UNIVERSITI TEKNOLOGI MARA
FAKULTI KEJURUTERAAN KIMIA
FLUIDS AND THERMODYNAMICS LABORATORY (CHE486)
No.
NAME Title Allocated Marks
: DK NUR SHANEEZ ASNEENA (%)
(2019545093) Marks
1. Abstract/Summary 5
FARAH NURHUSNA BINTI ZOOLKIPLI (2019362021)
2. Introduction 10
3. Aims RAIDAH BINTI ROZALI (2019353537)
5
4. Theory MUHAMMAD IMRAN KHAIZURAN 10 (2018653956)
5. Apparatus 5
GROUP : EH2203A
6. Methodology/Procedure 10
EXPERIMENT
7. Results : 6 (PROPERTIES MEASUREMENT)10
8. Calculations
DATE PERFORMED : 10
9. Discussion 20
SEMESTER : 3
10. Conclusion 5
PROGRAMME
11. Recommendations: EH220 5
12. Reference/Appendix 5
TOTAL MARKS 100
Remarks,
Checked by :
--------------------------------
Date :
TABLE OF CONTENT
CONTENT PAGE
1.0 ABSTRACT
2.0 INTRODUCTION
3.0 OBJECTIVE
4.0 THEORY
5.0 APPARATUS
6.0 PROCEDURE
7.0 RESULTS
8.0 CALCULATIONS
9.0 DISCUSSION
10.0 CONCLUSION
11.0 RECOMMENDATION
12.0 REFERENCES
13.0 APPENDICES
1.0 ABSTRACT
2.0 INTRODUCTION
3.0 OBJECTIVE
4.0 THEORY
5.0 APPARATUS
6.0 PROCEDURE
1.
7.0 RESULTS
Table 7.1: Boyle’s Law
Pressure (kpa abs) Before expansion After expansion
Pressure to PT1 = 149.2 PT1 = 132.7
atmosphere PT2 = 103.3 PT2 = 131.3
Atmospheric to PT1 = 104.6 PT1 = 90.6
vacuum PT2 = 53.4 PT2 = 89.3
Pressurized to PT1 = 146.4 PT1 = 115.3
vacuum PT2 = 60.4 PT2 = 114.0
Table 7.2: Gay Lussac’s Law
Pressure Trial 1 Trial 2 Trial 3
Temperature (℃) Temperature (℃) Temperature (℃)
Pressurize Depressurize Pressurized Depressurize Pressurize Depressurized
(kpa abs)
d vessel d vessel vessel d vessel d vessel vessel
114 28.3 27.3 28.2 32.4 27.3 31.7
124 28.9 28.3 28.5 31.6 27.7 30.4
134 29.7 29.3 29.2 30.6 28.6 29.3
144 30.4 30.3 30.0 29.6 29.7 28.1
154 31.3 31.0 30.8 28.5 30.6 27.3
164 32.0 31.3 31.7 28.0 31.0 26.9
Table 7.3: Average value
Pressure (kpa abs) Average (℃)
114 29.2
124 29.2
134 29.5
144 29.7
154 29.9
164 30.2
Table 7.4: Determination of Ratio of Heat Capacity
Initial Final
PT1 (kPa abs) 153.5 111.5
T1 (℃) 30.6 28
8.0 CALCULATIONS
Experiment 1
Given; V1 = 0.025 m3, V2 = 0.01237 m3
1. Pressure to atmosphere
P1 V 1=P2 V 2
( PT 1before ×V 1 ) + ( PT 2before ×V 2 )=( PT 1 after ×V 1 ) + ( PT 2after ×V 2 )
( 149.2× 0.025 ) + ( 103.3 ×0.01237 )=( 132.7 ×0.025 ) +(131.3× 0.01237)
5.01=4.94
The difference is 0.07 kpa⸱m3, so the Boyle’s law is verified
2. Atmospheric to vacuum
P1 V 1=P2 V 2
( PT 1before ×V 1 ) + ( PT 2before ×V 2 )=( PT 1 after ×V 1 ) + ( PT 2after ×V 2 )
( 104.6 ×0.025 )+ ( 53.4 × 0.01237 ) =( 90.6 ×0.025 ) +(89.3 ×0.01237)
3.28=3.37
The difference is 0.09 kpa⸱m3, so the Boyle’s law is verified
3. Pressurized to vacuum
P1 V 1=P2 V 2
( PT 1before ×V 1 ) + ( PT 2before ×V 2 )=( PT 1 after ×V 1 ) + ( PT 2after ×V 2 )
( 146.4 × 0.025 )+ ( 60.4 ×0.01237 )=( 115.3 × 0.025 )+(114 ×0.01237)
4.41=4.29
The difference is 0.12 kpa⸱m3, so the Boyle’s law is verified
Experiment 2
Pressurized Vessel at Trial 1
33.0
32.0
31.0
Temperature (℃)
30.0
29.0
28.0
27.0
26.0
114 124 134 144 154 164
Pressure (kpa)
Figure 8.1: Pressurized Vessel at Trial 1
Depressurized Vessel at Trial 1
32.0
31.0
30.0
Temperature (℃)
29.0
28.0
27.0
26.0
25.0
114 124 134 144 154 164
Pressure (kpa)
Figure 8.2: Depressurized Vessel at Trial 1
Pressurized Vessel at Trial 2
32.0
31.0
30.0
Temperature (℃)
29.0
28.0
27.0
26.0
114 124 134 144 154 164
Pressure (kpa)
Figure 8.3: Pressurized Vessel at Trial 2
Depressurized Vessel at Trial 2
33.0
32.0
31.0
Temperature (℃)
30.0
29.0
28.0
27.0
26.0
25.0
114 124 134 144 154 164
Pressure (kpa)
Figure 8.4: Depressurized Vessel at Trial 2
Pressurized Vessel at Trial 3
33.0
32.0
31.0
Temperature (℃)
30.0
29.0
28.0
27.0
26.0
25.0
24.0
114 124 134 144 154 164
Pressure (kpa)
Figure 8.5: Pressurized Vessel at Trial 3
Depressurized Vessel at Trial 3
33.0
32.0
31.0
Temperature (℃)
30.0
29.0
28.0
27.0
26.0
25.0
24.0
114 124 134 144 154 164
Pressure (kpa)
Figure 8.6: Depressurized Vessel at Trial 3
Average
30.4
30.2
30
Temperature (℃)
29.8
29.6
29.4
29.2
29
28.8
28.6
114 124 134 144 154 164
Pressure (kpa)
Figure 8.7: Average Temperature
Experiment 3
Cv T 2 V
ln =−ln 2
R T1 V1
V 2 P1 T 1
where =
V 1 P2 T 2
Cv 301 ( 153.5 )( 303.6 )
8.314
L. kPa
ln
303.6
=−ln [ ] [
(111.5 ) ( 301 ) ]
K . mol
L. kPa
C v =317.33
K . mol
C p=C v + R
¿ 317.33+8.314
L. kPa
¿ 325.64
K . mol
The ratio of the heat capacity,
C p 325.64
= =1.026
C v 317.33
9.0 DISCUSSION
10.0 CONCLUSION
11.0 RECOMMENDATION
12.0 REFERENCES
13.0 APPENDICES