The change in the entropy of 1 mole of an ide | vedclass

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(D) For an isothermal process,the temperature $T$ remains constant.For $n$ moles of an ideal gas,the heat exchanged $\Delta Q$ during an isothermal ex

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$R \ln \frac{V_2}{V_1}$

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(D) For an isothermal process,the temperature $T$ remains constant.For $n$ moles of an ideal gas,the heat exchanged $\Delta Q$ during an isothermal expansion from $V_1$ to $V_2$ is equal to the work done $W$,because the change in internal energy $\Delta U = 0$.$W = \int_{V_1}^{V_2} P \, dV = \int_{V_1}^{V_2} \frac{nRT}{V} \, dV = nRT \ln \frac{V_2}{V_1}$.Given $n = 1$ mole,the heat exchanged is $\Delta Q = RT \ln \frac{V_2}{V_1}$.The change in entropy $\Delta S$ is defined as $\Delta S = \frac{\Delta Q}{T}$.Substituting the value of $\Delta Q$,we get $\Delta S = \frac{RT \ln \frac{V_2}{V_1}}{T} = R \ln \frac{V_2}{V_1}$.

The change in the entropy of $1$ mole of an ideal gas which undergoes an isothermal process from an initial state $(P_1, V_1, T)$ to the final state $(P_2, V_2, T)$ is equal to

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