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Q.

One mole of ideal monoatomic gas at initial state a with pressure P0  and volume  V0 is taken to a final state 'd'  with pressure e3P0  and volume V0e  through the path  abcd (figure)  ab and  cd are adiabatic paths where as bc  is isothermal with temperature  T0. Find the work done by the gas in the process bc.   (Given ln(e)=1 ) 
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a

2RT0

b

53RT0

c

32RT0

d

4RT0

answer is A.

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Detailed Solution

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 P0V0γ=PbVbγ.....(i) (e3P0)(V0e)γ=PCVCγ.....(ii)
Dividing
PCPb(VCVb)γ=e3(1e)γ.....(iii)
But  PbVb=PCVC
(VCVb)γ1=e3(1e)γ (VCVb)2/3=e3(1e)5/3=e4/3 VCVb=e2 VbC=(1)RT0lnVCV=RT0lne2=2RT0

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