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Question
An ideal gas is taken in a cyclic process as shown in the figure. Calculate
- work done by the gas
- work done on the gas
- Net work done in the process
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Solution
a. Work done by the gas (along with AB)
W = P × ΔV = 600 × 3 = 1800 J
W = 1.8 kJ
b. Work is done on the gas (along with BC)
W = −P × ΔV = −400 × 3 = −1200 J
W = −1.2 kJ
c. Net work done in the process = Area under the curve AB
= Rectangle area + triangle area
= `("l" xx "b") + (1/2 xx "b" xx "h")`
= `(400 xx 3) + (1/2 xx 3 xx 200)`
= 1200 + 300
= 1500 J
W = 1.5 kJ
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