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Question

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

A
T = 174.8 K, V = 64 L
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B
T = 153 K, V = 57 L
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C
T = 165.4 K, V = 78.8 L
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D
T = 161.2 K, V = 68.3 L
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Solution

The correct option is A T = 174.8 K, V = 64 L
This is adiabatic irreversible process,
W=(P2V2P1V1γ1)....(i)
And work done against pressure is given as:
W=Pext(V2V1)....(ii)

P1=10 atm, P2=Pext=1 atm
T1=273 K, V1=10 L
γ=5/3 (for monoatomic gas)

Also we know:
P1V1=nRT110×10=n×0.082×273n=4.47 mol

Equating eq. (i) and (ii):

Pext(V2V1)=(P2V2P1V1γ1)1×(V210)=1×V210×101.671
(10V2)=V21000.676.70.67V2=V2100
106.7=1.67 V2V2=64 L

now using;
P2V2=nRT21×64=4.47×0.082×T2T2=174.6 mol
Hence option (a) is correct.

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