perfect gas expansion

PERFECTGAS EXPANSION REPORT December 4, 2012

ABSTRACT/SUMMARY:

This experiment had been done on 30th 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 chamber to pressurized chamber and 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 and 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 brief 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

p1 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

p2 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 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 to its original value Ts 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 and then integrating gives

Cv ln( T iT s )=−R ln(V iV s )

Now, for an ideal gas

T iT s

=PiV iP sV s

Therefore,

Cv( ln PiPs+ lnV iV s )=−R ln

V iV s

Rearranging and substituting in from equation x,

lnP iPs

=−C pCvlnV iV s

During the return of the temperature to the starting value,

V iV s

=P sP f

Thus,


lnPsP i

=C pCvlnP sP f

Rearranging gives the relationship in its required form:

CpCv

=ln Ps−lnPiln Ps−lnP f

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.

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 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


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


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


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.


chamber to increase until about 160kPa. Then, switch off the pump


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.

9.


RESULTS:

Experiment 1

CONDITIONS PRESSURE, KPa TEMPERATURE,

°C

Pressure To Atmosphere BEFORE PT1= 147.2 TT1= 30.2

PT2= 101.3 TT2=26.6

AFTER PT1= 131.5 TT1= 28.6

PT2= 131.4 TT2= 28.5

Atmospheric To Vacuum BEFORE PT1= 101.6 TT1= 24.9

PT2= 54.0 TT2= 22.9

AFTER PT1= 87.0 TT1= 25.9

PT2= 86.3 TT2= 25.7

Pressurized To Vacuum BEFORE PT1= 101.6 TT1= 28.7

PT2= 101.8 TT2= 23.7

AFTER PT1= 157.3 TT1= 28.2

PT2= 64.7 TT2= 24.6

Experiment 2

- Increasing pressure

FIRST TRIAL

PRESSURE (kPa)TEMPERATU

RE (°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


RE (°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


RE (°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


RE (°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


RE (°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


RE (°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, KPaTEMPERATURE,

°C

Pressure To Atmosphere BEFORE PT1= 147.2 TT1= 30.2

PT2= 101.3 TT2=26.6

AFTER PT1= 131.5 TT1= 28.6

PT2= 131.4 TT2= 28.5

Atmospheric To Vacuum BEFORE PT1= 101.6 TT1= 24.9

PT2= 54.0 TT2= 22.9

AFTER PT1= 87.0 TT1= 25.9

PT2= 86.3 TT2= 25.7

Pressurized To Vacuum BEFORE PT1= 101.6 TT1= 28.7

PT2= 101.8 TT2= 23.7

AFTER 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 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


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


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

20.0 22.0 24.0 26.0 28.0 30.0 32.00.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

gas expansion

temperature

pres

sure

Decrease

20.0 22.0 24.0 26.0 28.0 30.0 32.0 34.00.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

gas expansion

temperature

pres

sure

Trial 2:

Increase


20.0 22.0 24.0 26.0 28.0 30.0 32.00.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

gas expansion

temperature

pres

sure

Decrease

20.0 22.0 24.0 26.0 28.0 30.0 32.0 34.00.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

gas expansion

temperature

pres

sure

Trial 3:

Increase


20.0 22.0 24.0 26.0 28.0 30.0 32.00.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

gas expansion

temperature

pres

sure

Decrease

20.0 22.0 24.0 26.0 28.0 30.0 32.00.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

gas expansion

temperature

pres

sure

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 0.619

k = 1.4318

Experiment 4

26.0 27.0 28.0 29.0 30.0 31.0 32.00.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

pressure

temperature

pres

sure

Experiment 5


20.0 22.0 24.0 26.0 28.0 30.0 32.00.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

gas expansion

temperature

pres

sure

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

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:


C vRlnT 2T 1

=−lnV 2

V 1

∴whereV 2

V 1

=P1T 1P2T 2

C v8.314 LkPaK−1mol−1

ln [ 300.85K304.25K ]=−ln [ 159.7kPa (304.25K )113.4 kPa(300.85K ) ]

C v=261.7836 LkPa K−1mol−1

C p=C v+R

¿261.7836 LkPa K−1mol−1+8.314 LkPa K−1mol−1

270.098 LkPaK−1mol−1

Ratio:

CpC v

= 270.098261.7836

=1.032

Theoretical value of CpC v

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

V1=16.83L then expend V2 =18.84L. In the second condition, atmospheric to

vacuum the volume are V1 =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,

pV k=constant

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, the Boyle’s law equation can be

manipulated to find the volume ratio of gas. From the equation P1V1 = P2V2,

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:

1 st condition:

Error = (1.119 – 0.6) / 1.119 x 100

= 46.38 %

2 nd condition:

Error = (0.621 – 0.6) / 0.621 x 100

= 3.38 %

3 rd 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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