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OBJECTIVES I.
Experiment 1 : Boyle’s Law To determine the relationship between pressure and volume of an ideal gas. To compare the experimental results with theoretical results.
II.
Experiment 2 : Gay-Lussac To determine the relationship between pressure and temperature of an ideal gas.
III.
Experiment 3 : Isentropic Expansion To demonstrate the isentropic expansion process.
IV.
Experiment 5: Brief Depressurization To study the response of the pressurized vessel following a brief depressurization.
V.
Experiment 7 : Determination of Ratio of Heat Capacity To determine the ratio of heat capacity.
DATA AND RESULT
I.
Experiment 1 a. Experiment 1.1 (Pressure to Atmosphere) Before expansion
After expansion
PT 1 (kPa abs)
151.5
134.7
PT 2 (kPa abs)
102.9
133.8
b. Experiment 1.2 (Atmospheric to Vacuum) Before expansion
After expansion
PT 1 (kPa abs)
106.9
91.1
PT 2 (kPa abs)
57.5
90.1
c. Experiment 1.3 ( Pressure to Vacuum) Before expansion
After expansion
PT 1 (kPa abs)
153.3
122.8
PT 2 (kPa abs)
60.7
121.7
II.
Experiment 2
Pressure (Kpa Abs)
III.
Trial 1
Trial 2
Trial 3
Temperature (°C)
Temperature (°C)
Temperature (°C)
Pressurized Vessel
Depressurized Vessel
Pressurize d Vessel
Depressurized Vessel
Pressurized Vessel
Depressurized Vessel
110
30
29.6
29.4
29.6
29.5
29.6
120
30.2
30.2
29.9
30.4
29.7
30.2
130
30.9
31.2
30.4
31.5
30.5
31.2
140
31.8
32.8
31.5
32.9
31.3
32.4
150
32.8
32.3
32.4
34.2
32.2
33.9
160
33.7
35.8
33.3
35.8
32.8
34.2
Experiment 3 Before Expansion
After Expansion
PT 1 (kPa abs)
160
34.9
TT 1 ( ͦC)
104
29.7
IV.
Experiment 5 PT 1 (kPa abs)
V.
Initial
After Brief Expansion
104.8
31.0
106.3
30.5
107.6
29.8
108.7
29.5
109.5
29.4
110.1
29.3
110.7
29.3
111.0
29.4
111.5
29.7
Experiment 7 Initial
Intermediate
Final
PT 1 (kPa abs)
158.4
104.8
111.5
TT 1 ( ͦC)
31.4
31.0
29.7
CALCULATIONS I.
EXPERIMENT 1: BOYLE’S LAW Ideal gas equation PV=RT. For Boyle’s Law, temperature is constant at room temperature. Where, R= 8.314 L kPa K-1mol-1, T= 25 @ 298 a) From pressurized chamber to atmospheric chamber P1= 151.5 kPa, P2= 134.7 kPa, V1 and V2 are calculated: V1= 𝑅𝑇/𝑃1 = (8.314 𝑥 298)/151.5 = 16.354 L V2= 𝑅𝑇/𝑃2 = (8.314 𝑥 298)/134.7 =18.393 L According to Boyle’s Law, P1V1=P2V2 P1V1= (151.5 kPa)(16.354L) = 2477.631 L kPa P2V2= (134.7 kPa)(18.393L) = 2477.537 L kPa
b) From atmospheric chamber to vacuum chamber P1= 106.9 kPa, P2= 91.1 kPa, V1 and V2 are calculated: V1= 𝑅𝑇/𝑃1 = (8.314 𝑥 298)/106.9 = 23.177 L V2= 𝑅𝑇/𝑃2 = (8.314 𝑥 298)/91.1 =27.196 L According to Boyle’s Law, P1V1=P2V2 P1V1= (106.9 kPa)(23.177) = 2477.621 L kPa P2V2= (91.1 kPa)(27.196L) = 2477.556 L kPa c) From pressurized chamber to vacuum chamber P1= 153.38 kPa, P2= 122.8kPa, V1 and V2 are calculated: V1= 𝑅𝑇/𝑃1 = (8.314 𝑥 298)/153.33 = 16.158 L V2= 𝑅𝑇/𝑃2 = (8.314 𝑥 298)/122.8 =20.176 L According to Boyle’s Law, P1V1=P2V2 P1V1= (153.38 kPa)(16.158L) = 2478.314 L kPa P2V2= (122.80 kPa)(20.176L) = 2477.61 L kPa
EXPERIMENT 2: GAY-LUSSAC LAW Average trial 1,2,3 Pressure, kPa abs 110
Temperature, °C Pressurise Vessel 29.63
Temperature, °C Depressurise Vessel 29.60
120
29.93
30.27
130
30.60
31.33
140
31.53
32.70
150
32.47
34.13
160
33.40
35.27
Pressure vs Temperature
180
Pressure,kPa abs
160 140 120 100 80 pressurized vessel
60
40 20 0 29
Pressure,kPa abs
II.
30
31 32 Temperature,°C
33
34
Pressure vs Temperature
180 160 140 120 100 80 60 40 20 0
Depressurized Vessel
29
30
31
32 33 Temperature,°C
34
35
36
III.
EXPERIMENT 3 : ISENTROPIC EXPANSION
𝑇2⁄𝑇1 = (𝑃2⁄𝑃1)(𝑘−1⁄𝑘) (29.7)/(34.9) = [(104)/(160)](𝑘−1⁄𝑘) 0.851 = (0.65)(𝑘−1⁄𝑘) In 0.851 = [(𝑘 − 1⁄𝑘)] In 0.65 k = 1.5988
IV.
EXPERIMENT 5: BRIEF DEPPRESURIZATION
Responses of pressurization vessel following of brief depressurization 112
Pressure,kPa abs
111 110
Responses of pressurization vessel following of brief depressurization
109 108 107 106 105 104 29
29.5
30
30.5
Temperature,°C
31
31.5
V.
EXPERIMENT 7: DETERMINATION OF RATIO OF HEAT CAPACITY
Calculate the value of heat capacity ratio, by the given formula of Cv:The expression of heat capacity ratio is:
𝐶𝑣 𝑅
𝐼𝑛
; 𝑤ℎ𝑒𝑟𝑒 𝐶𝑉 8.314 𝐿 𝑘𝑝𝐴
𝐾 −1 𝑚𝑜𝑙 −1
𝐼𝑛 [
𝑇2 𝑇1
= −𝐼𝑛
𝑉2 𝑉1
=
302.85 304.55
𝑉2 𝑉1
𝑃1𝑇1 𝑃2𝑇2
] = −𝐼𝑛[
158.4𝑘𝑃𝑎(304.55) 111.5(302.85)
]
Cv = 529.79 𝐿 𝑘𝑝𝐴 𝐾 −1 𝑚𝑜𝑙 −1
Cp= Cv +R = 529.79 𝐿 𝑘𝑝𝐴 𝐾 −1 𝑚𝑜𝑙 −1+ 8.314 𝐿 𝑘𝑝𝐴 𝐾 −1 𝑚𝑜𝑙 −1 = 538.10 𝐿 𝑘𝑝𝐴 𝐾 −1 𝑚𝑜𝑙 −1
Therefore the ratio of: 𝑐𝑝 𝑐𝑣
𝑐𝑝
=
538.10 529.79
The theoretical value of 𝑐𝑣 is 1.4
= 1.016
Experiment 1 : Boyle’s Law To determine the relationship between pressure and volume of an ideal gas. To compare the experimental results with theoretical results.
II.
Experiment 2 : Gay-Lussac To determine the relationship between pressure and temperature of an ideal gas.
III.
Experiment 3 : Isentropic Expansion To demonstrate the isentropic expansion process.
IV.
Experiment 5: Brief Depressurization To study the response of the pressurized vessel following a brief depressurization.
V.
Experiment 7 : Determination of Ratio of Heat Capacity To determine the ratio of heat capacity.
DATA AND RESULT
I.
Experiment 1 a. Experiment 1.1 (Pressure to Atmosphere) Before expansion
After expansion
PT 1 (kPa abs)
151.5
134.7
PT 2 (kPa abs)
102.9
133.8
b. Experiment 1.2 (Atmospheric to Vacuum) Before expansion
After expansion
PT 1 (kPa abs)
106.9
91.1
PT 2 (kPa abs)
57.5
90.1
c. Experiment 1.3 ( Pressure to Vacuum) Before expansion
After expansion
PT 1 (kPa abs)
153.3
122.8
PT 2 (kPa abs)
60.7
121.7
II.
Experiment 2
Pressure (Kpa Abs)
III.
Trial 1
Trial 2
Trial 3
Temperature (°C)
Temperature (°C)
Temperature (°C)
Pressurized Vessel
Depressurized Vessel
Pressurize d Vessel
Depressurized Vessel
Pressurized Vessel
Depressurized Vessel
110
30
29.6
29.4
29.6
29.5
29.6
120
30.2
30.2
29.9
30.4
29.7
30.2
130
30.9
31.2
30.4
31.5
30.5
31.2
140
31.8
32.8
31.5
32.9
31.3
32.4
150
32.8
32.3
32.4
34.2
32.2
33.9
160
33.7
35.8
33.3
35.8
32.8
34.2
Experiment 3 Before Expansion
After Expansion
PT 1 (kPa abs)
160
34.9
TT 1 ( ͦC)
104
29.7
IV.
Experiment 5 PT 1 (kPa abs)
V.
Initial
After Brief Expansion
104.8
31.0
106.3
30.5
107.6
29.8
108.7
29.5
109.5
29.4
110.1
29.3
110.7
29.3
111.0
29.4
111.5
29.7
Experiment 7 Initial
Intermediate
Final
PT 1 (kPa abs)
158.4
104.8
111.5
TT 1 ( ͦC)
31.4
31.0
29.7
CALCULATIONS I.
EXPERIMENT 1: BOYLE’S LAW Ideal gas equation PV=RT. For Boyle’s Law, temperature is constant at room temperature. Where, R= 8.314 L kPa K-1mol-1, T= 25 @ 298 a) From pressurized chamber to atmospheric chamber P1= 151.5 kPa, P2= 134.7 kPa, V1 and V2 are calculated: V1= 𝑅𝑇/𝑃1 = (8.314 𝑥 298)/151.5 = 16.354 L V2= 𝑅𝑇/𝑃2 = (8.314 𝑥 298)/134.7 =18.393 L According to Boyle’s Law, P1V1=P2V2 P1V1= (151.5 kPa)(16.354L) = 2477.631 L kPa P2V2= (134.7 kPa)(18.393L) = 2477.537 L kPa
b) From atmospheric chamber to vacuum chamber P1= 106.9 kPa, P2= 91.1 kPa, V1 and V2 are calculated: V1= 𝑅𝑇/𝑃1 = (8.314 𝑥 298)/106.9 = 23.177 L V2= 𝑅𝑇/𝑃2 = (8.314 𝑥 298)/91.1 =27.196 L According to Boyle’s Law, P1V1=P2V2 P1V1= (106.9 kPa)(23.177) = 2477.621 L kPa P2V2= (91.1 kPa)(27.196L) = 2477.556 L kPa c) From pressurized chamber to vacuum chamber P1= 153.38 kPa, P2= 122.8kPa, V1 and V2 are calculated: V1= 𝑅𝑇/𝑃1 = (8.314 𝑥 298)/153.33 = 16.158 L V2= 𝑅𝑇/𝑃2 = (8.314 𝑥 298)/122.8 =20.176 L According to Boyle’s Law, P1V1=P2V2 P1V1= (153.38 kPa)(16.158L) = 2478.314 L kPa P2V2= (122.80 kPa)(20.176L) = 2477.61 L kPa
EXPERIMENT 2: GAY-LUSSAC LAW Average trial 1,2,3 Pressure, kPa abs 110
Temperature, °C Pressurise Vessel 29.63
Temperature, °C Depressurise Vessel 29.60
120
29.93
30.27
130
30.60
31.33
140
31.53
32.70
150
32.47
34.13
160
33.40
35.27
Pressure vs Temperature
180
Pressure,kPa abs
160 140 120 100 80 pressurized vessel
60
40 20 0 29
Pressure,kPa abs
II.
30
31 32 Temperature,°C
33
34
Pressure vs Temperature
180 160 140 120 100 80 60 40 20 0
Depressurized Vessel
29
30
31
32 33 Temperature,°C
34
35
36
III.
EXPERIMENT 3 : ISENTROPIC EXPANSION
𝑇2⁄𝑇1 = (𝑃2⁄𝑃1)(𝑘−1⁄𝑘) (29.7)/(34.9) = [(104)/(160)](𝑘−1⁄𝑘) 0.851 = (0.65)(𝑘−1⁄𝑘) In 0.851 = [(𝑘 − 1⁄𝑘)] In 0.65 k = 1.5988
IV.
EXPERIMENT 5: BRIEF DEPPRESURIZATION
Responses of pressurization vessel following of brief depressurization 112
Pressure,kPa abs
111 110
Responses of pressurization vessel following of brief depressurization
109 108 107 106 105 104 29
29.5
30
30.5
Temperature,°C
31
31.5
V.
EXPERIMENT 7: DETERMINATION OF RATIO OF HEAT CAPACITY
Calculate the value of heat capacity ratio, by the given formula of Cv:The expression of heat capacity ratio is:
𝐶𝑣 𝑅
𝐼𝑛
; 𝑤ℎ𝑒𝑟𝑒 𝐶𝑉 8.314 𝐿 𝑘𝑝𝐴
𝐾 −1 𝑚𝑜𝑙 −1
𝐼𝑛 [
𝑇2 𝑇1
= −𝐼𝑛
𝑉2 𝑉1
=
302.85 304.55
𝑉2 𝑉1
𝑃1𝑇1 𝑃2𝑇2
] = −𝐼𝑛[
158.4𝑘𝑃𝑎(304.55) 111.5(302.85)
]
Cv = 529.79 𝐿 𝑘𝑝𝐴 𝐾 −1 𝑚𝑜𝑙 −1
Cp= Cv +R = 529.79 𝐿 𝑘𝑝𝐴 𝐾 −1 𝑚𝑜𝑙 −1+ 8.314 𝐿 𝑘𝑝𝐴 𝐾 −1 𝑚𝑜𝑙 −1 = 538.10 𝐿 𝑘𝑝𝐴 𝐾 −1 𝑚𝑜𝑙 −1
Therefore the ratio of: 𝑐𝑝 𝑐𝑣
𝑐𝑝
=
538.10 529.79
The theoretical value of 𝑐𝑣 is 1.4
= 1.016
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