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