# Why is enthalpy change = heat change only for constant pressure?

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.

• As a mathematician, would you say that $\Delta (PV)=\int{PdV}$ – Chet Miller May 15 at 19:09
• 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/… – theorist May 15 at 20:27
• @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... – Eriol May 16 at 10:20
• @Eriol Have you got an answer to accept ? – Srijan M.T May 25 at 6:56

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.
• 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. – Eriol May 16 at 10:33