Work done per mole in an isothermal change is

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(C) For an isothermal process,the temperature $T$ remains constant.From the ideal gas equation for $1$ mole of gas,$PV = RT$,which implies $P = \frac{

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$RT \log_{e} \frac{V_2}{V_1}$

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(C) For an isothermal process,the temperature $T$ remains constant.From the ideal gas equation for $1$ mole of gas,$PV = RT$,which implies $P = \frac{RT}{V}$.The work done $W$ during an isothermal expansion from volume $V_1$ to $V_2$ is given by the integral:$W = \int_{V_1}^{V_2} P \, dV$Substituting the value of $P$:$W = \int_{V_1}^{V_2} \frac{RT}{V} \, dV$Since $R$ and $T$ are constants,we can take them out of the integral:$W = RT \int_{V_1}^{V_2} \frac{1}{V} \, dV$$W = RT [\ln V]_{V_1}^{V_2}$$W = RT (\ln V_2 - \ln V_1)$$W = RT \log_{e} \left( \frac{V_2}{V_1} \right)$.

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$Assertion :$ In an isothermal process,the whole of the heat supplied to the body is converted into internal energy.
$Reason :$ According to the first law of thermodynamics,$\Delta Q = \Delta U + P\Delta V$.

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