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Question

1 mole of NH3 gas at 27oC is expanded in reversible adiabatic condition to make volume 64 times (γ=1.33). Final temperature and magnitude of work done respectively are

A
150 K, 900.00 cal
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B
75 K, 1363.63 cal
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C
75 K, 1263.63 cal
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D
150 K, 1363.63 cal
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Solution

The correct option is B 75 K, 1363.63 cal
For a reversible adiabatic process :
T1Vγ11=T2Vγ12
Given, T1 =300 K
V2=64V1
putting the values we get,
300×V1/31=T2(64V1)1/3T2=75 K
So, final temperature is 75 K

and work done will be ,
W=nCv(T2T1)
we know Cv=Rγ1
so,
W=1×Rγ1(75300)=21.331(225)=1363.63 cal

so, option (b) is correct.

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