# Why when deriving the Clapeyron equation can you assume that dH = dq, as surely the pressure is not constant?

Part of the Clapeyron equation involves replacing $$\Delta S$$ with $$\Delta H/T$$. I can see the step from $$\Delta S$$ to $$\Delta q/T$$ but I'm not sure why you can replace $$\Delta q$$ with $$\Delta H$$. From my understanding $$\mathrm dH = \mathrm dq_V +V\mathrm dp$$ when assuming no work other than $$pV$$ work is done. I don't understand how you can substitute $$\Delta H$$ for $$\Delta q$$ without assuming that the pressure is constant?

• Heat absorbed at constant volume is change in internal energy i.e dU=nCvdT =qv this relation is true even when volume is not constant similarly change in enthalpy is heat absorbed at constant pressure .im yet to find the reason how Above 2 facts are true , but im quite sure that they are true. I was thinking of asking this question myself. Nov 1, 2019 at 16:01
• It is also interesting to note that mayers relationship is derived by stating the above mentioned facts to be true from dH = dU + Vdp -> Qp =Qv + nRdT -> nCpdT = nCvdT + nRdT -> Cp =Cv -R Nov 1, 2019 at 16:09
• Thank you, I understand that for a fixed volume ∆u = q assuming no non PV work is done, but I don't think ∆H is always equal to ∆q, or if it is I am not sure why. All of the textbooks I've read state that for a phase transition ∆H is ∆q but I am not sure if this is always the case, or if it is the case I am not sure why Nov 1, 2019 at 17:03

At any given temperature, the change from a liquid to a vapor occurs at constant pressure. What we are trying to find with the Clapeyron equation is how this constant pressure changes as the given temperature changes.

At equilibrium at a given temperature and constant pressure, we have $$G_{vapor}=G_{liquid}$$ and $$S_{vapor}-S_{liquid}=\frac{(H_{vapor}-H_{liquid})}{T}$$

Now, if we want to find out how temperature and pressure change in tandem along the equilibrium line, we can write:$$dG_{vapor}=-S_{vapor}dT+V_{vapor}dP$$and $$dG_{liquid}=-S_{liquid}dT+V_{liquid}dP$$Since, along the equilibrium line, we must have $$dG_{vapor}=dG_{liquid}$$, we can write:$$-S_{vapor}dT+V_{vapor}dP=-S_{liquid}dT+V_{liquid}dP$$or equivalently:$$\frac{dP}{dT}=\frac{(S_{vapor}-S_{liquid})}{(V_{vapor}-V_{liquid})}=\frac{(H_{vapor}-H_{liquid})}{T(V_{vapor}-V_{liquid})}$$This is the Clapeyron equation.

From the definition of the Gibbs free energy $$G=H-TS$$, $$\Delta G = \Delta H - T \Delta S \tag{constant T}$$ For a pure substance, equilibrium between two phases (such as for a point on a coexistence line) reduces the degrees of freedom to one, such that p is constrained by the choice of T, or vice-versa.

Since the condition for equilibrium at constant T and p is $$G_{m,\textrm{phase 2}}-G_{m,\textrm{phase 1}} = \Delta G_m =0 \tag{constant T and p}$$ where the molar Gibbs free energy $$G_{m,i}$$ is equal to the chemical potential $$\mu_i$$, it follows that on any point on a two-phase coexistence line $$\Delta H_m = T \Delta S_m$$ Note also that only for a reversible process $$\Delta S = q/T$$, and the reversible phase transitions in question occur at constant T and p, such that $$\Delta H = q$$.

The derivation of the Clapeyron equation starts from the premise that two points on a two-phase coexistence line share the same difference in chemical potential ($$\Delta \mu =0$$ for all points on the line), therefore if you move between two neighboring points on the line you must satisfy $$d\mu_1=d\mu_2$$. From this equality you can derive the equation by noting that $$d\mu=V_mdp -S_mdT$$.

• Hi, thank you for answering my question, you explained it very clearly. Would you mind showing me how that fits in with the rest of the derivation of the Clapeyron equation as I am still unsure as to how the assumptions fit in when you change the pressure and temperature to find the gradient of the phase diagram? Nov 2, 2019 at 15:34