The table in Vapor Pressure and Boiling shows water's vapor pressure climbing faster and faster as it warms. The Clausius–Clapeyron equation turns that curve into a straight line and lets you predict vapor pressures and boiling points from two numbers. It uses The Number e and Continuous Growth and the properties of logarithms.
Key idea
$$\ln\frac{P_2}{P_1} = -\frac{\Delta H_\text{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ with temperatures in kelvins, $R = 8.314\ \text{J/(mol·K)}$ and $\Delta H_\text{vap}$ in J/mol. Equivalently, a graph of $\ln P$ against $1/T$ is a straight line with slope $-\Delta H_\text{vap}/R$.
Why it works
To escape, a molecule needs at least about $\Delta H_\text{vap}$ per mole of energy. The share of molecules with that much energy grows like $e^{-\Delta H_\text{vap}/RT}$ (the same Boltzmann factor behind the speed distributions you will meet with gases). Vapor pressure is proportional to that share, so $P = Ae^{-\Delta H_\text{vap}/RT}$, and taking logarithms gives $\ln P = \ln A - \dfrac{\Delta H_\text{vap}}{R}\cdot\dfrac{1}{T}$, a line in $1/T$. Writing it for two temperatures and subtracting removes the unknown $A$ and gives the equation above. (It assumes $\Delta H_\text{vap}$ stays constant, a good approximation over a modest range.)
Worked example: Vapor pressure at another temperature
Water's vapor pressure is $760\ \text{mmHg}$ at $100.0^\circ\text{C}$, and $\Delta H_\text{vap} = 40.7\ \text{kJ/mol}$. Estimate its vapor pressure at $80.0^\circ\text{C}$.
- Kelvins: $T_1 = 373.15\ \text{K}$, $T_2 = 353.15\ \text{K}$. Joules: $\Delta H_\text{vap} = 40{,}700\ \text{J/mol}$.
- $$\dfrac{1}{T_2} - \dfrac{1}{T_1} = \dfrac{1}{353.15} - \dfrac{1}{373.15} = 1.518 \times 10^{-4}\ \text{K}^{-1}$$
- $$\ln\dfrac{P_2}{760} = -\dfrac{40{,}700}{8.314}(1.518 \times 10^{-4}) = -0.743$$
- $P_2 = 760\,e^{-0.743} = 362\ \text{mmHg}$. (Measured: 355 mmHg; $\Delta H_\text{vap}$ is not quite constant.)
Answer: About $362\ \text{mmHg}$.
Worked example: Finding the enthalpy of vaporization
A liquid has a vapor pressure of $100.0\ \text{mmHg}$ at $25.0^\circ\text{C}$ and $400.0\ \text{mmHg}$ at $60.0^\circ\text{C}$. Find $\Delta H_\text{vap}$.
- $\ln\dfrac{400.0}{100.0} = \ln 4 = 1.386$.
- $\dfrac{1}{333.15} - \dfrac{1}{298.15} = -3.524 \times 10^{-4}\ \text{K}^{-1}$.
- $1.386 = -\dfrac{\Delta H_\text{vap}}{8.314}(-3.524 \times 10^{-4})$, so $$\Delta H_\text{vap} = \dfrac{1.386 \times 8.314}{3.524 \times 10^{-4}} = 3.27 \times 10^{4}\ \text{J/mol}$$
Answer: $32.7\ \text{kJ/mol}$
Worked example: A boiling point at lower pressure
At what temperature does water boil when the pressure is $0.700\ \text{atm}$? ($\Delta H_\text{vap} = 40.7\ \text{kJ/mol}$; normal boiling point $373.15\ \text{K}$.)
- At the new boiling point, $P_2 = 0.700\ \text{atm}$; $P_1 = 1.000\ \text{atm}$ at $373.15\ \text{K}$.
- $$\ln 0.700 = -0.3567 = -\dfrac{40{,}700}{8.314}\left(\dfrac{1}{T_2} - \dfrac{1}{373.15}\right)$$
- $$\dfrac{1}{T_2} = \dfrac{1}{373.15} + \dfrac{0.3567 \times 8.314}{40{,}700} = 0.0027528$$ so $T_2 = 363.3\ \text{K}$.
Answer: $363.3\ \text{K}$, about $90^\circ\text{C}$.
Common mistake
Using Celsius temperatures, or mixing kJ/mol with $R$ in J/(mol·K). $1/80 - 1/100$ is a completely different number from $1/353.15 - 1/373.15$, and $40.7$ in place of $40{,}700$ makes the exponent 1000 times too small. Check: before you compute, write every $T$ in kelvins and $\Delta H_\text{vap}$ in J/mol. A sensible answer has $P_2 > P_1$ when $T_2 > T_1$.
With calculus: the equation comes from integrating $\dfrac{d\ln P}{dT} = \dfrac{\Delta H_\text{vap}}{RT^2}$, a separable differential equation (Separable Equations).