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The work of a van der Waals gas during an isothermal transformation is readily found to be ($n=1$ mol)

\begin{equation} W =RT \ln\frac{V_1-b}{V_2-b}+a\left(\frac{1}{V_1}-\frac{1}{V_2}\right)\qquad (1) \end{equation}

where $V_1$ and $V_2$ the initial and final volume.

My question is given the initial conditions in the form $(P_1,T_1)$ and the final in the form $(P_2,T_2)$, that is initial and final pressure and temperature are considered known but not the initial and final volume, can we express (1) in terms of the given quantities without solving the involving cubic equation with respect to the volume?

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    $\begingroup$ No. You need to solve for the volume. $\endgroup$ – Chet Miller Mar 27 at 11:23

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