Question

# Consider the cyclic process ABCA as shown in the figure, performed on a sample of 2.0 mole of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the magnitude of the work done (in joules) by the gas during the part BC. (R=8.3 JK−1mol−1)

A
2580 J
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B
3625 J
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C
4520 J
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D
1550 J
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Solution

## The correct option is C 4520 JThe net change in internal energy for a cyclic process is zero. ΔQ=−1200 J Using First law of thermodynamics, ΔQnet=ΔUnet+Wnet ∴Wnet=WAB+WBC+WCA=−1200 J...(i) Lets study the graph now, graph AB is a straight line in T v/s V graph ⇒P is constant. Hence it is an isobaric process. WAB=nRΔT In graph from C to A volume is constant. Hence it is an isochoric process. WCA=0 Substituting these values in equation (i), we get n×8.3×(500−300)+WBC+0=−1200 WBC=−4520 J Hence the magnitude of workdone is 4520 J

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