One mole of ${O_2}$ gas having a volume equal to $22.4$ litres at ${0^o}C$ and $1$ atmospheric pressure in compressed isothermally so that its volume reduces to $11.2$ litres. The work done in this process is ...... $J$
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(d)$W = - \mu RT{\log _e}\frac{{{V_2}}}{{{V_1}}} = - 1 \times 8.31 \times (273 + 0){\log _e}\left( {\frac{{22.4}}{{11.2}}} \right)$
$ = - \,8.31 \times 273 \times {\log _e}2$$ = - 1572.5J$ [${\log _e}2 = 0.693$]
$ = - \,8.31 \times 273 \times {\log _e}2$$ = - 1572.5J$ [${\log _e}2 = 0.693$]
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