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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=nRγ1[T2T1]

                    =1×8.314751[T2T1]

                T1V1r1=T2V2γ1

                T1T2=[V1V2]γ1

                T2T11=[V1V2]γ11

                w = 1824 J

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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