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Q.

At N.T.P  one mole of diatomic gas is compressed adiabatically to half of its volume γ=1.41  . The work done on gas will be

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a

1280J

b

1610J

c

1824J

d

2025J

answer is C.

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Detailed Solution

w=nRγ−1[T2−T1]                     =1×8.31475−1[T2−T1]                 T1V1r−1=T2V2γ−1                 T1T2=[V1V2]γ−1                 T2T1−1=[V1V2]γ−1−1                 w = 1824 J
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