1.0 ABSTRACTThe experiment was conducted to determine the properties of measurement In this experiment, the equipment that we used is Perfect Gas Expansion. There are total 7 experiment were conducted. The first experiment is Boyles law experiment. The experiment carried out in three condition which is from atmosphere to pressurize, vacuum to atmosphere and pressurize to vacuum. Next experiment is Gay-Lussac law experiment. For this experiment, the relationship of pressure and temperature obtained by plotting the graph of pressurize and depressurize, this experiment being conducted for three times to get the average value of the temperature at pressurize and depressurize vessels. The 3rd experiment is isentropic expansion process which is to determination value of k. Next is stepwise depressurized experiment, this experiment needed to note all changes values and plotted the graph. The pressure in the graph will drastically dropped when valve was opened. The 5th experiment is brief depressurized. This experiment similar as previous experiment that needed to note every changes of pressure dropped. However, in this experiment, the valve only opened once. Experiment 6 is about ratio of volume using Boyles law equation to get the V2/V1 which is ratio of volume. And lastly, experiment 7 is about ratio of heat capacity. The Cv and Cp was determined in the experiment and percentage error was calculated for experiment 6 and 7.

2.0 INTRODUCTIONGas 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.This experiment is conducted to improved our understanding about First and Second Law of Thermodynamics and relationship of pressure, volume and temperature. The Perfect Gas Expansion Apparatus (Model: TH 11) was used in the experiment. The apparatus contain 2 chambers. First chamber with bigger volume is pressure vessel (PT 1) and another chamber is vacuum vessel (PT 2). Both of the chamber is made up from glass. The apparatus also contain 5 valves and one pressure relief valve if pressure inside the chamber exceed than 2 bar.There are digital indicator to monitor the changes of pressure and temperature. This experiment is about ideal gas or perfect gas that obeys all the laws in this experiment. Ideal gas obeys the PvT relationship. The equation for ideal gas is Pv = nRT. R is gas constant that depend on molar mass of the gas.

3.0 AIMSEXPERIMENT 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.

4.0THEORY4.1 PERFECT GAS First, the theory will start with ideal gas or known as perfect gas. An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and which there are no intermolecular attractive forces. One can visualize it as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other. In such a gas, all theinternal energyis in the form of kinetic energy and any change in internal energy is accompanied by a change intemperature. An ideal gas is an imaginary substance that obeys the ideal gas equation of state. J. Charles and J. Gay-Lussac find out that at low pressure, the volume of gas is proportional to its temperature.

where the constant of proportionality R is called the gas constant and is different for each gas which known as ideal gas equation of state. In this equation, any gas that obeys this equations law called as ideal gas. P is refer to absolute pressure, T is absolute temperature and is specific volume. The ideal gas equation also can be written as V = mvthusPV = mRTWe can relate the both equation for a fixed mass. The properties of ideal gas at two different state can be relate as

The ideal gas relation experimentally observed are approximately the P-v-T behavior of real gases at low density. When low pressure, high temperature, density of gas will decrease and the gas will behave as ideal gas. The ideal gas also obey the following law which is Boyles Law, Charles Law and Gay-Lussacs Law.4.2 BOYLES LAWFirst of all, we start with Boyles law. Boyles law is inversely proportional relationship between the absolute pressure and volume of gas in closed system with constant temperature. The equation of Boyles law is PV = kWhere:P = pressure of the systemV = volume of the gask = constant value representative of the pressure and volume of the systemAt same amount of energy and at constant temperature, the value of k will constant theoretically. However, with volume increase, the pressure must decrease proportionally. In short, the volume decreasing with pressure increasing. The Boyles law equation is to relate the volume and pressure at fixed amount of gas before and after expansion process with constant temperature. P1V1=P2V24.3 CHARLES LAW Next is Charles law. Charles law has stated that at constant pressure, the volume of given mass of ideal gas increases as the absolute temperature increases.

Where :V = volume of the gasT = temperature of the gas (measured in Kelvin)k = constantThe constant k must be maintain during the heating of gas at fixed pressure with volume increasing. In contrast, the volume of gas will decrease in cooling process. The volume increases as the temperature increase.

4.4 GAY-LUSSACS LAWThe pressure of a fixed quantity of gas at constant volume, is directly proportional to its temperature in Kelvin. The equation of Gay-lussacs law is

Where :P = pressure of the gasT = temperature of the gas (measured in Kelvin)k = constant4.5 FIRST LAW OF THERMODYNAMICSFirst law of thermodynamics state that the energy can be neither created nor destroyed. But it can change form. This law also known as the conservation of energy principle. It can be expressed as the net change of total energy in the system is equal to difference between total energy entering and leaving the system during the process. The equation for energy balance as below:Ein Eout = EsystemThe energy change of a system during process involves the energy change of the system at beginning and the end of the process which is:Energy change = energy at final state energy at initial state.Energy is exist as internal, potential, electrical, magnetic kinetic and many more. However, in simple compressible system, the change in total energy of system is the sum of change of energy in form if internal, kinetic and potential energy.

Where : = m (u2 u1) KE= = mg (z2-z1) Therefore energy can be in form of heat, work and mass flow. The boundary system in energy interaction indicate of energy is gained or lost during the process. The energy only involved heat and work in closed system. For the open system, the energy involved all form of energy which is work, heat and mass flow. The internal energy of system increase as the heat transfer increases when the heat into the system meanwhile the energy transfer of system decreases when energy is out as heat from the system. For the work, the energy is that involves rising the piston or rotating shaft. Work transfer into the system increase the energy while work out from system will decrease the energy of system. For the mass flow, the energy increases when mass entering the system and decreases when mass out from system. Ein-Eout = (Qin-Qout) - (Win-Wout) (Emass,in-Emass,out) =

4.6 SPECIFIC HEATSDefinition of specific heat is the energy required to raise the temperature of unit mass of a substance by one degree. In general, the energy depends on how process executed. There are two kinds of specific heat that we are interested in thermodynamics which is specific heat at constant volume cv and specific heat at constant pressure cp. The specific heat at constant volume can be define as energy required to raise temperature of unit mass of substance by one degree at maintain pressure meanwhile specific heat at constant volume can be view as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. Then, the cp is always larger that cv because at constant pressure the system allowed to expand and the energy for expansion must supply to system. From the equation, it shows that the Cv is a measure of the variation of internal energy of a substance with temperature, and Cp is a measure of the variation of enthalpy of a substance with temperature.Cv vCppFrom the equation, it shows that the Cv is a measure of the variation of internal energy of a substance with temperature and Cp is a measure of the variation of enthalpy of a substance with temperature.4.7 INTERNAL ENERGY, ENTHALPY AND SPECIFIC HEATS OF IDEAL GASESWe define an ideal gas as a gas which temperature, pressure and specific volume related byPv=RTIt has been demonstrated mathematically and experimentally by Joule for and ideal gas the internal energy is function of temperature only. u=u(T)In experiment, Joule has submerged two tanks connected with pipe and a valve I water bath. One tank contain air at high pressure and other is evacuated. Then he opened the valve when thermal equilibrium was attained. This is to let air pass from one tank to another tank until the pressure is same. Joule observed that no change in temperature of water bath. He assumed that no heat transfer during the process. And also no work done by or on the system. Because of that, he make a conclusion that internal energy did not change even the pressure and volume has changed. Then he make another conclusion state that internal energy is function of temperature and not function of pressure and specific volume. Using the definition of enthalpy and equation state of ideal gas, h = u + Pv and Pv = RTthen , the equation will become h = u + RTsince R is a constant and u= u(T), the enthalpy of an ideal gas is also a function of temperature only :h = h (T)Therefore, at a given temperature for an ideal gas, u, h, Cv and Cp will have fixed values regardless of the specific volume or pressure. Thus the differential changes in the internal energy and enthalpy of an ideal gas can be expressed as:du = Cv(T)dTdh = Cp(T)dT4.8 DETERMINATION OF THE HEAT CAPACITY RATIOThe heat capacity ratio, k is to determine for air near standard pressure and temperature by two steps process. First, an adiabatic reversible expansion from initial pressure, Pi to an intermediate pressure Pm. Second is a return of temperature to original value, To at constant volume with final pressure Pfk = For an ideal gas, Cp = Cv + RFor a non-ideal gas, a reversible adiabatic expantion dq = 0. According to the first law of thermodynamics, dU = dq + dWDuring the expansion process:dU = dWdU = -PdV (24)The heat capacity related the change in temperature to the change in internal energy when the volume is held constant. dU = CvdTsubstituting CvdT into equation dU = -PdV. Then , CvdT = -PdVBy substitute the equation into the ideal gas law,

Rearrange and substitute the equation,

During the return of the temperature to its initial value, the following relationship is known

Rearranging it to obtain a heat capacity ratio and compare the theoretical value with the experimental heat capacity ratios. Thus;

4.9 DETERMINATION OF RATIO OF VOLUMES USING AN ISOTHERMAL PROCESSThe ratio of volume using an isothermal process can be determined when one pressurized vessel allowed to leak slowly to another vessel of different size. At the end of the process, two vessels are equilibrate and final pressure become constant. The final equilibrium absolute pressure, Pabs , be determined by ideal gas equation. Pabsf = The process is under isothermal process therefore the initial and final temperature are same. By taking the ideal gas equation;m1 = m2 = By combines the equation;Pf =

4.10 STEPWISE DEPRESSURIZATIONStepwise depressurization can be explained by depressurizing the chamber or tank step by step slowly by release the gas. The gas will expand at every instant opened and closed to identify gradual change in pressure and temperature with the contrary decreases with expansion. 4.11 BRIEF DEPRESSURIZATIONBrief depressurization is reduced in terms of time. The time interval increased to a few seconds. Therefore, the effect on the pressure and temperature can be observes which can be compared later. Thus, thegraph should be higher in gradient.

5.0 APPARATUS1. Pressure transmitter1. Pressure relief valve1. Temperature sensor1. Vacuum chamber1. Pressure chamber1. Vacuum pump1. Electrode

6.0 PROCEDURES:6.1 GENERAL START-UP1. The equipment is connected to a single phase power supply and the unit is switched on.2. Then, all valves and the pressure reading panel is opened. After that, all the valves is closed.3. Next, the pipe from compressive port of the pump to pressure chamber is connected or the pipe from vacuum port of the pump to vacuum chamber is connected. Now, the unit is ready to use.

6.2 EXPERIMENT 11. The general start up procedure is performed. All valve is being make sure that is fully closed.2. Compressive pump is switched on and the pressure inside the chamber is allowed to increase up to about 150kPa. Then, the pump is switched off and the hose is removed from the chamber.3. The pressure reading inside the chamber is being monitored until the reading stabilizes.4. The pressure reading for both chambers is recorded before expansion.5. V02 is fully opened and the pressurized air is allowed to 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, the PV value is calculated and the Boyles Law is being proven.6.3 EXPERIMENT 21. The general start up is being performed. All valves is being make sure to fully closed.2. The hose from the compressive pump is connected to pressurized chamber.3. The compressive pump is switched 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, valve V 01 is opened and the pressurized air is allowed to flow out. The temperature reading for every decrement of 10kPa is being recorded.5. The experiment is stopped 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 is plotted.6.4 EXPERIMENT 31. The general start up is performed and all valve is being make sure to fully closed.2. The hose form compressive pump is connected to pressurized chamber.3. The compressive pump is switched on and allowed the pressure inside the chamber to increase until about 160kPa. Then, the pump is switched off and the hose is removed from the chamber.4. The pressure reading inside is monitored until it is stabilizes. The pressure reading PT1 and temperature reading TT1 are recorded.5. Then, the valve V 01 slightly opened and the air is allowed to 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.6.5 EXPERIMENT 41. The general start up procedures is performed. All valve are make sure to fully closed.2. The hose is connected from the compressive pump to the pressurized chamber.3. The compressive pump is switched on and allowed the pressure inside the chamber to increase until about 160kPa. Then, the pump is switched off and the hose is removed from the chamber.4. The pressure reading is monitored until it is stabilizes. Recorded the pressure reading PT1.5. The valves V 01 is fully opened and bring it back to the closed position instantly. The pressure reading PT1 is monitored and recorded until it became stable.6. Step5 is repeated for at least 4 times.7. The pressure is plotted on the graph and being discussed.6.6 EXPERIMENT 51. The general start up procedure is performed. 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, the pump is switched off and the hose is removed from the chamber.4. The reading inside the chamber is monitored until it is stabilizes. The pressure reading PT1 is recorded.5. Valve V 01 is fully opened and bring it back to the closed position after few second. The pressure reading PT1 is recorded and monitored until it becomes stable.6. The pressure reading is display on the graph and discuss.

6.7 EXPERIMENT 61. The general start up procedure is performed. Make sure all valve is close2. The compressive pump is switched on and the pressure inside the chamber is allowed increase up to 150kPa. Then, switched off the pump and the hose is removed from the chamber.3. The pressure reading inside the chamber is monitored until it stabilizes.4. The pressure reading for both chambers before the expansion is recorded.5. The V 02 is opened and the pressure air is allowed 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 conditiona) 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.6.8 EXPERIMENT 71. The general start up is performed. Make sure all valve is fully close.2. The compressive pump is connected to pressurized chamber.3. The compressive pump is switched on and the pressure inside the chamber allowed 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 monitored until is stabilized. The pressure reading PT1 and temperature TT1 is recorded.5. The valve V 01 is fully opened and bring it to close until after a few seconds. The reading PT1 and temperature TT1 is monitored and recorded until it become stable.6. The ratio of the heat capacity is determined and then it being compared with the theoretical value.9.

7.0 RESULTS7.1 EXPERIMENT 1: BOYLES LAW EXPERIMENTPressure (kPa abs)Before ExpansionAfter Expansion

Pressure to Atmospherea) PT 1 = 153.8b) PT 2 = 102.8a) PT 1 = 136.7b) PT 2 = 136.7

Atmospheric to Vacuuma) PT 1 = 106.1b) PT 2 = 59.0a) PT 1 = 90.5 b) PT 2 = 90.0

Pressurized to Vacuuma) PT 1 = 154.1b) PT 2 = 56.9a) PT 1 = 122.2b) PT 2 = 121.8

7.2 EXPERIMENT 2: GAY-LUSSAC LAW EXPERIMENTPressure (kPa abs)Trial 1Trial 2Trial 3

Temperature(oC)Temperature(oC)Temperature(oC)

Pressurise vesselDepressurise vesselPressurise vesselDepressurise vesselPressurise vesselDepressurise vessel

11025.324.624.128.125.825.1

12025.724.824.528.926.125.7

13026.624.825.229.326.926.8

14027.525.226.229.627.728.2

15028.325.727.329.628.629.6

16029.229.028.329.729.430.3

Pressure (kPa abs)Average reading temperature

Temperature (oC)

Pressurise vesselDepressurise vessel

11025.125.9

12025.426.5

13026.227.0

14027.128.1

15028.128.3

16029.029.7

7.3 EXPERIMENT 3: ISENTROPIC EXPANSION PROCESSBefore expansionAfter expansion

PT 1 (kPa abs)161.5103.3

TT 1 (oC)28.924.8

7.4 EXPERIMENT 4: STEPWISE DEPRESSURIZATIONPT 1 (kPa abs)

InitialAfter first expansionAfter second expansionAfter third expansionAfter fourth expansion

159.1122.9115.6108.1103.5

123.7115.7108.2103.6

124.8115.8108.4103.7

125.0115.9108.5103.8

125.2116.0108.6103.9

125.3116.1108.7104.0

125.5116.2108.8104.1

125.6116.3108.9104.2

125.7116.4109.0104.3

125.8116.5109.1104.4

125.9116.6109.2104.5

126.0116.7109.3104.6

126.5116.8109.4104.7

126.6116.9109.5104.8

126.7117.0109.6104.9

126.8117.1109.7105.0

126.9117.2109.8105.1

117.3110.0

117.4

7.5 EXPERIMENT 5: BRIEF DEPRESSURIZATIONPT 1 (kPa abs)

InitialAfter brief expansion

161.7105.3

106.5

107.1

107.9

108.4

108.7

109.0

109.3

109.5

109.6

109.7

109.8

109.9

110.0

110.1

110.2

110.3

110.4

110.5

7.6 EXPERIMENT 6: DETERMINATION OF RATIO OF VOLUMEPressure (kPa abs)Before ExpansionAfter Expansion

Pressure to Atmospherea) PT 1 = 150.0b) PT 2 = 102.7a) PT 1 = 133.8b) PT 2 = 133.4

Atmospheric to Vacuuma) PT 1 = 103.5b) PT 2 = 58.3a) PT 1 = 88.8b) PT 2 = 88.4

Pressurized to Vacuuma) PT 1 = 150.7b) PT 2 = 60.0a) PT 1 = 120.6b) PT 2 = 120.2

7.7 EXPERIMENT 7: DETERMINATION OF RATIO OF HEAT CAPAITYInitialIntermediateFinal

PT 1 (kPa abs)159.1103.5110.1

TT 1 (oC)29.428.526.5

8.0 CALCULATION8.1 EXPERIMENT 1Ideal gas equation, PV=RT. For Boyles law, temperature is constant at room temperatureHence, R= 8.314 L kPa K-1mol-1, T= 298.15 @ 25Ci) From pressurized chamber to atmospheric chambera) For PT 1P1= 153.8 kPa, P2= 136.7 kPa. Then V1 and V2 is calculated

V1= RT/P1= (8.314 L kPa K-1mol-1) (298.15 K) / (153.8 kPa)V1 =10.71 LV2 = (8.314 L kPa K-1mol-1) (298.15 K) / (136.7 kPa)V2 =12.05 LAccording to Boyles law: P1V1=P2V2P1V1= (153.8 kPa) (10.71 L) = 1647.20 kPa.LP2V2= (136.7 kPa) (12.05 L) = 1647.24 kPa.LP1V1 P2V2 (proved)

b) For PT 2

P1= 102.8 kPa, P2= 136.7 kPa. Then V1 and V2 is calculatedV1= RT/P1= (8.314 L kPa K-1mol-1) (298.15 K) / (102.8 kPa)V1 =16.03 LV2 = (8.314 L kPa K-1mol-1) (298.15 K) / (136.7 kPa)V2 =12.05 LAccording to Boyles law: P1V1=P2V2P1V1= (102.8 kPa) (16.03 L) = 1647.88 kPa.LP2V2= (136.7 kPa) (12.05 L) = 1647.24 kPa.LP1V1 P2V2 (proved)

ii) From the atmospheric chamber to vacuum chambera) For PT 1P1= 106.1 kPa, P2= 90.5 kPa. Then V1 and V2 is calculated

V1= RT/P1= (8.314 L kPa K-1mol-1) (298.15 K) / (106.1 kPa)V1 =23.36 LV2 = (8.314 L kPa K-1mol-1) (298.15 K) / (90.5 kPa)V2 =27.39 LAccording to Boyles law: P1V1=P2V2P1V1= (106.1 kPa) (23.36 L) = 2478.50 kPa.LP2V2= (90.5 kPa) (27.39 L) = 2478.80 kPa.LP1V1 P2V2 (proved)

b) For PT 2P1= 59.0 kPa, P2= 90.0 kPa. Then V1 and V2 is calculatedV1= RT/P1= (8.314 L kPa K-1mol-1) (298.15 K) / (59.0 kPa)V1 =42.01 LV2 = (8.314 L kPa K-1mol-1) (298.15 K) / (90.0 kPa)V2 =27.54 LAccording to Boyles law: P1V1=P2V2P1V1= (59.0 kPa) (42.01 L) = 2478.59 kPa.LP2V2= (90.0 kPa) (27.54 L) = 2478.60 kPa.LP1V1P2V2 (proved)

iii) From pressure chamber to vacuum chambera) For PT 1P1= 154.1 kPa, P2= 122.1 kPa. Then V1 and V2 is calculatedV1= RT/P1= (8.314 L kPa K-1mol-1) (298.15 K) / (154.1 kPa)V1 =16.09 LV2 = (8.314 L kPa K-1mol-1) (298.15 K) / (122.1 kPa)V2 =20.30 LAccording to Boyles law: P1V1=P2V2P1V1= (154.1 kPa) (16.09 L) = 2479.47 kPa.LP2V2= (122.1 kPa) (20.30 L) = 2478.63 kPa.LP1V1 P2V2 (proved)

b) For PT 2P1= 56.9 kPa, P2= 121.8 kPa. Then V1 and V2 is calculated

V1= RT/P1= (8.314 L kPa K-1mol-1) (298.15 K) / (56.9 kPa)V1 = 43.56 LV2 = (8.314 L kPa K-1mol-1) (298.15 K) / (121.8 kPa)V2 =20.35 LAccording to Boyles law: P1V1=P2V2P1V1= (56.9 kPa) (43.56 L) = 2478.56 kPa.LP2V2= (121.8 kPa) (20.35 L) = 2478.63 kPa.LP1V1 P2V2 (proved)

8.2 EXPERIMENT 2

8.3 EXPERIMENT 3

T2/T1 = (P2 / P1)(k-1 / k)(24.8) / (28.9) = [(103.3) / (161.5)](k-1 / k)0.8581 = (0.6396) (k-1 / k)ln 0.8581 = [ (k-1)/ k] ln 0.6396k = 1.5207

8.4 EXPERIMENT 4

8.5 EXPERIMENT 58.6 EXPERIMENT 6Theoretical value of ratio of volume,V2/V1 = 12.37/25= 0.495Percentage error = [(Theoretical value - Actual value) / Actual value] 1001. From pressurize chamber to atmospheric chambera) PT 1 P1V1 = P2V2V2/V1 = P1/P2V2/V1 = 150.0/133.8V2/V1 = 1.121Percentage error = [(0.495-1.121)/1.121] 100 = -55.84% = 55.84%

b) PT 2P1V1 = P2V2V2/V1 = P1/P2V2/V1 = 102.7/133.4V2/V1 = 0.770Percentage error = (0.495-0.770)/0.770 = -35.71% = 35.71%

1. From atmospheric chamber to vacuum chambera) PT 1P1V1 = P2V2V2/V1 = P1/P2V2/V1 = 103.5/88.8V2/V1 = 1.166Percentage error = (1.166-0.495)/1.166 = 57.55%

b) PT 2P1V1 = P2V2V2/V1 = P1/P2V2/V1 = 58.3/88.4V2/V1 = 0.660Percentage error = (0.660-0.495)/0.660 = 25.00%

1. From pressurize chamber to vacuum chambera) PT 1P1V1 = P2V2V2/V1 = P1/P2V2/V1 = 150.7/120.6V2/V1 = 1.250Percentage error = (1.250-0.495)/1.250 = 60.40%

b) PT 2P1V1 = P2V2V2/V1 = P1/P2V2/V1 = 60.0/120.2V2/V1 = 0.499Percentage error = (0.499-0.495)/0.499 = 0.80%

8.7 EXPERIMENT 7Calculate the value of heat capacity ratio, by the given formula of Cv,The expression of heat capacity ratio is: = = = 1.1679The theoretical value of is 1.4The percentage error = [(Theoretical value Actual value) / Actual value] 100= [(1.4 1.1679) / 1.1679] 100= 19.87 %

9.0 DISCUSSIONThis experiment involved First Law of Thermodynamics, Second Law of Thermodynamics and relationship between P-v-T. The experiment also involve several law such Boyles law, Gay-Lussacs law and Charles law. For the experiment 1, Boyles Law experiment. Boyles law state that absolute pressure and volume of given mass are inversely proportional with constant temperature. The relationship of Boyles law can be expressed as P1V1=P2V2. For from pressurized chamber to atmospheric chamber, the initial pressure is 13.8kPa and after expansion 136.7kPa for PT 1 and for PT 2 is 102.8kPa for initial and 136.7kPa for after expansion. The volume is calculated by using equation PV=RT. Volume for V1 is 10.71 L, V2 is 12.05 L and V1 is 16.03L, V2 is 12.05L for PT 1 and PT 2 respectively. And for from atmospheric chamber to vacuum chamber, the initial pressure is 106.1kPa and after expansion is 90.5kPa for PT 1 and for PT 2 is 59.0kPa for initial and 90.0kPa for after expansion. The V1 is 23.36 L, V2 is 27.39L and V1 is 42.01L, V2 is 27.54L for PT 1 and PT 2 respectively. And lastly for from pressure chamber to vacuum chamber, the pressure before expansion for PT 1 is 154.1 kPa and 122.1 kPa after expansion for PT 1 and for PT 2 is 56.9kPa for initial and 121.8kPa for after expansion. V1 is 16.09 L, V2 is 20.30 L and V1 is 43.56L, V2 is 20.35L for PT 1 and PT 2 respectively. The Boyles law was proven with equation of P1V1=P2V2. The value obtain from both side is approximately equal.Gay-Lussacs Law stated that the pressure is directly proportional to the temperature which is means if the pressure increase the temperature also increase with constant volume. Depressurize means reduction of air pressure in vessel or procedure that allow air to flow out. The experiment was repeated thrice so that we can get the average reading of pressure. From the data recorded and graph plotted, it can be said that the Gay-Lussacs Law is verified.Experiment 3 is about Isentropic Expansion Process. This experiment determine the ratio of heat capacity which is k. If compression or expansion of gas take place with no flow of heat energy either into or out, the process called as isentropic or adiabatic process. The k is the ratio of both type specific heat capacity which is cp/cv. The equation of isentropic also as pvk=constant. No heat is added to the flow and no energy form due to friction or dissipative effects. From the result, the pressure decrease proportionally to the temperature. This is due to air flow out from the chamber. The value for k in this experiment is 1.5207 calculating by using T2/T1 = (P2 / P1)(k-1 / k) equation. Which is P is absolute pressure and T is absolute temperature.Stepwise depressurization is a strategy to adopt an equal time-stepwise depressurization approach. In this study yield more reliable result for an example in the production sector in industries. The molecule in the chamber affected when the number of the decreasing slowly as they do not have to collide between them more often. The depressurization shown that pressure decrease with time and also affecting the temperature. For experiment 5, brief depressurization shown in graph plotted in result which is decrease linearly compared to stepwise. The expansion occur when the pressure of gas increase. Expansion of gas decrease as the gas is free to flow out time by time.Next is experiment 6 which is determination the ratio of volume. By using Boyles law equation, P1V1=P2V2the ratio of volume is solved. After arranging the equation, the ratio of volume is V2/V1=P1/P2. This experiment carried out in three condition which is from pressurized chamber to atmospheric chamber, from vacuum chamber to atmospheric chamber and lastly from pressurized chamber to vacuum chamber. The theoretically value ratio of volume also can be determine which is 0.495. For first condition (pressurize to atmosphere), the ratio of volume is 1.121 and the percentage error is 55.84% for PT 1, while for PT 2 the ratio of volume is 0.770 and percentage error is 35.71%. Second condition (vacuum to atmosphere), the ratio of volume is 1.166 with 57.55% percentage error for PT 1 and for PT 2 the ratio of volume is 0.660 and the percentage error is 25%. And lastly, the last condition for PT 1 the ratio of volume is 1.250 with 60.40% percentage error and for PT 2 is 0.499 with 0.80% percentage error. The percentage error is high due to some error during conducting the experiment. Some of air probably left from chamber due to not properly close the valve or before the experiment, the gas did not left out completely from the chamber.From the result the determination of ratio of heat capacity is 1.1679. The theoretical value is 1.4. The deviation is 19.87%. The deviation is due to the measurement error. The actual intermediate pressure supposed to be lowered and from the data obtained the intermediate pressure is lowered that the initial and final reading. Since the percentage difference is more than 10% the experiment can be declare as failed.

10.0 CONCLUSIONIn the conclusion, the experiment was to determine the properties of measurement /PVT according to the Boyles law, Gay-Lussacs Law, isentropic expansion, and heat capacity equation. We have proven the Boyles law and Gay-Lussacs law based on their law. Although there is fail experiment but we have the reason behind the failure. For experiment 7, the failure is due to the intermediate pressure not be taken after the valve is closed. Finally, the experiment is successfully done and the objective of the experiment is achieved.

11.0 RECOMMENDATIONSThere are several recommendation has to be taken during conduct this experiment. First during experiment, we must always concentrate observed pressure reading whether the pressure is exceed 200 kPa and the change of pressure and temperature. If the pressure is too high inside the chamber, the glass probably will break even though there are pressure relief valve. Pressure relief valve function as release pressure when the pressure inside the chamber very high. Next is the procedure of general start-up and shut-down must properly followed so that no gas left inside the chambers. And lastly for experiment 2, the average value must be taken by repeated the experiment three times. So that the result become more accurate. Lastly, for safety student has to wear google because of possibility for glass vessel to break.

12.0 APPENDIX1. Irfan, M. H. (2013). The Perfect Gas Expansion Experiment (TH11). Muhammad Haidharul Irfan