Phase equilibrium — Clausius-Clapeyron equation: Exercises

Constants used throughout: gas constant $R = 8.314 J mo l^{- 1} K^{- 1}$ .

The two equations we lean on:

Level 1 — Recognition

Goal: pick the right equation and read off signs. No heavy algebra.

L1.1

A liquid and its vapour coexist. Which quantities of the two phases are equal at equilibrium?

Recall Solution

==Temperature $T$ , pressure $P$ , and chemical potential $μ$ == (the molar Gibbs free energy, see Gibbs Free Energy). WHY: if $μ$ differed, particles would flow from the higher- $μ$ phase to the lower- $μ$ one to lower total $G$ — so no true equilibrium. Equal $T$ and $P$ are thermal and mechanical equilibrium; equal $μ$ is chemical equilibrium.

L1.2

For the water→vapour transition (boiling), is the slope $d P / d T$ of the coexistence curve positive or negative? Explain using the sign of $Δ v$ and $L$ .

Recall Solution

Both $L > 0$ (heat absorbed to boil) and $Δ v = v_{g a s} - v_{l i q} > 0$ (vapour is far bigger). So $\frac{d P}{d T} = \frac{L}{T Δ v} > 0.$ The coexistence curve rises — higher $T$ needs higher $P$ to keep the two phases in balance. That matches everyday life: a pressure cooker (high $P$ ) boils water at a higher temperature.

Level 2 — Application

Goal: plug numbers into one equation cleanly.

L2.1

Ammonia has $L_{v a p} = 23.3 kJ/mol$ and boils at $P_{1} = 101.3 kPa$ , $T_{1} = 239.8 K$ . Estimate its vapour pressure at $T_{2} = 250 K$ .

Recall Solution

WHY the vapour form: one phase is a gas, $Δ v \approx v_{g a s} = R T / P$ (see Ideal Gas Law), so we use $ln \frac{P _{2}}{P _{1}} = - \frac{L}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Compute the bracket: $\frac{1}{250} - \frac{1}{239.8} = 4.000 \times 1 0^{- 3} - 4.170 \times 1 0^{- 3} = - 1.702 \times 1 0^{- 4} K^{- 1} .$ Then $ln \frac{P _{2}}{P _{1}} = - \frac{23300}{8.314} \times (- 1.702 \times 1 0^{- 4}) = + 0.4771.$ So $P_{2} = 101.3 e^{0.4771} = 101.3 \times 1.611 = 163.2 kPa$ . Sanity: higher $T$ → higher vapour pressure. ✓

L2.2

The melting curve of a metal has measured slope $d P / d T = 3.0 \times 1 0^{7} Pa/K$ at its melting point $T = 600 K$ , with molar volume change on melting $Δ v = 6.0 \times 1 0^{- 7} m^{3} /mol$ . Find $L_{f u s}$ .

Recall Solution

WHY the exact form: neither phase is a gas, so the ideal-gas shortcut is illegal — use $\frac{d P}{d T} = \frac{L}{T Δ v}$ . Rearrange: $L = \frac{d P}{d T} \cdot T \cdot Δ v = (3.0 \times 1 0^{7}) (600) (6.0 \times 1 0^{- 7}) = 1.08 \times 1 0^{4} J/mol .$ So $L_{f u s} = 10.8 kJ/mol$ . Positive $Δ v$ and positive slope are consistent — this metal contracts on freezing like most solids (unlike water).

Level 3 — Analysis

Goal: reason with the structure of the equation, not just plug in.

L3.1

A student measures vapour pressures of a liquid and plots $ln P$ against $1/ T$ , getting a straight line of slope $- 4200 K$ and intercept $+ 22.5$ . (a) Find $L_{v a p}$ . (b) Predict $P$ at $T = 350 K$ (with $P$ in Pa).

Recall Solution

WHY a straight line: integrating the vapour form gives $ln P = - \frac{L}{R} \cdot \frac{1}{T} + C$ . Comparing to $y = m x + c$ with $x = 1/ T$ : the slope is $m = - L / R$ and $c$ is the intercept. This linearisation is the whole point of the experiment.

Figure — Phase equilibrium — Clausius-Clapeyron equation

(a)

L = - R \times (slope) = - 8.314 \times (- 4200) = 3.492 \times 1 0^{4} J/mol = 34.9 kJ/mol

. (b) At

T = 350

1/ T = 2.857 \times 1 0^{- 3}

ln P = - 4200 \times 2.857 \times 1 0^{- 3} + 22.5 = - 12.00 + 22.5 = 10.50.

P = e^{10.50} = 3.63 \times 1 0^{4} Pa \approx 36.3 kPa .

Sanity check on the graph: a steeper (more negative) slope would mean a bigger latent heat — the vapour pressure climbs faster with temperature. The red line in the figure is exactly this fit.

L3.2

Explain, using $\frac{d P}{d T} = \frac{L}{T Δ v}$ , why the sublimation curve (solid→vapour) is steeper than the vapourisation curve (liquid→vapour) near the triple point.

Recall Solution

Both curves have $Δ v \approx v_{g a s}$ (the gas dwarfs both the solid and the liquid), and both share nearly the same $T$ near the triple point (see Phase Diagrams and Triple Point). So the only differing factor is $L$ : $(\frac{d P}{d T})_{sub} / (\frac{d P}{d T})_{vap} \approx \frac{L _{sub}}{L _{vap}} .$ Because $L_{sub} = L_{fus} + L_{vap}$ (to go solid→gas you both melt and boil, energetically — see Latent Heat), we have $L_{sub} > L_{vap}$ . Therefore the sublimation slope is larger: the sublimation curve is steeper. This is a real, checkable feature of every phase diagram's triple point.

Level 4 — Synthesis

Goal: combine CC with entropy, Gibbs energy, or two-step reasoning.

L4.1

Water boils at $373 K$ at $101.3 kPa$ with $L_{v a p} = 40.7 kJ/mol$ . (a) Using $Δ s = L / T$ , find the molar entropy of vapourisation. (b) Using the exact form with $Δ v \approx v_{g a s} = R T / P$ , find $d P / d T$ at the boiling point. (c) Use that slope to estimate the boiling pressure at $375 K$ (linear approximation) and compare to the full log formula.

Recall Solution

(a) The transition is isothermal and reversible, so from Entropy and the Second Law, $Δ s = \frac{L}{T} = \frac{40700}{373} = 109.1 J mo l^{- 1} K^{- 1} .$ (This is close to Trouton's rule, $\approx 88$ ; water is above it because of hydrogen bonding.) (b) First the gas molar volume: $v_{g a s} = \frac{R T}{P} = \frac{8.314 \times 373}{101300} = 0.03061 m^{3} /mol .$ Then $\frac{d P}{d T} = \frac{L}{T Δ v} = \frac{40700}{373 \times 0.03061} = 3565 Pa/K .$ (c) Linear estimate over $Δ T = 2 K$ : $P (375) \approx 101300 + 3565 \times 2 = 108430 Pa = 108.4 kPa .$ Full log formula for comparison: $ln \frac{P _{2}}{101.3} = - \frac{40700}{8.314} (\frac{1}{375} - \frac{1}{373}) = - 4895 \times (- 1.430 \times 1 0^{- 5}) = 0.07001,$ $P_{2} = 101.3 e^{0.07001} = 108.6 kPa .$ Agreement: $108.4$ vs $108.6$ kPa. The linear tangent and the exact curve match over a small 2 K step — exactly what "the slope is $d P / d T$ " promises. They'd diverge over a large step because $ln P$ is curved.

L4.2

Show that the exact Clausius-Clapeyron equation follows from requiring $d μ_{1} = d μ_{2}$ along the curve, then explain in one sentence which thermodynamic identity guarantees $d g = - s d T + v d P$ .

Recall Solution

Along coexistence $μ_{1} = μ_{2}$ everywhere, so their changes match: $d μ_{1} = d μ_{2}$ . With $d μ = d g = - s d T + v d P$ for each phase (from First Law of Thermodynamics plugged into $d G = d U - T d S - S d T + P d V + V d P$ , where $d U = T d S - P d V$ cancels the middle terms): $- s_{1} d T + v_{1} d P = - s_{2} d T + v_{2} d P .$ Collect: $(v_{1} - v_{2}) d P = (s_{1} - s_{2}) d T$ , hence $\frac{d P}{d T} = \frac{s _{2} - s _{1}}{v _{2} - v _{1}} = \frac{Δ s}{Δ v} = \frac{L}{T Δ v} .$ One-sentence identity: $d g = - s d T + v d P$ is the statement that $g$ has natural variables $T$ and $P$ , which is exactly the setup from which the Maxwell Relations $(\frac{\partial s}{\partial P})_{T} = - (\frac{\partial v}{\partial T})_{P}$ is derived by equality of mixed second derivatives of $g$ .

Level 5 — Mastery

Goal: full multi-step problems with sign subtleties and interpretation.

L5.1 — The ice-skating myth, quantitatively

Ice melts at $T = 273.15 K$ at $101.3 kPa$ , with $L_{f u s} = 6010 J/mol$ , $v_{w a t er} = 18.00 \times 1 0^{- 6}$ , $v_{i ce} = 19.66 \times 1 0^{- 6} m^{3} /mol$ . (a) Compute $d T / d P$ (how much the melting temperature shifts per pascal). (b) A skater exerts about $Δ P = 5.0 \times 1 0^{6} Pa$ under the blade. Estimate the melting-point drop $Δ T$ . (c) Is pressure-melting a convincing explanation for skating? Comment.

Recall Solution

(a) $Δ v = v_{w a t er} - v_{i ce} = (18.00 - 19.66) \times 1 0^{- 6} = - 1.66 \times 1 0^{- 6} m^{3} /mol$ (negative — ice is less dense). $\frac{d P}{d T} = \frac{L}{T Δ v} = \frac{6010}{273.15 \times ( - 1.66 \times 1 0 ^{- 6} )} = - 1.326 \times 1 0^{7} Pa/K .$ Invert: $\frac{d T}{d P} = \frac{1}{- 1.326 \times 1 0 ^{7}} = - 7.54 \times 1 0^{- 8} K/Pa .$ (b) $Δ T \approx \frac{d T}{d P} Δ P = (- 7.54 \times 1 0^{- 8}) (5.0 \times 1 0^{6}) = - 0.377 K$ . (c) A drop of only $\approx 0.38 K$ — melting temperature falls to about $- 0.38 °C$ . Skating happens far colder than that, so pressure-melting is too weak to explain skating. The real culprit is frictional heating plus a naturally pre-melted surface layer. But the sign is the physics gem: because $Δ v < 0$ , pressure lowers the melting point at all — a direct fingerprint of water's anomaly.

L5.2 — Two-point latent heat from field data

A mountaineer measures water boiling at $T_{1} = 373.15 K$ ( $P_{1} = 101.3 kPa$ ) at sea level and $T_{2} = 344.15 K$ ( $P_{2} = 31.0 kPa$ ) on a summit. Assuming the vapour form with constant $L$ , extract $L_{v a p}$ .

Recall Solution

WHY the vapour form: liquid↔vapour, gas volume dominates. Solve $ln \frac{P _{2}}{P _{1}} = - \frac{L}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ for $L$ : $L = - R \frac{l n ( P _{2} / P _{1} )}{\frac{1}{T _{2}} - \frac{1}{T _{1}}} .$ Numerator: $ln \frac{31.0}{101.3} = ln (0.3060) = - 1.1841$ . Bracket: $\frac{1}{344.15} - \frac{1}{373.15} = 2.9057 \times 1 0^{- 3} - 2.6799 \times 1 0^{- 3} = 2.258 \times 1 0^{- 4} K^{- 1}$ . $L = - 8.314 \times \frac{- 1.1841}{2.258 \times 1 0 ^{- 4}} = 8.314 \times 5244 = 4.36 \times 1 0^{4} J/mol = 43.6 kJ/mol .$ Sanity: the accepted $L_{v a p}$ for water is $\approx 40.7 kJ/mol$ ; our two-point estimate of $43.6$ is close — the small excess comes from $L$ drifting with temperature over a 29 K span. Two pressures, two temperatures → a latent heat, entirely from CC.

Recall One-line recap of the whole page

Exact form for any transition; log form only when one phase is a gas; the slope of $ln P$ vs $1/ T$ is $- L / R$ ; and signs matter — a negative $Δ v$ (water) flips the melting curve. ::: If you can classify a problem into "gas phase present or not" and track the sign of $Δ v$ , every CC exercise falls out.