Water's solid-liquid coexistence curve has a negative slope in the P-T phase diagram, meaning that increasing pressure lowers ice's melting point. What thermodynamic property of water produces this anomaly?
AWater has an unusually large latent heat of fusion, which makes the numerator L in the Clausius-Clapeyron equation dominate.
BIce is less dense than liquid water, so ΔV = V_liquid − V_solid is negative, giving dP/dT = L/(T·ΔV) a negative slope.
CThe melting temperature of ice lies below the triple point, inverting the normal sign convention.
DWater's entropy change at melting is negative, unlike most substances.
The Clausius-Clapeyron equation dP/dT = L/(T·ΔV) shows that the sign of the slope is determined by the sign of ΔV (volume change upon phase transition). For most substances, the solid is denser than the liquid, so ΔV > 0 and the slope is positive. Ice is anomalous: it is less dense than liquid water (due to its open hydrogen-bonded crystal structure), so ΔV < 0, and the slope is negative. L > 0 (melting absorbs heat) and T > 0 always, so the sign of ΔV alone flips the slope.
Question 2 Multiple Choice
The Clausius-Clapeyron equation is derived by requiring which condition to hold as you move along a phase coexistence boundary?
AThe entropy of the two phases must remain equal at every point on the boundary.
BThe temperature must remain constant as pressure varies along the boundary.
CThe Gibbs free energies of the two coexisting phases must remain equal as T and P change together.
DThe volume change between phases must remain constant along the boundary.
Two phases coexist when they are in thermodynamic equilibrium: G_1 = G_2. Moving along the coexistence curve means staying on this equilibrium condition, so dG_1 = dG_2 as well. Using dG = −S dT + V dP for each phase and equating gives (S_2 − S_1)dT = (V_2 − V_1)dP, which rearranges to dP/dT = ΔS/ΔV = L/(T·ΔV). The constraint is equal Gibbs free energies, not equal entropies — in fact, the entropy difference ΔS between phases is what produces the latent heat L = TΔS.
Question 3 True / False
A pressure cooker cooks food faster because the elevated pressure inside lowers the boiling point of water, so the water boils at a lower temperature and less energy is needed.
TTrue
FFalse
Answer: False
This is exactly backwards. For liquid-vapor transitions, ΔV > 0 (vapor is much larger than liquid) and L > 0, so the Clausius-Clapeyron equation gives dP/dT > 0: increasing pressure RAISES the boiling point. A pressure cooker works by sealing steam inside, raising the pressure above atmospheric, which raises the boiling point above 100°C to roughly 120°C. This allows cooking at higher temperatures, speeding up chemical reactions in the food. Lower pressure (like at high altitude) lowers the boiling point, slowing cooking.
Question 4 True / False
A plot of ln P vs. 1/T for the vapor pressure of a liquid should be approximately linear, with a slope that gives a direct measure of the latent heat of vaporization.
TTrue
FFalse
Answer: True
Integrating the approximate Clausius-Clapeyron equation d(ln P)/dT = L/RT² gives ln P = −L/(RT) + const, so ln P is linear in 1/T with slope −L/R. This is the standard method for measuring latent heats experimentally: measure vapor pressure at several temperatures, plot ln P vs. 1/T, and extract L from the slope. The approximation assumes ideal gas vapor and negligible liquid volume, which holds well far from the critical point.
Question 5 Short Answer
Explain where the Clausius-Clapeyron equation comes from. What physical condition defines a coexistence curve, and how does that condition lead to the relation dP/dT = L/(T·ΔV)?
Think about your answer, then reveal below.
Model answer: A coexistence curve is the locus of (T, P) points where two phases are in thermodynamic equilibrium, meaning their Gibbs free energies are equal: G_1(T,P) = G_2(T,P). Moving along the curve means staying on this equality, so dG_1 = dG_2. Using dG = −S dT + V dP, equality of differentials gives −S_1 dT + V_1 dP = −S_2 dT + V_2 dP, which rearranges to dP/dT = (S_2 − S_1)/(V_2 − V_1) = ΔS/ΔV. Since latent heat L = T·ΔS at a first-order transition, this becomes dP/dT = L/(T·ΔV).
The key insight is that the coexistence condition (equal Gibbs energies) must be maintained as you move along the boundary. The equation is not an empirical fit — it is an exact thermodynamic consequence of that equilibrium condition.