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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 PV 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 PV graph
=12×(P0+P0/2)×(2V0V0)=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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