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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

a
1280 J
b
1610 J
c
1815 J
d
2025 J

detailed solution

Correct option is C

T2=T1V1V2γ−1=273(2)0.41=273×1.328=363KW=RT1−T2γ−1=8.31(273−363)1.41−1=−1824⇒|W|≈1815J

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