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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, 2019 at 11:23
  • $\begingroup$ what cubic equation? just replace $V_i$ by $nRT/P_i$ $\endgroup$
    – user65081
    Mar 26, 2020 at 22:38
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    $\begingroup$ @Wolphramjonny But it isn't an ideal gas so you cannot do it. $\endgroup$
    – user258881
    Apr 3, 2020 at 10:25
  • $\begingroup$ @FakeMod right!! $\endgroup$
    – user65081
    Apr 3, 2020 at 17:05

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