One of the solution guides to a question I was working on said that pressure and temperature is constant for a phase change. I understand why temperature is always constant for a phase change, but don't understand why pressure is. If the heat coming in and out of the system and the volume is changing, shouldn't the pressure change too, or does the solution approximate that pressure is constant because it changes by a marginal amount?

The question is:

1.00 mol of steam is reversibly compressed to form water(l) at 100 degrees celcius ($\Delta H_\text{vap}(\text{water}) = \pu{2.259 kJ/g}$). Calculate the entropy of the phase change.

I attached the solutions on the answer key:

Constant pressure and isothermal. Only a change of state occurs. As steam condenses, heat is lost from the system.

$\begin{align}&\Delta H_{\text{condensation}} = \pu{-2.259 kJ/g}=q_{\text{rev}}\\ &\Delta S_{\text{condensation}} =q_{\text{rev}}/T = q_p/T = \Delta H_{\text{condensation}}/T_{\text{condensation}} \\ &m_{\text{steam}}=\pu{1.00 mol}\times \pu{18.0 g/mol} = \pu{18.0 g} \\ &\Delta H_{\text{condensation}} = q_p=(\pu{-2259 J/g})(\pu{18.0 g})=\pu{-40700 J}\\&\Delta S_{\text{condensation}} = q_p/T=\pu{-40700 J}/\pu{373 K}=\pu{-109 J/K}\end{align}$

  • $\begingroup$ Is system constrained by a closed container, or, does surrounding pressure change ? If not, pressure is constant. Note that $\Delta H$ means change of internal energy at constant pressure. $\endgroup$
    – Poutnik
    Sep 20, 2019 at 7:02
  • $\begingroup$ @Poutnik I think what you meant to say is that it is the change in enthalpy at constant pressure. $\endgroup$ Sep 20, 2019 at 11:59
  • $\begingroup$ This has nothing to do with any process you apply to a substance. You're looking at the wrong side of the equation. It has to do with the thermodynamic equilibrium states of a substance. $\endgroup$ Sep 20, 2019 at 12:01
  • $\begingroup$ @Chet Miller. Yes, sure. :-) $\endgroup$
    – Poutnik
    Sep 20, 2019 at 12:18
  • $\begingroup$ sorry the question asked to calculate the entropy of the phase change. The problem didn't say whether it was in a closed container @Poutnik $\endgroup$
    – Timothy
    Sep 20, 2019 at 17:40

1 Answer 1


For a pure substance, the phase rule implies that at coexistence of two phases only one intensive variable (pressure or temperature) can be altered independently, the other being dependent on the choice of the independent variable. So the answer is yes and no, you can choose to alter one of the two intensive variables and still retain two-phase equilibrium, but the choice of one intensive variable determines the value of the other. This fundamental principle derives from the conditions of thermodynamic equilibrium necessary for coexistence of the two phases. The set of $(p,T)$ points on the coexistence line can be described in terms of the Clapeyron equation $dp/dT = \Delta S/\Delta V$ and a single reference point $(p_{ref},T_{ref})$ on the line.

Consider a pure substance in a diathermal container whose volume we can alter at will, and kept at a fixed temperature. Assume we begin with the container completely filled with a condensed phase (say liquid) of the substance, that is, the volume is too small (or alternately, the pressure too high) to allow vapor to form. We have however chosen a temperature that corresponds to a point on the coexistence line at a lower pressure (or greater volume) than that currently in the vessel. You can bring the substance to an equilibrium between two phases by changing the volume until the pressure in the container equals the equilibrium vapour pressure - a fixed pressure at that temperature. You can then continue to change the volume (all the while keeping the temperature constant). The proportion of the two phases (their volumes) will readjust to accommodate the new total volume, with heat exchanging with the surroundings to keep the temperature constant, and the pressure returning to the value at equilibrium - the same vapour pressure - provided the volume change is not too big. Naturally if you increase or decrease the volume too much you will end up with a single phase and a different pressure below or above the vapour pressure.

Remember: you don't have to have two phases. But if you want to observe coexistence at a chosen temperature, you have no choice wrt the pressure that will ensure coexistence. What's cool is that the proportions of the two phases can be altered within a certain limited range of total volume so that p and T remain constant. I recommend consulting a p-V-T phase diagram and inspecting it carefully to see this graphically. The following is a slice in the p-V dimensions which I borrowed from this site:

enter image description here

  • $\begingroup$ Sorry, I'm still a bit confused. Does pv=nrt still apply here or do we have to use the van der waals equation? If P and T are constant then how could v change? $\endgroup$
    – Timothy
    Sep 20, 2019 at 17:45
  • $\begingroup$ The equation of state for the gas will only apply to the gas component. The total volume will be a sum of the volume of gas and volume of liquid (which, being usually poorly compressible, will remain approx constant). Maybe this answer chemistry.stackexchange.com/questions/119024/… to a related post will help. $\endgroup$
    – Buck Thorn
    Sep 20, 2019 at 17:48
  • $\begingroup$ v can change because the proportions of liquid and gas change. $\endgroup$
    – Buck Thorn
    Sep 20, 2019 at 17:49
  • $\begingroup$ Does this happen for every reversible phase change? $\endgroup$
    – Timothy
    Sep 22, 2019 at 2:48
  • $\begingroup$ for a pure substance, yes. The phase rule dictates that p/T are mutually dependent, so if you pick T, P is also fixed. $\endgroup$
    – Buck Thorn
    Sep 22, 2019 at 10:58

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