4A-5 : | Quasi-Equilibrium Expansion of a Gas | 4 pts |
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Estimate the work done by a gas during an unknown process. Data obtained that relates pressure and volume are : | |||||||||
P (kPa) | 200 |
250 |
300 |
350 |
400 |
450 |
500 |
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V (cm3) | 800 |
650 |
550 |
475 |
415 |
365 |
360 |
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Read : | The key concept here is that boundary or PV work is represented by the area under the process path curve on a PV Diagram. So, once we plot the given data on a PV Diagram, all we need to do is numerically integrate to determine the area under the curve and we will have the work ! | ||||||||
Given: | P (kPa) |
V (L) |
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200 |
0.800 |
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250 |
0.650 |
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300 |
0.550 |
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350 |
0.475 |
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400 |
0.415 |
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450 |
0.365 |
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500 |
0.360 |
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Find: | W | ??? |
J | ||||||
Assumptions: | - Each state in the data table is an equilibrium state. | ||||||||
- The process is a quasi-equilibrium process. | |||||||||
- The system is a closed system. | |||||||||
- The trapezoidal rule gives an acceptable estimate of the area under the process path in the PV Diagram. | |||||||||
Solution : | The area of each trapezoid under the process path in the PV Diagram is the product of the average pressure for that trapezoid and the change in volume across the trapezoid. | ||||||||
Trapezoid |
Pavg (kPa) |
ΔV (L) |
W (J) |
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A |
475 |
-0.005 |
-2 |
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B |
425 |
-0.050 |
-21 |
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C |
375 |
-0.060 |
-23 |
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D |
325 |
-0.075 |
-24 |
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E |
275 |
-0.100 |
-28 |
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F |
225 |
-0.150 |
-34 |
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Answer: | Wtotal = | -132 |
J |