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Please help me, a mathematician doing his elective course on Physical Chemistry, out with this very simple question: why does the relation between enthalpy and heat $$\Delta H = \Delta U + p\,\Delta V = \Delta Q$$ hold only under constant pressure? With the integral definition of work, couldn't one just as well write $$\Delta H = \Delta U + \int_{V_i}^{V_f} p(V)\,\mathrm dV = \Delta Q $$ in the case where the pressure is not constant during expansion?

I'm still having some trouble with chemical notation of derivatives/infinitesimals as well as with the physical realities behind the formulae, so there might any number of basic things I'm missing here.

Thanks in advance. Also as this is my first question here, apologies if I did anything wrong or if I missed a similar question.

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    $\begingroup$ As a mathematician, would you say that $\Delta (PV)=\int{PdV}$ $\endgroup$ – Chet Miller May 15 at 19:09
  • $\begingroup$ See my answer here: chemistry.stackexchange.com/questions/126218/… Also, using $\Delta Q$ is incorrect; please see my answer here: chemistry.stackexchange.com/questions/136237/… $\endgroup$ – theorist May 15 at 20:27
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    $\begingroup$ @Chet Miller Thanks, I believe that is the critical error I was making: if $\Delta (PV)$ is defined as $P(V_2) V_2 - P(V_1) V_1$, then of course it can only be equal to the integral if the pressure is in fact constant. What an embarrassing mistake to make as a maths major, but here we are... $\endgroup$ – Eriol May 16 at 10:20
  • $\begingroup$ @Eriol Have you got an answer to accept ? $\endgroup$ – Srijan M.T May 25 at 6:56
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Firstly, $\int_{V_i}^{V_f} p\, dV\neq p \Delta V$ unless $p$ is constant. Hence the formula holds in that case only. Also, the definition of $H=U+pV$, hence if we wanted $\Delta H$, we should take:$$\Delta H=\Delta U +\int_{V_i}^{V_f}\,d(pV)$$ Here, of course, $d(pV)=pdV+Vdp$ using the product rule.

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  • $\begingroup$ Thank you, as mentioned above my main error was indeed confusing $\Delta (pV) $ and $\int_{V_i}^{V_f} p dV$, but I appreciate your additional clarifications. $\endgroup$ – Eriol May 16 at 10:33
  • $\begingroup$ @Eriol If you believe, this answered your question than consider upvoting and accepting the answer. $\endgroup$ – Nisarg Bhavsar May 18 at 17:26
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    $\begingroup$ @NisargBhavsar Oh sorry, first post, didn't know I could to do that...I accepted it now, I think. $\endgroup$ – Eriol May 26 at 11:21
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The enthalpy may be calculated even when the pressure is not constant, for example if the volume is constant. But it is much easier to measure a heat effect at constant pressure, usually at atmospheric pressure. You don't need complicated apparatus like an autoclave. You just need a thermometer. This is why you find enthalpies of formation or enthalpies of reaction in tables, and not the corresponding values of internal energies.

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    $\begingroup$ While you can calculate the enthlapy change for any process, it will only be equal to the heat flow under three conditions: (1) closed system; (2) the only type of work is pV-work; (3) constant pressure. $\endgroup$ – theorist May 15 at 20:31
  • $\begingroup$ OP the comment by @theorist is likely the most straightforward answer. The formulas aren't necessarily general definition but those of convenient practical use. That is why in chemistry more often you encounter Gibbs energy and less often Helmotz energy, too. $\endgroup$ – Alchimista May 16 at 8:11

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