1
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Question

# An ideal gas shows Pâˆ’V work done in a cyclic process as shown in a figure. The net work done by the gas during the cycle is:

A
12P1V1
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B
6P1V1
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C
9P1V1
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D
5P1V1
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Solution

## The correct option is D 5P1V1For a cyclic process, change in internal energy is zero. The work done during the process is equal to the area under the graph. The area under the graph is equal to the area of the triangle.Area=12×(Base)×(Height) =12×(6P1−P1)×(3V1−V1)=5P1V1Hence, the correct option is D.

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