5.3 Clausius-Clapeyron Equation

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation connects vapor pressure to temperature along a phase boundary. It gives you a quantitative way to predict how vapor pressure changes with temperature, and to extract thermodynamic quantities like $\Delta H_{vap}$ from experimental data. For Physical Chemistry II, you need to understand where this equation comes from, how to use it for calculations, and where it breaks down.

Clausius-Clapeyron Equation Derivation

Thermodynamic Principles

The starting point is the Clapeyron equation, which describes the slope of any phase boundary on a P-T diagram:

$\frac{dP}{dT} = \frac{\Delta S}{\Delta V} = \frac{\Delta H}{T \Delta V}$

This comes directly from the condition that, at equilibrium between two phases, the chemical potentials must be equal: $\mu_{\alpha} = \mu_{\beta}$. If you move along the coexistence curve (changing both T and P while staying at equilibrium), the changes in chemical potential for each phase must match. Setting $d\mu_{\alpha} = d\mu_{\beta}$ and substituting the fundamental relation $d\mu = -S_m dT + V_m dP$ gives you the Clapeyron equation above.

From Clapeyron to Clausius-Clapeyron

The exact Clapeyron equation applies to any phase transition, but it's hard to use directly because $\Delta V$ and $\Delta H$ both depend on T and P. For liquid-vapor or solid-vapor transitions, two simplifying assumptions turn it into something much more practical:

1. The molar volume of the condensed phase is negligible compared to the vapor: $\Delta V \approx V_{m,\text{gas}}$
2. The vapor behaves as an ideal gas: $V_{m,\text{gas}} = RT/P$

Substituting these into the Clapeyron equation:

$\frac{dP}{dT} = \frac{\Delta H_{vap} \cdot P}{RT^2}$

Rearranging gives the differential form of the Clausius-Clapeyron equation:

$\frac{d(\ln P)}{dT} = \frac{\Delta H_{vap}}{RT^2}$

This is the exact relationship you'll use for calculations and plotting.

Vapor Pressure Calculation with Clausius-Clapeyron

Equation and Parameters

If you assume $\Delta H_{vap}$ is constant over your temperature range, you can integrate the differential form between two states $(T_1, P_1)$ and $(T_2, P_2)$ to get the integrated Clausius-Clapeyron equation:

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

where:

• $P_1, P_2$ are vapor pressures at temperatures $T_1, T_2$ (in K)
• $\Delta H_{vap}$ is the enthalpy of vaporization (J/mol)
• $R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}$

To find an unknown vapor pressure, you need three of the four quantities ($P_1, P_2, T_1, T_2$) plus $\Delta H_{vap}$.

Example: Water has a vapor pressure of 1.00 atm at 373 K and $\Delta H_{vap} = 40.7 \text{ kJ/mol}$. To find the vapor pressure at 350 K:

$\ln\!\left(\frac{P_2}{1.00}\right) = \frac{40700}{8.314}\left(\frac{1}{373} - \frac{1}{350}\right) = 4894 \times (-1.77 \times 10^{-4}) = -0.866$

$P_2 = e^{-0.866} = 0.42 \text{ atm}$

Application and Plotting

A plot of $\ln(P)$ vs. $1/T$ should yield a straight line with slope $= -\Delta H_{vap}/R$. This is the Clausius-Clapeyron plot, and it's the standard way to extract $\Delta H_{vap}$ from experimental vapor pressure data. A linear fit confirms that $\Delta H_{vap}$ is roughly constant over that temperature range; curvature indicates it's changing.

Be careful with units: temperatures must be in kelvin, and your choice of pressure units doesn't affect the slope (since $\ln(P_2/P_1)$ is dimensionless), but it does affect the intercept if you're writing the equation in the form $\ln P = -\Delta H_{vap}/RT + C$.

Enthalpy Estimation with Clausius-Clapeyron

Rearranging the Equation

Given vapor pressure measurements at two or more temperatures, you can solve for $\Delta H_{vap}$:

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

Steps for a two-point calculation:

1. Convert both temperatures to kelvin.

2. Compute $\ln(P_2/P_1)$.

3. Compute $1/T_2 - 1/T_1$.

4. Divide and multiply by $-R$.

Data Requirements and Assumptions

• You need vapor pressure data at a minimum of two temperatures. With more data points, plot $\ln P$ vs. $1/T$ and extract $\Delta H_{vap}$ from the slope, which averages out measurement error.
• The result you get is an average $\Delta H_{vap}$ over the temperature range used. If $\Delta H_{vap}$ varies significantly with temperature (as it does over wide ranges or near the critical point), this average may not represent the value at any single temperature well.
• Accuracy depends on the quality of your pressure measurements. Manometer readings, for instance, introduce uncertainty that propagates directly into your enthalpy estimate. With only two data points, there's no way to assess scatter, so more measurements are always better.

Limitations of the Clausius-Clapeyron Equation

Assumptions and Constraints

Every simplification that made the equation tractable also limits where it works. The three key assumptions are:

• Constant $\Delta H_{vap}$: Valid over small temperature ranges. Over large ranges, $\Delta H_{vap}$ decreases with increasing temperature and drops to zero at the critical point. If you see curvature in your $\ln P$ vs. $1/T$ plot, this assumption is failing.
• Ideal gas behavior of the vapor: Reasonable at low pressures and temperatures well below the critical temperature. At high pressures or near the critical point, intermolecular interactions in the vapor become significant and the ideal gas law breaks down.
• Negligible liquid molar volume: The molar volume of a typical liquid is roughly 1000× smaller than its vapor at 1 atm, so this is usually fine. But as pressure increases and you approach the critical point, liquid and vapor densities converge, and this approximation fails.

Situations Where It Does Not Apply

• Near the critical point: All three assumptions break down simultaneously. Liquid and vapor become indistinguishable, $\Delta H_{vap} \to 0$, and the vapor is far from ideal.
• Non-equilibrium conditions: The derivation assumes a reversible phase transition at equilibrium. Kinetic effects like superheating, supercooling, or nucleation barriers aren't captured.
• Other phase transitions: The equation is derived specifically for liquid-vapor and solid-vapor equilibria. It does not apply to solid-liquid transitions (melting) or solid-solid transitions, because those don't involve a gas phase where the ideal gas and negligible-volume approximations make sense. For those transitions, you need the full Clapeyron equation.
• Multicomponent systems: The equation as written applies to a pure substance. Mixtures require modified treatments (e.g., Raoult's law combined with Clausius-Clapeyron for each component).