MCQ
Work done per mol in an isothermal change is
  • A
    $RT{\log _{10}}\frac{{{V_2}}}{{{V_1}}}$
  • B
    $RT{\log _{10}}\frac{{{V_1}}}{{{V_2}}}$
  • $RT{\log _e}\frac{{{V_2}}}{{{V_1}}}$
  • D
    $RT{\log _e}\frac{{{V_1}}}{{{V_2}}}$

Answer

Correct option: C.
$RT{\log _e}\frac{{{V_2}}}{{{V_1}}}$
c
(c)For isothermal process $PV = RT \Rightarrow P = \frac{{RT}}{V}$
$\therefore $W$ = PdV = \int_{\,{V_1}}^{\,{V_2}} {\frac{{RT}}{V}dV = RT} {\log _e}\frac{{{V_2}}}{{{V_1}}}$

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