# Clarification on the derivation of Clausius-Clapeyron equation

I have some trouble understanding what is going on in some steps of the Clausius-Clapeyron equation derivation. When we have a substance in two phases the Gibbs energy will be a function $$G=G(T,P,n_1,n_2)$$. At equilibrium (constant $$T$$ and $$P$$):

\begin{align} dG&=-SdT + VdP + μ_1dn_1 + μ_2dn_2=0\\ μ_1dn_1&= -μ_2dn_2 \end{align}

and because $$dn_1=-dn_2$$ the chemical potentials must be equal:

$$μ_1(T,P)=μ_2(Τ,P)$$

so must be their differentials and the derivation goes like this (taking temperature as independent variable):

$${\left(\frac{\partial{μ_1}}{\partial{T}}\right)_P}dP + {\left(\frac{\partial{μ_1}}{\partial{P}}\right)_T}dT = {\left(\frac{\partial{μ_2}}{\partial{T}}\right)_P}dP + {\left(\frac{\partial{μ_2}}{\partial{P}}\right)_T}dT$$

Because we move on the coexisting curve (taking temperature as independent variable):

$$-S_1+V_1\frac{dP}{dT}=-S_2 + V_2\frac{dP}{dT}$$

With a little manipulation one gets:

$$\frac{dP}{dT}=\frac{S_2-S_1}{V_2-V_1}$$

where both $$S$$ and $$V$$ denote molar quantities.

My first question is why we are able to substitute $${\left(\frac{\partial{μ_1}}{\partial{T}}\right)_P}$$ with $$S_1$$. When I tried to derive it I thought the following:

$$dG=-VdP + SdT + μ_1dn_1 + μ_2dn_2 \tag{1}$$

and because Gibbs is homogeneous with respect to the extensive variables:

$$dG=μ_1dn_1 + μ_2dn_2 + n_1dμ_1 + n_2dμ_2 \tag{2}$$

we can write:

$$n_1dμ_1 + n_2dμ_2=-VdP + SdT \Rightarrow \left(\frac{\partial{μ_1}}{\partial{T}}\right)=-\frac{S}{n_1}$$

Why $$S/n_1$$ must be equal to $$S_1$$ (molar). Doesn't the entropy of the system depend both on $$n_1$$ and $$n_2$$ whereas the molar entropy for a given species is only a function of $$n_1$$? Can we do this because the species are in different phases so we can treat them seperately that is by defining $$G^1=G(T,P,n_1)$$ and $$G^2=G(T,P,n_2)$$ (where the superscripts denote the corresponding phases)?

My next trouble is why we are able to substitute $$ΔS=S_2-S_1$$ with $$\frac{1}{T}ΔH$$

In some derivations it is justified because the process is reversible so we can use the classical definition of entropy $$dS=\frac{dq}{T}$$ and because the process happens at constant pressure $$q=ΔH$$. But $$S$$ in the classical definition should correspond to the total entropy of the system that is $$S=S_1+S_2$$, same gos for enthalpy. How does it make sense to say that $$ΔS_{system}=S_2-S_1$$? Change in system's entropy means:

$$ΔS=Δ(S_1+S_2)=S^{f}_1 + S^{f}_2 - S^{i}_1 - S^{i}_2$$

So to sum it up my questions aer why it is valid to say $$S/n_1=S_1$$ and why we are able to substitute $$S_2-S_1$$ with $$q/T$$.

Why [must $$S/n_1$$] be equal to $$S_1$$ (molar)[?]

By definition. $$S$$ depending on a variety of other parameters doesn't preclude such a definition. We can conclude that the molar entropy of species 1 in this system also depends on those parameters.

why we are able to [replace] $$\Delta S=S_2−S_1$$ with $$\frac{1}{T}ΔH$$[?]

$$\Delta G = 0$$ during a phase transition, and $$G\equiv H-TS$$, so $$\Delta H=T_{\mathrm{transition}}\Delta S$$ at that transition temperature.

I suspect there is some confusion here because you are considering the entropy of the total system. There is nothing wrong with as long as the total quantities and the quantities of the separate phases are not confused.

In your last equation $$$$n_1\,d\mu_1 + n_2\,d\mu_2 = -S\,dT +V\,dP$$$$ You should have $$$$\left(\frac{\partial\mu_1}{\partial T}\right)_{P} = -\frac{S}{n_1} - \frac{n_2}{n_1}\left(\frac{\partial\mu_2}{\partial T}\right)_{P}$$$$ Whereas the Gibbs-Duhem equation for phases 1 and 2 give $$$$\left(\frac{\partial\mu_1}{\partial T}\right)_{P} = -\frac{S^1}{n_1},\quad \left(\frac{\partial\mu_2}{\partial T}\right)_{P} = -\frac{S^2}{n_2}$$$$ If you substitute this into the previous equation you see that they are consistent

For the last part of the question it seems that you are confused about the meaning of the operator $$\Delta$$. Here it means the excess of the value in the second phase over the value in the first.