Phase equilibria

Counting freedom with the Gibbs phase rule. For a system with $C$ independent components and $P$ phases coexisting in equilibrium, the number of intensive variables (like $T$ and $P$) that can still be varied independently without disturbing which phases are present is

where: $C$ is the number of independent components, $P$ is the number of coexisting phases, and $2$ counts temperature and pressure as the two intensive variables available in a pure $PVT$ system. For a one-component system ($C=1$), a single phase gives $F=2$ — a full 2-D region of the phase diagram where both $T$ and $P$ roam freely. Two phases in equilibrium give $F=1$ — a 1-D curve, where choosing $T$ fixes $P$. Three phases together give $F=0$ — a single unmovable point.

Deriving the boundary slope. Along any two-phase coexistence curve, $\mu^{(\alpha)}(T,P)=\mu^{(\beta)}(T,P)$ holds at every point on the curve, not just one. Differentiating this equality using $d\mu=-S_m\,dT+V_m\,dP$ per mole (from $dG=-S\,dT+V\,dP$) for each phase, and using $\Delta S_{trs}=\Delta H_{trs}/T_{trs}$ for a reversible transition, gives the Clausius–Clapeyron equation:

where: $\Delta H_{trs}$ and $\Delta V_{trs}$ are the molar enthalpy and volume change on crossing the boundary (fusion, vaporisation, or sublimation), and $T$ is the coexistence temperature. This single formula is valid for any of the three boundaries on a one-component phase diagram — only the transition being described changes.

Integrating for the vapour branches. When a gas phase is involved, $\Delta V_{trs}\approx V_m^{(gas)}=RT/P$ (the condensed-phase volume is negligible by comparison), so the Clausius–Clapeyron equation becomes a separable differential equation in $P$ and $T$ that integrates, for roughly constant $\Delta H_{vap}$, to

where: $P_1,T_1$ is one known vapour-pressure/temperature point, $P_2,T_2$ is the point you want, and $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$. This lets you predict vapour pressure at a new temperature from one reference point, or extract $\Delta H_{vap}$ from two measured points.

Water's normal boiling point is $373.15\ \text{K}$ at $1\ \text{atm}$, and its molar enthalpy of vaporisation is $\Delta H_{vap}=40.7\ \text{kJ mol}^{-1}$. Estimate the vapour pressure of water at $363.15\ \text{K}$ ($90^\circ\text{C}$), then use the Gibbs phase rule to explain why this vapour-pressure curve is a line, not a region, on the phase diagram.

Step 1. Apply the integrated Clausius–Clapeyron equation: $\ln\dfrac{P_2}{P_1} = -\dfrac{40700}{8.314}\left(\dfrac{1}{363.15}-\dfrac{1}{373.15}\right) = -4895.6\times7.39\times10^{-5} = -0.3618$.

Step 2. Exponentiate: $\dfrac{P_2}{P_1}=e^{-0.3618}=0.6965$, so $P_2 = 0.6965\times101{,}325\ \text{Pa} = 70{,}590\ \text{Pa}\approx\mathbf{0.697\ atm}$. This is close to the experimental $\approx0.692\ \text{atm}$ at $90^\circ\text{C}$; the small gap comes from treating $\Delta H_{vap}$ as constant and the vapour as ideal.

Step 3. Apply the Gibbs phase rule to the liquid-vapour boundary: two phases coexist ($P_{phases}=2$) for a one-component system ($C=1$), so $F=1-2+2=1$. One degree of freedom means choosing $T$ fixes $P$ — you cannot pick both independently and stay on the boundary — which is exactly why the vaporisation curve traced by the equation above is a 1-D curve, not a 2-D region. The vapour pressure at $90^\circ\text{C}$ is about $0.697\ \text{atm}$, and $F=1$ on that curve confirms $T$ alone determines it.

Competitive stretch

At the triple point of water, three phases coexist simultaneously ($P_{phases}=3$), so $F=1-3+2=0$: neither $T$ nor $P$ can be varied at all without losing a phase, pinning the triple point to a unique $(273.16\ \text{K},\ 611.657\ \text{Pa})$ — the reason it is used as a fixed calibration reference for temperature scales.

• Applying $\Delta V_{trs}\approx V_m^{(gas)}$ (valid for vaporisation/sublimation) to the fusion curve — both phases there are condensed, so $\Delta V_{fus}$ is small and can even be negative, which is why ice's melting curve has a negative slope via $dP/dT=\Delta H_{fus}/(T\Delta V_{fus})$.
• Confusing "$F$ degrees of freedom" with a count of phases or components — $F$ counts independently variable intensive variables like $T$ and $P$, not how many substances or phases are present.
• Treating $\Delta H_{vap}$ as truly constant over a wide temperature range in the integrated equation — it shrinks toward zero as $T$ approaches the critical point, where the phase distinction itself disappears.
• Trying to extend this one-component treatment to binary or ternary alloy phase diagrams — CY11001 scopes this topic to pure substances; multi-component phase diagrams need a different framework.

Ask the tutor: if the step from $\mu^{(\alpha)}=\mu^{(\beta)}$ along a curve to the Clausius–Clapeyron slope formula felt rushed, ask the AI tutor to expand $d\mu^{(\alpha)}=d\mu^{(\beta)}$ term by term.