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Question

Three moles of an ideal gas are taken through a cyclic process ABCA as shown in T−V indicator diagram. The gas loses 2510 J of heat in the complete cycle. If TA=100 K and TB=200 K, the work done by the gas during the process BC is (line AB if extended passes through origin)

A
5000 J
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B
5000 J
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C
4000 J
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D
2500 J
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Solution

The correct option is B 5000 J
From the diagram we can deduce that, cyclic process is in anti-clockwise direction. So, the work done by the gas during the cyclic process is negative.
In process AB, the volume V increases linearly with temperature T. Hence process AB is isobaric.
Work done during process AB is given by
WAB=PΔV=nRΔT (PV=nRT)
From the data given in the question,
WAB=3×8.3×(200100)=+2490 J
From the figure we can say that, process CA is isochoric. Hence, work done in this process WCA=0.
Since, the whole process ABCA is cyclic, the change in internal energy in complete cycle is zero, i.e. ΔU=0.
Now, from the first law of thermodynamics,
Q=ΔU+Wnet
Since, gas loses heat energy , Heat energy lost by gas over the complete cycle is Q=2510 J
Q=ΔU+WAB+WBC+WCA
Substituting the values we get, 2510=0+2490+WBC+0
WBC=25102490=5000 J
Thus, the correct choice is option (b).

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