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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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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:
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
.
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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).
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
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,
PERFECTGAS EXPANSION REPORT December 4, 2012
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,
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
APPARATUS AND EQUIPMENT:
Perfect gas expansion apparatus, model TH 11
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
9.
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
SECOND TRIAL
PRESSURE (kPa)TEMPERATU
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
PRESSURE (kPa)TEMPERATU
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
PERFECTGAS EXPANSION REPORT December 4, 2012
FIRST TRIAL
PRESSURE (kPa)TEMPERATU
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
PRESSURE (kPa)TEMPERATU
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
PRESSURE (kPa)TEMPERATU
RE (°C)
101.7 26.5
111.7 28.2
121.7 29.1
131.7 30.3
141.7 31.1
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
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
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
PERFECTGAS EXPANSION REPORT December 4, 2012
According to Boyle’s law: P1V1=P2V2
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
According to Boyle’s law: P1V1=P2V2
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:
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
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)
PERFECTGAS EXPANSION REPORT December 4, 2012
(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
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
(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:
PERFECTGAS EXPANSION REPORT December 4, 2012
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:
PERFECTGAS EXPANSION REPORT December 4, 2012
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
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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.
PERFECTGAS EXPANSION REPORT December 4, 2012
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.