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Question

An ideal gas with adiabatic exponent γ , is according to the law P =αV where α is a constant. The initial volume of the gas is V0. As a result increases η times. Find the increment in internal energy and work done.

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Solution

The process is given as P=αV
PV1=α=constant
Comparing with PVm=constant we get m=1
Given : V1=Vo V2=ηVo
P1=αVo and P2=αηVo

Change in internal energy ΔU=Rγ1×n(T2T1)=P2V2P1V1γ1
ΔU=[αηVo(ηVo)αVo(Vo)]γ1=αV2o(η21)γ1

Work done W=P2V2P1V11m

W=[αηVo(ηVo)αVo(Vo)]1(1)=αV2o(η21)2

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