Effect of pressure on transition temperature

Question

I was wondering if someone can explain why the answer to the following question is volume instead of the others.

If pressure has no effect on the transition temperature between two crystalline forms of matter the two forms have the same molar
(a) volume
(b) energy
(c) enthalpy
(d) entropy
(e) heat capacity

I know the answer is volume but I'm not sure which equation that comes from. I was thinking maybe the law of reduced properties but at this point I'm just guessing based on the answer and I know that that particular equation is for gases. Does anyone know if there's an analog for crystalline forms of matter or if there's a more intuitive explanation? Thanks!

Answer 1

The mathematical answer is based on Clapeyron's equation (written in what follows in form inverted relative to the more usual):

$$\frac{dT}{dp}= \frac{\Delta V}{\Delta S}$$

which must hold on any point on the phase coexistence line.

If $\frac{dT}{dp}=0$ then it follows that $\Delta V=0$.

Answer 2

You can understand this using Le Chatelier's principle. In transitioning from a smaller-volume form to a larger-volume form, the substance takes up more space, and in doing so has to push against the external pressure to make more room for itself. The greater the external pressure, the more unfavorable this transition becomes.

Conversely, if there is no difference in volume, the two forms are equally favorable with respect to external pressure. Hence changes in pressure have no effect on the transition.