Work done per mol in an isothermal change is

R T {l o g}_{10} \frac{V_{2}}{V_{1}}

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R T {l o g}_{10} \frac{V_{1}}{V_{2}}

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R T {l o g}_{e} \frac{V_{2}}{V_{1}}

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R T {l o g}_{e} \frac{V_{1}}{V_{2}}

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Solution

The correct option is C $R T {l o g}_{e} \frac{V_{2}}{V_{1}}$
Work done in isothermal process $= P . d V$

We have the equation,

$P V = n R T$

and,

$P = \frac{n R T}{V}$

Then, work done $= \int_{V_{1}}^{V_{2}} [\frac{n R T}{V}] d V$

$= n R T l o g_{e} (\frac{V_{2}}{V_{1}})$

$= R T l o g_{e} \frac{V_{2}}{V_{1}}$

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Work done per mol in an isothermal change is

The correct option is C RTlogeV2V1Work done in isothermal process =P.dVWe have the equation,PV=nRTand,P=nRTVThen, work done =∫V2V1[nRTV]dV=nRTloge(V2V1)=RTlogeV2V1