JEE PHYSICS MCQ: THERMODYNAMICS

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Work done per mole in an isothermal change is

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Correct option: C

RT{\log _e}\frac{{{V_2}}}{{{V_1}}}

Option C (

RT{\log _e}\frac{{{V_2}}}{{{V_1}}}

) is correct because it represents the work done in an isothermal process derived from integrating

PdV

with

P = \frac{RT}{V}

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0.25\,\,{m^3}

{10^3}N/{m^2},

T

P

V

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