The work done in an isothermal reversible pro | vedclass

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(A) For an isothermal reversible process,the temperature $T$ is constant. From the ideal gas law,$PV = nRT$,so $P = \frac{nRT}{V}$. The work done $w$

Vedclass · Accepted Answer

$-nRT \ln \frac{V_2}{V_1}$

Vedclass · Answer

(A) For an isothermal reversible process,the temperature $T$ is constant. From the ideal gas law,$PV = nRT$,so $P = \frac{nRT}{V}$. The work done $w$ is given by the integral: $w = -\int_{V_1}^{V_2} P \, dV = -\int_{V_1}^{V_2} \frac{nRT}{V} \, dV$. Since $n$,$R$,and $T$ are constants,we pull them out of the integral: $w = -nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV = -nRT [\ln V]_{V_1}^{V_2}$. Therefore,$w = -nRT \ln \frac{V_2}{V_1}$.

The work done in an isothermal reversible process for an ideal gas is-

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