During a phase transition, I have read that phase transitions occur at constant pressure and temperature. The explanations I have found both in books and online for this is that all the energy supplied to the system is used to overcome intermolecular bonds, and so no at constant pressure conditions, no temperature change occurs.

My main question regarding this is that why does all the breaking of intermolecular bonds occur at a single temperature? Is there a theoretical reason for this, or is this just an experimental fact?

For example, consider a system which starts its phase transition at a temperature $T_0$, finishes at $T_1$, and is $\frac{T-T_0}{T_1-T_0}\%$ complete at a temperature $T_0<T<T_1$. Why is this not possible?

| cite | improve this question | | | | |

It doesn't have to, it only happens when the process is performed in a slow, reversible manner. That is because usually there's only one specific temperature at which different phases may coexist.

Let's for example have a mixture of ice and liquid water. Liquid water is unstable in temperature lower than $0^\circ C$ and while it may be overcooled, at the slightest stimuli it freezes, turning into ice. Freezing process releases energy. This released energy heats the created ice and remaining water, raising their themperature to $0^\circ C$. Similarily, ice is unstable at temperature greater than $^\circ C$, and if we try to heat it above $0^\circ C$, it will melt, changing its structure to liquid water. Melting process absorbs the heat energy, lowering the temperature of both the created water and the remainign ice to $0^\circ C$.

If we dperform the heating/cooling slowly, then this happens continuously, and the temperature remains constant $0^\circ C$ until all of the water melts/freezes. However, as I've shown in examples above, if the prces is rapid and sudden, the temperature may change - when overcooled water freezes, its temperature grows.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.