One mole of a diatomic gas undergoes a process P=P0/1+V/V03 where P0 and V0 are constants. The translational kinetic energy of the gas when V=V0 is given by

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5P0V0/4

3P0V0/4

3P0V0/2

5P0V0/2

answer is B.

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Detailed Solution

P=P01+V/V03=P02T=P0V02RTherefore translational kinetic energy is equal to 32RT=3R2P0V02R=3P0V04

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