In the equation $$\Delta H = \Delta U + \Delta PV,$$ under what circumstances can we simplify it into $$\Delta H = \Delta U + P\,\Delta V + V\,\Delta P$$ by product rule?

In this article (pdf slides) it is mentioned that it isn't true. In one question I got a wrong answer using $$\Delta H = \Delta U + \Delta PV + V\,\Delta P,$$ but the correct answer using $$\Delta H = \Delta U + \Delta PV = \Delta U + P_\mathrm f V_\mathrm f -P_\mathrm i V_\mathrm i.$$

The question was:

one mole of a non ideal gas undergoes a change of state from ($\pu{2 atm}$, $\pu{3 L}$, $\pu{95 K}$) to ($\pu{4 atm}$, $\pu{5 L}$, $\pu{245 K}$) with a change in internal energy, $\Delta U = \pu{30.0 L atm}$. Calculate $\Delta H$ in the process.

Using $$\Delta H = \Delta U + P\,\Delta V + V\,\Delta P$$ you get $\Delta H = \pu{40 L atm}$, which is wrong. And using $$\Delta H = \Delta U + \Delta PV$$ you get $\Delta H = \pu{44 L atm}$, which is correct.

Is it something to do with the non ideal gas?

  • 2
    $\begingroup$ It's nothing to do with ideality or lack thereof. The "product rule" works for infinitesimals, i.e. $\mathrm d(pV) = p\,\mathrm dV + V\,\mathrm dp$, or $\mathrm d(pV)/\mathrm dx = p(\mathrm dV/\mathrm dx) + V(\mathrm dp/\mathrm dx)$ where $x$ is some other variable. For finite quantities that doesn't make sense, because you have two pressures $p_1, p_2$ and two volumes $V_1, V_2$; so exactly which pressure should $p$ in $p\Delta V$ be, and which volume should $V$ in $V\Delta p$ be? $\endgroup$ – orthocresol Dec 30 '18 at 17:56
  • $\begingroup$ @orthocresol - the product rule works just fine in this case, if you add back in the term that's dropped out for the infinitesimals. See my answer for details. The answer to your question is that Pi should be in PDV and Vi in VDP. $\endgroup$ – Andrew Feb 21 '19 at 14:33
  • $\begingroup$ @Andrew you're correct. This is bordering on splitting hairs, but I think the only thing we're disagreeing on is how one defines a product rule. OP seems to associate it with the one used in basic calculus: $(fg)' = f'g + g'f$, and so I went along with that; that doesn't work with finite changes, of course. If you add back in the $\Delta p \Delta V$ term, the expansion of $\Delta (pV)$ no longer explicitly has the same form as the 'product rule' $d(pV) = pdV + Vdp$; but if you define $\Delta (pV) = p_i\Delta V + V_i\Delta p + \Delta p \Delta V$ to also be a 'product rule', then yes, it works. $\endgroup$ – orthocresol Feb 21 '19 at 15:18

The answer is clear if you pay attention to how the "product rule" comes about.

Let's write out $\Delta (PV)$ as $P_fV_f-P_iV_i$, where subscript $f$ means final and $i$ means initial.

Now substitute $P_f=P_i+\Delta P$ and $V_f=V_i+\Delta V$:

\begin{align} \Delta (PV) &= (P_i+\Delta P)(V_i+\Delta V)-P_iV_i\\ &=P_iV_i+P_i\Delta V + V_i\Delta P + \Delta P \Delta V -P_iV_i\\ &=P_i\Delta V + V_i\Delta P + \Delta P \Delta V \end{align}

When we are dealing with infinitesimals in derivatives, that last term is $\mathrm{d}P\mathrm{d}V$ and is assumed to be negligible. But when $\Delta V$ and $\Delta P$ are not infinitesimal, you need to include it.


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