How to find and use the Clausius-Clapeyron equation

Question

I know how to get the equation from the Clapeyron equation but I have a question regarding a the integration along a phase boundary and a small step in the derivation that I will make clear when I reach that step. Firstly, the Clapeyron equation:

$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta S}{\Delta V}$$

Or alternatively, by recognizing that $\Delta G=0$ when two phases are in equilibrium; $\Delta G =\Delta H - T\Delta S$ can be rearranged to give:

$$\Delta S = \frac{\Delta H}{T}$$

Substituting this into to the the first equation gives:

$$\frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\Delta H}{T\Delta V}$$

At a solid/gas or liquid/gas phase phase boundary, it is a reasonable to approximate that $\Delta V \approx V_{m,\text{gas}}$ as $\Delta V = V_{m,\text{gas}}-V_{m,\text{condensed}}$ since $V_{m,\text{gas}} \gg V_{m,\text{condensed}}$

Consequently, using the equation of state for a perfect gas and substituting for $\Delta V$ in the second form of the Clapeyron equation, the following result is obtained:

$$\frac1p \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{d}\ln p}{\mathrm{d}T}=\frac{\Delta H}{RT^2}$$

However, why is $\frac1p \frac{\mathrm{d}p}{\mathrm{d}T}=\frac{\mathrm{d}\ln p}{\mathrm{d}T}$?

The following step is the step that I don’t understand In my lecture handout this step is shown as:

$$\int_{p_1}^{p_2}\frac1p\mathrm{d}p=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\mathrm{d}T$$

Why can the left hand side be integrated with respect to $p$, yet the right hand side be integrated with respect to $T$. I cannot make sense of this. Other instances in thermodynamics when integration is used like this, both sides are integrated with respect to the same variable. For example, finding how $H$ varies with $p$. I will quickly go through what I mean without much verbal explanation:

$$\mathrm{d}H=T\mathrm{d}S+V\mathrm{d}p=\left(\frac{\partial H}{\partial S}\right)_p \mathrm{d}S+\left(\frac{\partial H}{\partial p}\right)_S \mathrm{d}p$$

Consequently, for one mole of a perfect gas:

$$V=\left(\frac{\partial H}{\partial p}\right)_S=\frac{RT}p$$

Thus:

$$\int_{p_1}^{p_2}\left(\frac{\partial H}{\partial p}\right)_S\mathrm{d}p=\int_{p_1}^{p_2}\frac{RT}p\mathrm{d}p$$

Since $\left(\frac{\partial H}{\partial p}\right)_S\mathrm{d}p=\mathrm{d}H$ at a constant $T$ (I think this is right – please clarify)

$$\int_{p_1}^{p_2}\mathrm{d}H=RT\int_{p_1}^{p_2}\frac{1}p\mathrm{d}p$$

This can obviously be integrated quite easily but this only illustrates my point regarding the variable that the integration is carried out with respect to. Surely it must be the same on both the RHS and the LHS of the equation?

Answer 1

See Integration by substitution, see also this question and another question at math.SE. This is the rule we have to apply here:

$$\int_{x_0}^{x_1} f(u(x))u'(x)\,\mathrm dx = \int_{u(x_0)}^{u(x_1)} f(u)\,\mathrm du $$

So we go from this equation: $$\frac1p \frac{\mathrm dp}{\mathrm dT}=\frac{\mathrm d\ln p}{\mathrm dT}=\frac{\Delta H}{RT^2},$$ and integrate over temperature: $$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{T_1}^{T_2} \frac{\Delta H}{RT^2}\,\mathrm dT,$$ Now we use the first equation, with $f=\frac{1}{p}$, $p=u$, and $x=T$: $$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{p_1(T_1)}^{p_2(T_2)}\frac1p\,\mathrm dp=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$

Bonus question: Its just the application of the chain rule, namely $$\frac {\mathrm d}{\mathrm dx}z = \frac {\mathrm dz}{\mathrm dy} \frac {\mathrm dy}{\mathrm dx} $$ which, if applied leads to $$\frac{\mathrm d}{\mathrm dT}\ln p = \frac{\mathrm d\ln(p)}{\mathrm dp} \frac{\mathrm dp}{\mathrm dT} = \frac{1}{p}\frac{\mathrm dp}{\mathrm dT} .$$