JEE PHYSICS MCQ: THERMODYNAMICS

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One mole of an ideal mono atomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature,

{27^0}C

. The work done on the gas will be

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Correct option: D

300 R ln 2

Option D (300 R ln 2) is correct because the work done in an isothermal compression is calculated using

W = nRT \ln(\frac{{P_1}}{{P_2}})

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\left( {\gamma = \frac{{{C_p}}}{{{C_v}}}} \right)