# Constant Pressure and Temperature during Phase Change

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}

• 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. Sep 20, 2019 at 7:02
• @Poutnik I think what you meant to say is that it is the change in enthalpy at constant pressure. Sep 20, 2019 at 11:59
• 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. Sep 20, 2019 at 12:01
• @Chet Miller. Yes, sure. :-) Sep 20, 2019 at 12:18
• 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 Sep 20, 2019 at 17:40

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.