the Clausius-Clapeyron equation
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Why does water boil at a lower temperature high on a mountain, and why does a pressure cooker cook faster? Both are answered by the slope of the line that separates two phases on a pressure-temperature diagram. The Clausius-Clapeyron equation gives that slope exactly, telling you how much the boiling (or melting) temperature shifts when you change the pressure.
Along a coexistence curve the two phases have equal chemical potentials, mu_1 = mu_2, so their molar Gibbs energies stay equal as you move along the line: dmu_1 = dmu_2. Applying the Gibbs-Duhem relation dmu = -s dT + v dp to each phase gives the Clapeyron equation dp/dT = (s_2 - s_1)/(v_2 - v_1) = L/(T delta v), where L = T delta s is the latent heat and delta v the molar volume change. For a liquid-vapour boundary, approximating the vapour as an ideal gas and neglecting the liquid volume gives the Clausius-Clapeyron form dp/dT = L p/(R T^2), i.e. ln p = -L/(R T) + const.
It underlies phase diagrams, weather (the water-vapour saturation curve), pressure cookers and freeze-drying. Watch the sign of delta v: for most substances the solid is denser than the liquid, so the melting line slopes up, but for water the solid (ice) is less dense, so delta v < 0 on melting and the ice-water line slopes backwards — which is why increasing pressure lowers water's melting point.
Water's latent heat of vaporization is about L = 4.07 x 10^4 J/mol. Using dp/dT = L p/(R T^2) near 373 K and 101 kPa gives a slope of roughly 3.6 kPa/K — so climbing to where the pressure is about 10 kPa lower drops the boiling point by nearly 3 K, and at altitude water boils appreciably below 100 C.
The Clausius-Clapeyron slope quantifies why boiling temperature tracks ambient pressure.
The exact Clapeyron form dp/dT = L/(T delta v) is general; the tidy exponential ln p ~ -L/(RT) is the Clausius-Clapeyron approximation, valid only for a gas phase treated as ideal with the condensed-phase volume neglected and L roughly constant.