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Question

# An ideal monoatomic gas is at pressure P0 and volume V0. It is taken to final volume 2V0 and final pressure P02 in a process which is straight line on P−V diagram.

A
Final temperature is greater than initial temperature
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B
Internal energy increases
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C
Work done by the gas is +P0V04
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D
Heat is absorbed by the gas
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Solution

## The correct option is D Heat is absorbed by the gas Initial temperature, Ti=P0V0nR and Final temperature, Tf=P02×2V0nR=P0V0nR So, Ti=Tf As, Ti=Tf, ΔU=0 Workdone = area under P−V graph =12×(P0+P0/2)×(2V0−V0)=3P0V04 As, ΔU=0 and W>0 ⇒ΔQ>0 [ From first law of thermodynamics ] Therefore, heat is absorbed in the process.

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