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Question

20 litres of a monoatomic ideal gas at 0oC and 10 atm pressure is suddenly released to 1 atm pressure and the gas expands adiabatically against this constant pressure. The volume of the gas respectively are

A
V=164 L
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B
V=57 L
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C
V=123.7 L
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D
V=68.3 L
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Solution

The correct option is C V=123.7 L
This is an adiabatic irreversible process, so for this process
PVγ= Constant, is not applicable
W=Pext(V2V1)
V2 is final volume,V1 is initial volume,W is work done
T2 is final temperature, T1 is initial temperature
But for adiabatic process,

W=dU=(P2V2P1V1γ1)
Also, we know;
PV=nRT10×20=n×0.082×273n=8.92. moles
Also, Comparing the work done, We get
Pext(V2V1)=(P2V2P1V1γ1)
1×(V210)=1×V210×201.671
(10V2)=V22000.67
6.70.67 V2=V2200
206.7=1.67 V2
V2=123.75 L

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