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Question


An ideal gas shows PV work done in a cyclic process as shown in a figure. The net work done by the gas during the cycle is:
577010.png

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 5P1V1
For 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×(6P1P1)×(3V1V1)=5P1V1

Hence, the correct option is D.

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