the same time. 5. Now, the unit is ready to use. Experiment 1 1. The general start up procedure is performed. Make sure all valve are fully closed. 2. Compressive pump is switch on and allowed the pressure inside the chamber to increase up to about 150kPa. Then, switch off the pump and remove the hose from the chamber. 3. The pressure reading inside the chamber is monitor until the reading stabilizes. 4. The pressure reading for both chambers is recorded before expansion. 5. Open V02 fully and allowed the pressurized air flow into the atmospheric chamber. 6. The pressure reading for both chambers after expansion is recorded. 7. The experiment is repeated under difference condition: a) From atmospheric chamber to vacuum chamber. b) From pressurized chamber to vacuum chamber. 8. Then, calculated the PV value and prove the Boyles’ Law.


Experiment 2 1. Perform the general start up. Make sure all e valves are fully closed. 2. The hose from the compressive pump is connected to pressurized chamber. chamber. 3. The compressive pump is switch on and the temperature for every increment of 10kPa I the chamber is recorded. The pump stop went the pressure PT1 reaches about 160kPa. 4. Then, open valve V 01 and allowed the pressurized air to flow out. Recorded the temperature temperature reading for every decrement of 10kPa. 5. Stop the experiment when the pressure reaches atmospheric pressure. 6. The experiment is repeated for 3 times to get the average value. 7. The graph of the pressure versus temperature Plot.


Experiment 3 1. The general start up is perform make sure all valve are fully closed. 2. The hose form compressive pump is connected to pressurized chamber. 3. The compressive pump is switch on and allowed the pressure inside the chamber to increase until about 160kPa. Then, switch off the pump and remove the hose from the chamber. 4. The pressure reading inside is monitor until it is stabilizes. The pressure reading PT1 and temperature reading TT1 are recorded. 5. Then, open the valve V 01 slightly and allow the air flow out slowly until it reach atmospheric pressure. 6. The pressure of the reading and the temperature reading after the expansion process are recorded. 7. The isentropic expansion process is discussed. Experiment 4 1. Perform the general start up procedures. Make sure all valve are fully closed. 2. The hose is connected from the compressive pump to the pressurized chamber. 3. The compressive pump is swatch on and allowed the pressure inside the chamber to increase until about 160kPa. Then, switch off the pump and remove the hose from the chamber. 4. The pressure reading is monitor until it is stabilizes. Recorded the pressure reading PT1. 5. The valves V 01 is open fully and bring it back to the closed position instantly. Monitor and recorded the pressure reading PT1 until it became stable. 6. Repeated step5 for at least 4 times. 7. The pressure is display on the graph and discuss. Experiment 5 1. Perform the general start up procedure. Make sure all valve is closed.


2. The compressive pump is connected to the pressurized chamber. 3. The compressive pump is switch on and allows the pressure inside the chamber to increase until 160kPa. Then, switch off the pump and remove the hose from the chamber. 4. The reading inside the chamber is monitor until it is stabilizes. The pressure reading PT1 is recorded. 5. Open valve V 01 fully and bring it back to the closed position after few second. Monitor and recorded the pressure reading PT1 until it becomes stable. 6. The pressure reading is display on the graph and discuss. Experiment 6 1. Perform the general start up procedure. Make sure all valve is close 2. The compressive pump is switch on and allows the pressure inside the chamber increase up to 150kPa. Then, switch off the pump and remove the hose from the chamber. 3. The pressure reading inside the chamber is monitor until it stabilizes. 4. The pressure reading for both chambers before the expansion is recorded. 5.

The V 02 is open and allows the pressure air flow into the atmospheric chamber slowly.

6. The pressure reading for both chambers after the expansion is recorded. 7. The experiment procedure is repeated for difference condition a) From atmospheric chamber to vacuum chamber. b) From pressurized chamber to vacuum chamber. 8. Then, the ratio of the volume is calculated and compare with the theoretical value. Experiment 7 1. The general start up is performs. Make sure all valve is fully close. 2. The compressive pump is connected to pressurized chamber.


3. The compressive pump is switch on and allows the pressure inside the chamber to increase until about 160kPa. Then, switch off the pump and remove the hose from the chamber. 4. The pressure reading inside the chamber is monitor until is stabilized. The recorded the pressure reading PT1 and temperature TT1. 5. Open the valve V 01 fully and bring it to close until after a few seconds. Monitor and recorded the reading PT1 and temperature TT1 until it become stable. 6. The ratio of the heat capacity is determines and then compare with the theoretical value.


RESULTS: Experiment 1 CONDITIONS Pressure To Atmosphere

PRESSURE, KPa BEFORE

AFTER

Atmospheric To Vacuum

BEFORE

AFTER

Pressurized To Vacuum

BEFORE

AFTER

PT1= 147.2

TT1= 30.2

PT2= 101.3

TT2=26.6

PT1= 131.5

TT1= 28.6

PT2= 131.4

TT2= 28.5

PT1= 101.6

TT1= 24.9

PT2= 54.0

TT2= 22.9

PT1= 87.0

TT1= 25.9

PT2= 86.3

TT2= 25.7

PT1= 101.6

TT1= 28.7

PT2= 101.8

TT2= 23.7

PT1= 157.3

TT1= 28.2

PT2= 64.7

TT2= 24.6

Experiment 2 -

Increasing pressure

FIRST TRIAL

PRESSURE (kPa)

TEMPERATURE, TEMPERATURE , °C

TEMPERATURE (°C)

101.7

25.4

111.7

26.6

121.7

27.1

131.7

27.8

141.7

29.0

151.7

30.0

161.7

31.0


SECOND TRIAL

PRESSURE (kPa)

TEMPERATURE (°C)

101.6

26.2

111.6

26.3

121.6

26.9

131.6

27.8

141.6

28.6

151.6

29.7

161.6

30.6

THIRD TRIAL

PRESSURE (kPa)

TEMPERATURE (°C)

101.7

26.6

111.7

26.7

121.7

27.2

131.7

28.2

141.7

29.0

151.7

30.1

161.7

31.1


-

Decreasing pressure

FIRST TRIAL

PRESSURE (kPa)

TEMPERATURE (°C)

101.7

26.0

111.7

27.2

121.7

28.0

131.7

28.4

141.7

28.8

151.7

29.0

161.7

29.7

SECOND TRIAL

PRESSURE (kPa)

TEMPERATURE (°C)

101.6

26.6

111.6

28.1

121.6

29.1

131.6

30.1

141.6

31.1

151.6

31.9

161.6

32.2

THIRD TRIAL

PRESSURE (kPa)

TEMPERATURE (°C)

101.7

26.5

111.7

28.2


121.7

29.1

131.7

30.3

141.7

31.1

151.7

31.6

161.7

32.0

Experiment 3 BEFORE

PT1= 101.6 TT1= 26.6

AFTER COMPRESSION

PT1= 164.1

T1

TT1= 29.7

AFTER EXPANSION

PT1= 101.6

T2

TT1= 25.7

Experiment 4 PT1

TT1

INITIAL

101.7

25.9

BEFORE

162.1

28.7

OPEN V1 INSTANTLY

131.6

27.0

2ND OPENING

107.6

26.0

3TH OPENING

103.4

26.3

4TH OPENING

102.0

26.6


Experiment 5 PT1

TT1

INITIAL

101.7

25.8

BEFORE

159.7

31.1

OPEN V1 FOR A FEW SECONDS

113.4

27.7

2NDOPENING

103.5

26.9

3RD OPENING

101.8

26.8

Experiment 6 CONDITIONS Pressure To Atmosphere

BEFORE

AFTER

Atmospheric To Vacuum

BEFORE

AFTER

Pressurized To Vacuum

BEFORE

AFTER

PRESSURE, KPa

TEMPERATURE, TEMPERATURE , °C

PT1= 147.2

TT1= 30.2

PT2= 101.3

TT2=26.6

PT1= 131.5

TT1= 28.6

PT2= 131.4

TT2= 28.5

PT1= 101.6

TT1= 24.9

PT2= 54.0

TT2= 22.9

PT1= 87.0

TT1= 25.9

PT2= 86.3

TT2= 25.7

PT1= 101.6

TT1= 28.7

PT2= 101.8

TT2= 23.7

PT1= 157.3

TT1= 28.2

PT2= 64.7

TT2= 24.6

Experiment 7 CONDITION

PT1 (kPa)

TT1 (°C)

INITIAL

101.8

25.4

BEFORE VALVE OPEN

168.5

30.2

OPEN VALVE FOR 3 SECOND

110.2

27.0


Calculation Experiment 1: Boyle’s law

Ideal gas equation, PV=RT. For Boyle’s law, temperature temperature is constant at room temperature

Hence, R= 8.314 L kPa K-1mol-1, T= 298 @ 25°C i)

From atmospheric chamber to pressurized chamber P1= 147.2kPa, P2= 131.5kPa. Then V1 and V2 is calculated

V1= RT/P1 = (8.314 L kPa K-1mol-1) (298.15 K) / (147.2kPa) =16.83L V2 = (8.314 L kPa K-1mol-1) (298.15 K) / (131.5kPa) =18.84L

According to Boyle’s law: P1V1=P2V2

P1V1= (147.2kPa) (16.83L) = 2477.38L kPa P2V2= (131.5kPa) (18.84L) = 2477.46 L kPa

ii)

From the atmospheric chamber to vacuum chamber P1= 54.0kPa, P2= 87.0kPa. Then V1 and V2 is calculated

V1= RT/P1 = (8.314 L kPa K-1mol-1) (298.15 K) / (54kPa) =45.90L


V2 = (8.314 L kPa K-1mol-1) (298.15 K) / (87.0kPa) =28.49L


P1V1= (54.0kPa) (45.90L) = 2478.60 L kPa P2V2= (87.0kPa) (28.49L) = 2478.63 L kPa

iii)

From pressure chamber to vacuum chamber P1= 101.6kPa, P2= 157.3kPa. Then V1 and V2 is calculated

V1= RT/P1 = (8.314 L kPa K-1mol-1) (298.15 K) / (101.6kPa) =24.39L V2 = (8.314 L kPa K-1mol-1) (298.15 K) / (157.3kPa) =15.76L


P1V1= (101.6kPa) (24.39L) = 2478.02 L kPa P2V2= (157.3kPa) (15.76L) = 2479.05 L kPa


Experiment 2 INCREASING AND DECREASING PRESSURE Trial 1: Increase 180.0 160.0 140.0 120.0 e r u s s e r p

100.0 80.0

gas expansion

60.0 40.0 20.0 0.0 20.0

22.0

24.0

26.0

28.0

30.0

32.0

temperature

Decrease 180.0 160.0 140.0 120.0 e r u s s e r p

100.0 80.0

gas expansion

60.0 40.0 20.0 0.0 20.0

22.0

24.0

26.0

28.0

temperature

30.0

32.0

34.0


Trial 2: Increase 180.0 160.0 140.0 120.0 e r u s s e r p

100.0 80.0

gas expansion

60.0 40.0 20.0 0.0 20.0

22.0

24.0

26.0

28.0

30.0

32.0

temperature


100.0 80.0

gas expansion

60.0 40.0 20.0 0.0 20.0

22.0

24.0

26.0

28.0

temperature

30.0

32.0

34.0


Trial 3: Increase 180.0 160.0 140.0 120.0 e r u s s e r p

100.0 80.0

gas expansion

60.0 40.0 20.0 0.0 20.0

22.0

24.0

26.0

28.0

30.0

32.0

temperature


100.0 80.0

gas expansion

60.0 40.0 20.0 0.0 20.0

22.0

24.0

26.0 temperature

28.0

30.0

32.0


Experiment 3 T2/T1 = (P2 / P1)(k-1 / k) (25.7) / (29.7) = [(101.6) / (164.1)](k-1 / k) 0.8653 = (0.619) (k-1 / k) ln 0.8653 = [ (k-1)/ k] ln l n 0.619 k = 1.4318 Experiment 4 180.0 160.0 140.0 120.0 e r u s s e r p

100.0 80.0

pressure

60.0 40.0 20.0 0.0 26.0

27.0

28.0

29.0 temperature

30.0

31.0

32.0


Experiment 5 180.0 160.0 140.0 120.0 e r u s s e r p

100.0 80.0

gas expansion

60.0 40.0 20.0 0.0 20.0

22.0

24.0

26.0

28.0

30.0

temperature

Experiment 6 (i)From atmospheric chamber to pressurized chamber P1V1 = P2V2 V2/ V1 = P1/ P2 V2/ V1 = 147.2 / 131.5 V2/ V1 =1.119 (ii)From atmospheric chamber to vacuum chamber P1V1 = P2V2 V2/ V1 = P1/ P2

32.0


V2/ V1 = 54.0 / 87.0 V2/ V1 = 0.621 (iii)From pressurized chamber to vacuum chamber P1V1 = P2V2 V2/ V1 = P1/ P2 V2/ V1 = 101.6/ 157.3 V2/ V1 = 0.645 In vacuum chamber: P1V1 = P2V2 V2/ V1 = P1/ P2 V2/ V1 = 64.7 / 101.8 V2/ V1 = 0.636 Theoretical value V 2/ V1 = 15 / 25 = 0.6


Experiment 7 The expression of heat capacity ratio is:

             ][ ] [  [                            Ratio:

      Theoretical value of

 is 1.4 


DISCUSSION:

The pressure of the gas is inversely proportional to the volume it occupies according to Boyle’s law. From the ideal gas equation, PV=RT the volume is calculated for each of the

pressure of the experiment 1. In first condition, the pressurized to the atmospheric the value of volume are V 1=16.83L then expend V2 =18.84L. In the second condition, atmospheric to vacuum the volume are V 1 =45.90L then expend to V2 =28.49L. For the last condition pressurized to vacuum, the reading is taken separately for pressure chamber and vacuum chamber. In pressure chamber, V1= 24.39L before expansion while V2= 15.79L after expansion.. Follow the Boyle’s law P1V1=P2V2. From the calculation, we can see that the

P1V1 is near to the value of P2V2 this prove there are same error happened during the experiment. Hence, we can say that the experiment to prove Boyle’s law is successful. The isentropic expansion process happen went both reversible and adiabatic, there will be no heat transferred within the system, and no energy transformation occurs. Given that,

   Where, k is constant. Given the value of temperature and pressure before and after expansion, we can find the value of k. Thus, the calculated value of k in this experiment is 1.4318. In experiment 4, the Stepwise Depressurization show by the graph above shows the relationship between pressure and temperature. From this graph, we can conclude that the pressure increase accordingly with temperature. Thus, the pressure is directly relation to the temperature. In experiment 5, Brief Depressurization show from this graph, it shows that the gas


expands as the temperature rises. The expansion of gas can only occur when the pressure increase. Since temperature increases, pressure also increases and therefore gas expansion takes place. In determination of ratio of volume, t he Boyle’s law equation can be manipulated to find the volume ratio of gas. From the equation P 1V1 = P 2V2, the volume ratio of gas is then: V2/ V1 = P1/P2. There are also three conditions in this experiment. For the first condition (atmospheric to pressurize) the volume ratio of the gas is 1.119. For second condition (atmospheric to vacuum), the volume ratio is 0.621 while for the third condition (pressurized to vacuum), are 0.645 and 0.636 in pressure chamber and vacuum chamber respectively. The theoretical value for the volume ratio of gas is given as 0.6. Hence, the percentage errors are calculated as follows: 1st condition: Error = (1.119 – 0.6) / 1.119 x 100 = 46.38 % 2nd condition: Error = (0.621 – 0.6) / 0.621 x 100 = 3.38 % 3rd condition: Pressure chamber: Error = (0.645 – 0.6) / 0.645 x 100 = 6.98 % Vacuum chamber: Error = (0.636 – 0.6) / 0.636 x 100 = 5.66 % not valid Since the percentage error is too large (less than 10%), then we conclude that this experiment is successful.


Experiment 7, the determination of ratio of heat capacity using the expression of the heat capacity ratio, the heat capacity ratio is calculated to be 1.032. This value deviated a little from the theoretical value which is 1.4. Hence, the percentage errors calculated are as follows: Percentage error = (theoretical value – actual value) / theoretical value x 100 = (1.4 – 1.032) / 1.4 x 100 = 26.29 % Since the percentage error is too large (more than 10%), this experiment is considered not successful. This is may be because of the error while handling this kind of equipment.


CONCLUSION: In the conclusion, we can concluded that the experiment was to determining the properties measurement/PVT according to Boyle’s Law, Gay-Lussac Law, heat capacity equation and

isentropic expansion process.. Even we make some parallax error we still manage to get the result of what we want such as in experiment one which me manage to prove the Boyle’s

law that is when pressure decrease the volume will increase and vice versa. We also manage to prove the Gay-Lussac law that is pressure is proportional to temperature.In conclusion, this experiment is successfully done and the objective of the experiment is achieved.


RECOMMENDATIONS: There are four experiments must be done under properties measurement/PVT. Each experiment we must do the start-up and shut-down experiment first in order to make sure there are no gas are left in the chamber. We must ovoid the parallax error during taking the reading of pressure and temperature.Repeat the experiment three time to get the average and more accurate result.Open and close the valve carefully according to the procedure given.The experiment should be conducted at the stable and unshaken place. All the data must be recorded into a table.

REFERENCES: Yusus A. Cengel, M. A. (2011). second low of thermodynamics. In Thermodynamics an engineering apploach (pp. 274-309). New York: Mc Graw Hill.

perfect gas expansion

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