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We have to calculate delta H for a gas that follows the Van der Waals equation of state:

$$\left(p + \frac{a}{V_\mathrm m^2}\right)\left(V_\mathrm m-b\right)=RT$$

I know that $H = U + pV$ but I don't know how to approach this question.

Since I also know that $\Delta H = \Delta U + \Delta (pV)$, does that mean that I can calculate this simply by calculating $\Delta U$ (which I know how to do) and then $\Delta (pV) = p_2V_2 - p_1V_1$? Is that an appropriate method?

Because previously in other questions I have also seen it done as $\mathrm{d}H = \mathrm{d}U + V\mathrm{d}p + p\mathrm{d}V + \mathrm{d}V\mathrm{d}p$ so I am wondering when is which equation applicable?

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  • $\begingroup$ You seem certain that you know how to calculate $\Delta U$. So please tell us how you would do it for a Van der Waals gas. $\endgroup$
    – Chet Miller
    Oct 13, 2015 at 10:43
  • $\begingroup$ I've edited your post with LaTeX to make it look more presentable - you can learn how to do this here. Your vdW equation of state was incorrect; you should have used the molar volume $V_m$ instead of $V$. I corrected it. Also, in the last equation, the product of two infinitesimals should not be in the equation. $\endgroup$
    – orthocresol ♦
    Oct 13, 2015 at 12:58
  • $\begingroup$ By the product rule, $\mathrm{d}(fg)/\mathrm{d}x = f(\mathrm{d}g/\mathrm{d}x) + g(\mathrm{d}f/\mathrm{d}x)$; you can "multiply throughout by $\mathrm{d}x$" (it is mathematically incorrect, but it gives you the right result) to get: $\mathrm{d}(fg) = f \mathrm{d}g + g \mathrm{d}f$. It can be more properly derived by considering the properties of differentials but that is another story. $\endgroup$
    – orthocresol ♦
    Oct 13, 2015 at 13:00

1 Answer 1

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Are you familiar with the following equation?: $$dH=C_pdT+\left(V_m-T\left(\frac{\partial V_m}{\partial T}\right)_P\right)dP$$

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  • $\begingroup$ Ok, but in order to calculate ∂Vm/∂T don't I need to rewrite the equation of state in terms of V? And how can I do this? $\endgroup$
    – user130996
    Oct 13, 2015 at 18:08
  • $\begingroup$ Good point. A better choice of an equation to start with is:$$dU=C_vdT-\left(P-T\left(\frac{\partial P}{\partial T}\right)_V_{m}\right)dV_m$$ $\endgroup$
    – Chet Miller
    Oct 13, 2015 at 18:55
  • $\begingroup$ A better choice is: $$dU=C_vdT-\left(P-T\left(\frac{\partial P}{\partial T}\right)_{V_m}\right)dV_m$$ $\endgroup$
    – Chet Miller
    Oct 13, 2015 at 19:06

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