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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1280 J

1610 J

1815 J

2025 J

answer is C.

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Detailed Solution

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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