12.2 Clapeyron equation

The Clapeyron equation connects vapor pressure, temperature, and phase transitions by starting from the equilibrium condition between two coexisting phases. It lets you calculate vapor pressures at new temperatures, predict how boiling points shift with pressure, and understand the shape of phase diagrams.

Clapeyron Equation and Assumptions

The General Clapeyron Equation

At a phase boundary, the two phases are in thermodynamic equilibrium, meaning their specific Gibbs functions are equal. If you move along the phase boundary (changing both $T$ and $P$ together), the Gibbs functions must remain equal, which leads to the Clapeyron equation:

$\frac{dP}{dT} = \frac{\Delta h}{\,T\,\Delta v\,}$

where $dP/dT$ is the slope of the saturation (phase-boundary) curve, $\Delta h$ is the enthalpy change of the phase transition (e.g., $h_{fg}$ for vaporization), $T$ is the absolute temperature, and $\Delta v$ is the specific volume change between the two phases.

This is an exact thermodynamic relation. It applies to any two-phase equilibrium: solid-liquid, liquid-vapor, or solid-vapor. No approximations have been made yet.

From Clapeyron to Clausius-Clapeyron

The Clausius-Clapeyron equation is an approximate form used specifically for liquid-vapor (or solid-vapor) transitions. It introduces three simplifying assumptions:

1. The vapor behaves as an ideal gas: $v_g = RT/P$ (on a molar basis).
2. The specific volume of the liquid (or solid) is negligible compared to the vapor: $\Delta v \approx v_g$.
3. The enthalpy of vaporization $\Delta h_{vap}$ is approximately constant over the temperature range of interest.

Substituting assumption 1 and 2 into the Clapeyron equation gives:

$\frac{dP}{dT} = \frac{P\,\Delta h_{vap}}{RT^2}$

which can be separated and integrated (using assumption 3) to yield:

$\ln\!\left(\frac{P_2}{P_1}\right) = -\frac{\Delta h_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Keep the distinction clear: the Clapeyron equation is exact; the Clausius-Clapeyron equation is approximate and only valid when those three assumptions hold reasonably well (typically at low-to-moderate pressures, far from the critical point).

Slope of Phase Transition Lines

What the Slope Tells You

The sign and magnitude of $dP/dT$ along a phase boundary reveal how the equilibrium pressure responds to a temperature change.

• Positive slope (most substances): the phase-boundary pressure rises with temperature. Every liquid-vapor and solid-vapor curve has a positive slope because $\Delta h > 0$ and $\Delta v > 0$ for these transitions.
• Negative slope (rare): the solid-liquid line for water has a negative slope because ice is less dense than liquid water, so $\Delta v_{fus} < 0$ while $\Delta h_{fus} > 0$. This is why increasing pressure on ice slightly lowers its melting point.

Calculating the Slope: Step-by-Step

1. Identify the phase transition and look up $\Delta h$ at the temperature and pressure of interest.
2. Find $\Delta v$ for the transition. For liquid-vapor transitions at moderate pressures, you can approximate $\Delta v \approx v_g \approx RT/P$ (ideal gas). For solid-liquid transitions, you need both specific volumes from property tables.
3. Plug into the Clapeyron equation: $dP/dT = \Delta h / (T\,\Delta v)$.

Example: For water vaporizing at 373 K and 1 atm, with $\Delta h_{vap} = 40.7 \text{ kJ/mol}$ and $\Delta v_{vap} \approx 30.2 \text{ L/mol}$:

$\frac{dP}{dT} = \frac{40{,}700 \text{ J/mol}}{373 \text{ K} \times 0.0302 \text{ m}^3/\text{mol}} \approx 3.61 \times 10^3 \text{ Pa/K} \approx 0.0356 \text{ atm/K}$

The positive value confirms that raising the temperature increases the saturation pressure, which matches everyday experience.

Clapeyron Equation Terms

Slope of the Saturation Curve ($dP/dT$)

This is the rate of change of saturation pressure with temperature along the phase boundary. At higher temperatures, $\Delta v_{vap}$ shrinks (the vapor becomes denser) and $T$ increases, so the slope's behavior depends on the competing effects. Near the critical point, the liquid and vapor volumes converge and the Clausius-Clapeyron approximation breaks down entirely.

Enthalpy of Phase Change ($\Delta h$)

This is the energy per unit mass (or per mole) absorbed during the transition at constant temperature and pressure. Stronger intermolecular forces mean a larger $\Delta h$.

• Water: $\Delta h_{vap} = 40.7 \text{ kJ/mol}$ at 100°C (strong hydrogen bonding).
• Ethanol: $\Delta h_{vap} = 38.6 \text{ kJ/mol}$ at 78.4°C.
• $\Delta h$ is not truly constant; it decreases as temperature rises and drops to zero at the critical point.

Absolute Temperature ($T$)

$T$ appears in the denominator, so for a given $\Delta h$ and $\Delta v$, the slope $dP/dT$ decreases at higher temperatures. Physically, this means the saturation curve flattens out as you move toward the critical point.

Volume Change ($\Delta v$)

$\Delta v = v_{\text{vapor}} - v_{\text{liquid}}$ for vaporization. At low pressures the vapor volume dominates (for water at 100°C: $v_g \approx 30.2 \text{ L/mol}$ vs. $v_f \approx 0.018 \text{ L/mol}$). As pressure increases toward the critical point, $v_g$ shrinks and $v_f$ grows until they become equal, and $\Delta v \to 0$.

A larger $\Delta v$ in the denominator reduces the slope, so the common statement "larger $\Delta v$ gives a steeper slope" is only true if you're comparing substances at conditions where $\Delta h$ differences dominate. Always check the full ratio.

Phase Equilibria Applications

Vapor Pressure Calculations

Using the integrated Clausius-Clapeyron equation, you can find the saturation pressure at a new temperature if you know it at a reference state:

$\ln\!\left(\frac{P_2}{P_1}\right) = -\frac{\Delta h_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Example: Find the vapor pressure of water at 90°C, given $P_1 = 1 \text{ atm}$ at $T_1 = 373 \text{ K}$ and $\Delta h_{vap} = 40.7 \text{ kJ/mol}$.

1. Convert: $T_2 = 363 \text{ K}$, $R = 8.314 \text{ J/(mol·K)}$.

2. Compute the right side: $-\frac{40{,}700}{8.314}\left(\frac{1}{363} - \frac{1}{373}\right) = -4893 \times (-7.39 \times 10^{-5}) = 0.362$

   Wait, let's redo this carefully. $\frac{1}{363} - \frac{1}{373} = \frac{373 - 363}{363 \times 373} = \frac{10}{135{,}399} = 7.39 \times 10^{-5} \text{ K}^{-1}$

   The equation has a negative sign in front, and the parenthetical is positive, so: $\ln(P_2/P_1) = -4893 \times 7.39 \times 10^{-5} = -0.362$

3. Solve: $P_2 = 1 \text{ atm} \times e^{-0.362} = 0.70 \text{ atm}$.

This matches steam table data reasonably well (the actual value is about 0.692 atm).

Boiling Point Prediction

You can rearrange the same equation to solve for $T_2$ when you know the desired pressure.

Example: At what temperature does water boil at 0.5 atm?

1. Set $P_2 = 0.5 \text{ atm}$, $P_1 = 1 \text{ atm}$, $T_1 = 373 \text{ K}$.

2. $\ln(0.5) = -0.693 = -\frac{40{,}700}{8.314}\left(\frac{1}{T_2} - \frac{1}{373}\right)$

3. $\frac{1}{T_2} - \frac{1}{373} = \frac{0.693}{4893} = 1.416 \times 10^{-4}$

4. $\frac{1}{T_2} = 2.681 \times 10^{-3} + 1.416 \times 10^{-4} = 2.822 \times 10^{-3}$

5. $T_2 = 354 \text{ K} \approx 81°\text{C}$

This is why water boils at lower temperatures at high altitude, where atmospheric pressure is reduced.

Other Phase Transitions

The exact Clapeyron equation applies to any two-phase boundary, not just liquid-vapor.

• Solid-liquid (melting): For ice at 0°C and 1 atm, $\Delta h_{fus} = 6.01 \text{ kJ/mol}$ and $\Delta v_{fus} = -1.64 \times 10^{-6} \text{ m}^3/\text{mol}$ (negative because ice is less dense than water). The slope is approximately $-134 \text{ atm/K}$. That steep magnitude means you need a very large pressure change to shift the melting point even 1 K.
• Solid-vapor (sublimation): The same form applies, using $\Delta h_{sub}$ and $\Delta v_{sub}$. Since $\Delta h_{sub} = \Delta h_{fus} + \Delta h_{vap}$, the sublimation curve is steeper than the vaporization curve on a $\ln P$ vs. $1/T$ plot.

Phase Diagrams

The Clapeyron equation defines the slopes of every phase-boundary line on a $P$-$T$ phase diagram.

• The triple point is where all three phase-boundary curves meet. For water, this is at 0.01°C and 611.7 Pa (0.00604 atm). All three phases coexist here.
• The critical point terminates the liquid-vapor curve. For water, this is at 374°C and 22.06 MPa (approximately 218 atm). Beyond this point, there is no distinct phase transition between liquid and vapor.
• The negative slope of water's solid-liquid line (due to $\Delta v_{fus} < 0$) is an anomaly compared to most substances, whose solid-liquid lines slope to the right (positive slope).