Worked examples — Phase equilibrium — Clausius-Clapeyron equation

This child page of Phase equilibrium — Clausius-Clapeyron equation does one thing: it drills every kind of situation the equation can hand you. Before we compute anything, let us lay out the full landscape so you never meet a case we did not show.

---

The scenario matrix

We will use two master forms throughout:

If any of these is unfamiliar revisit the parent note, Latent Heat and Ideal Gas Law first.

#	Case class	Which phase pair	Form to use	Sign of $\Delta v$	Sign of $dP/dT$	Covered by
A	Normal vaporisation	liquid → vapour	vapour (log)	$+$ (huge)	$+$	Ex 1
B	Sublimation	solid → vapour	vapour (log)	$+$ (huge)	$+$	Ex 2
C	Normal melting	solid → liquid, solid denser	exact	$+$	$+$	Ex 3
D	Anomalous melting	ice → water, solid less dense	exact	$-$	$-$	Ex 4
E	Measuring $L$ from a graph	any vapour line	slope of log form	—	—	Ex 5
F	Degenerate: $\Delta v\to 0$	near a critical point	exact (limiting)	$\to 0$	$\to \infty$	Ex 6
G	Real-world word problem	cooking at altitude	vapour (log)	$+$	$+$	Ex 7
H	Exam twist: mixed units / find $L$	vaporisation, solve for $L$	vapour (log)	$+$	$+$	Ex 8

The two "axes" of the matrix are: (i) is a vapour involved? (chooses the form) and (ii) what is the sign of $\Delta v$? (chooses the sign of the slope). Every other detail is just plugging numbers.

Look at the figure: the amber curves rise steeply (vapour cases A, B, G — small slope in $P$ but huge relative change), the cyan solid–liquid lines are nearly vertical (cases C, D) because $\Delta v$ is tiny, and case D leans left (negative slope). Keep this picture in mind — each example lives on one of these lines.

---

Example 1 — Case A: normal vaporisation

1. Pick the form. One phase is vapour and $v_{gas}\gg v_{liquid}$, so we use the log form. Why this step? Cell A of the matrix says vapour → integrated form; the exact form would need $v_{gas}(T)$ explicitly.
2. Write the equation with numbers. $\ln \frac{60}{101.3}=-\frac{40700}{8.314}\left(\frac{1}{T_2}-\frac{1}{373}\right)$ Why this step? Substituting is the whole computation; keep $L$ in J/mol so it matches $R$.
3. Evaluate the left side. $\ln (0.5923)=-0.5238$. Why this step? $\ln$ converts the pressure ratio into the linear quantity that the formula predicts.
4. Solve for $1/T_2$. With $L/R=4895.4\text{K}$: $\frac{1}{T_2}-\frac{1}{373}=\frac{0.5238}{4895.4}=1.0700\times 10^{-4}$ $\frac{1}{T_2}=2.6810\times 10^{-3}+1.0700\times 10^{-4}=2.7880\times 10^{-3}$ Why this step? We isolate $1/T_2$ because the formula is linear in $1/T$, not in $T$.
5. Invert. $T_2=358.7\text{K}$.

Verify: Units — the argument of $\ln$ is the ratio $P_2/P_1$, which is already dimensionless; separately the exponent factor $\frac{L}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$ must also be dimensionless. Check it: $\frac{L}{R}$ is $\frac{\text{J/mol}}{{\text{J mol}}^{-1}{\text{K}}^{-1}}=\text{K}$, times $\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$ in ${\text{K}}^{-1}$, gives $\text{K}\cdot {\text{K}}^{-1}=$ dimensionless. ✓ Sanity — $358.7<373$: lower pressure gives a lower boiling point, matching the forecast. ✓

---

Example 2 — Case B: sublimation

1. Pick the form. Solid → vapour still has one gas phase with dominant volume, so the log form applies exactly as for boiling. Why this step? The matrix treats vaporisation and sublimation identically — the only difference is which condensed phase we started from.
2. Rearrange to find $P_2$. $\ln \frac{P_2}{P_1}=-\frac{L}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$ Why this step? Here the unknown is a pressure, not a temperature, so we solve for $P_2$ instead.
3. Compute the bracket. $\frac{1}{205}-\frac{1}{195}=4.8780\times 10^{-3}-5.1282\times 10^{-3}=-2.5019\times 10^{-4}{\text{K}}^{-1}$. Why this step? Note it is negative (we raised $T$, so $1/T$ dropped) — the two minus signs will combine to a positive exponent.
4. Multiply. $-\frac{26100}{8.314}\times (-2.5019\times 10^{-4})=-3139.3\times (-2.5019\times 10^{-4})=0.7855$. Why this step? This is $\ln (P_2/P_1)$; positive value means $P_2>P_1$, as forecast.
5. Exponentiate. $P_2=100\times e^{0.7855}=100\times 2.1935=219.4\text{kPa}$.

Verify: $P_2>P_1$, consistent with heating toward the gas phase. ✓ If we had accidentally flipped a sign we would have got $P_2\approx 45\text{kPa}$ (pressure falling on heating) which is physically absurd — a good sign-check.

---

Example 3 — Case C: normal melting (positive $\Delta v$)

1. Choose the exact form. No gas is present, so the log form is forbidden. Why this step? Cell C: solid–liquid means we must use $dP/dT=L/(T\Delta v)$ with real volumes.
2. Compute $\Delta v$ as final minus initial (solid → liquid). $\Delta v=v_{liquid}-v_{solid}=(8.20-8.00)\times 10^{-6}=+2.0\times 10^{-7}{\text{m}}^3/\text{mol}$ Why this step? The direction of $L$ (heat into melting) must match the direction of $\Delta v$ (solid→liquid), else the sign is wrong.
3. Plug in. $\frac{dP}{dT}=\frac{9000}{600\times 2.0\times 10^{-7}}=\frac{9000}{1.2\times 10^{-4}}=7.5\times 10^7\text{Pa/K}$ Why this step? This is the raw evaluation of the exact formula.

Verify: Units — J/mol ÷ (K · m³/mol) = J/(K·m³) = Pa/K. ✓ Sign — positive, matching a "normal" denser solid. ✓ Magnitude — tens of MPa per kelvin: melting curves are steep (nearly vertical) because $\Delta v$ is tiny, exactly the cyan lines in Figure s01.

---

Example 4 — Case D: the anomaly (negative $\Delta v$)

1. $\Delta v$ for melting (ice → water). $\Delta v=v_{water}-v_{ice}=(18.00-19.66)\times 10^{-6}=-1.66\times 10^{-6}{\text{m}}^3/\text{mol}$ Why this step? The sign of $\Delta v$ is the entire point of this cell — keep the direction solid→liquid.
2. Slope. $\frac{dP}{dT}=\frac{6010}{273\times (-1.66\times 10^{-6})}=\frac{6010}{-4.5318\times 10^{-4}}=-1.326\times 10^7\text{Pa/K}$ Why this step? Negative, as forecast — the melting curve leans left (the cyan D-line in Figure s01).
3. Invert to get $dT/dP$ and scale. $dT/dP=1/(dP/dT)=-7.54\times 10^{-8}\text{K/Pa}$. To drop $T$ by $1\text{K}$: $\Delta P=\frac{-1}{-7.54\times 10^{-8}}=1.326\times 10^7\text{Pa}\approx 131\text{atm}$ Why this step? This converts the abstract slope into a tangible "how hard must I squeeze?" number.

Verify: $131\text{atm}$ to shift the melting point by just $1\text{K}$ shows the effect is real but small — consistent with the parent note's remark that pressure-melting under a skate blade is a tiny effect. ✓ Sign negative ✓.

---

Example 5 — Case E: extracting $L$ from a straight-line fit

1. Compute the two axis values. $\ln P_1=\ln 40000=10.5966$, $\ln P_2=\ln 90000=11.4076$. And $1/T_1=2.8571\times 10^{-3}$, $1/T_2=2.6316\times 10^{-3}$. Why this step? We must plot in the linearising coordinates $(1/T,\ln P)$; raw $P$ vs $T$ would curve.
2. Slope of the line. $\text{slope}=\frac{\ln P_2-\ln P_1}{1/T_2-1/T_1}=\frac{11.4076-10.5966}{2.6316\times 10^{-3}-2.8571\times 10^{-3}}=\frac{0.8110}{-2.2556\times 10^{-4}}=-3595.4\text{K}$ Why this step? The formula $\ln P=-\frac{L}{R}\cdot \frac{1}{T}+C$ means this rise-over-run is $-L/R$.
3. Recover $L$. $L=-R\times \text{slope}=-8.314\times (-3595.4)=29.90\times 10^3\text{J/mol}\approx 29.9\text{kJ/mol}$ Why this step? Multiplying by $-R$ undoes the slope definition.

Verify: Units — slope in K, $R$ in J/(mol·K), product in J/mol. ✓ Magnitude $\sim 30\text{kJ/mol}$ is a believable vaporisation heat. ✓

---

Example 6 — Case F: degenerate limit $\Delta v\to 0$

1. Evaluate at the given point (exact form only). $\frac{dP}{dT}=\frac{2000}{600\times 5.0\times 10^{-6}}=\frac{2000}{3.0\times 10^{-3}}=6.667\times 10^5\text{Pa/K}$ Why this step? The log form is invalid here — near criticality the vapour is far from ideal and $v_{liquid}$ is not negligible.
2. Take the limit. As $\Delta v\to 0$ with $L$ staying finite relative to it, $dP/dT=L/(T\Delta v)\to \infty$. Why this step? This is the degenerate cell — the formula does not "break", it predicts a diverging slope, which experiment confirms as the curve turns vertical then ends at the critical point.
3. State the boundary of validity. Exactly at the critical point both $L=0$ and $\Delta v=0$, so $dP/dT$ becomes the indeterminate $0/0$ — Clausius-Clapeyron alone can no longer decide the slope, and the coexistence line simply terminates.

Verify: At the given numbers, $6.667\times 10^5\text{Pa/K}$ is finite and positive. ✓ The $\Delta v\to 0$ divergence and the $0/0$ endpoint together explain why phase diagrams show the vapour curve stopping at a point — see Phase Diagrams and Triple Point.

---

Example 7 — Case G: real-world word problem

1. Pick the form and write it with symbols, then substitute. A gas phase is present, so use the vapour log form and solve for $P_2$: $\ln \frac{P_2}{P_1}=-\frac{L}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=-\frac{40700}{8.314}\left(\frac{1}{361}-\frac{1}{373}\right)$ Why this step? Cell G of the matrix is Case A dressed as a story — vapour involved ⇒ integrated log form; writing the symbolic equation first keeps the substitution pattern identical to Examples 1 and 2.
2. Bracket. $\frac{1}{361}-\frac{1}{373}=2.7701\times 10^{-3}-2.6810\times 10^{-3}=8.9086\times 10^{-5}{\text{K}}^{-1}$. Why this step? Positive because we lowered the temperature ($1/T$ rises).
3. Exponent. $-\frac{40700}{8.314}\times 8.9086\times 10^{-5}=-4895.4\times 8.9086\times 10^{-5}=-0.43613$. Why this step? This is $\ln (P_2/P_1)$; negative means $P_2<P_1$.
4. Solve. $P_2=101.3\times e^{-0.43613}=101.3\times 0.6466=65.5\text{kPa}$.

Verify: $65.5\text{kPa}$ corresponds to roughly $3500\text{m}$ altitude — a realistic high mountain. ✓ Lower pressure ↔ lower boiling point, matching the forecast, and explaining why rice takes longer to cook up high (cooler boiling water).

---

Example 8 — Case H: exam twist, solve for $L$ with mixed units

1. Recognise you only need the ratio. $P_2/P_1=2$, so $\ln (P_2/P_1)=\ln 2=0.6931$. Why this step? Cell H of the matrix is a vaporisation problem solved backwards for $L$; the log form depends only on the pressure ratio, so absolute pressures are never needed — that is the trap.
2. Write and rearrange for $L$. $\ln 2=-\frac{L}{R}\left(\frac{1}{320}-\frac{1}{300}\right)\Rightarrow L=\frac{-R\ln 2}{\frac{1}{320}-\frac{1}{300}}$ Why this step? $L$ is the only unknown; isolate it algebraically.
3. Compute the bracket. $\frac{1}{320}-\frac{1}{300}=3.1250\times 10^{-3}-3.3333\times 10^{-3}=-2.0833\times 10^{-4}{\text{K}}^{-1}$. Why this step? Negative because $T$ increased — this negative combines with the leading minus to give positive $L$.
4. Finish. $L=\frac{-8.314\times 0.6931}{-2.0833\times 10^{-4}}=\frac{-5.7625}{-2.0833\times 10^{-4}}=2.766\times 10^4\text{J/mol}\approx 27.7\text{kJ/mol}$

Verify: Positive $L\approx 27.7\text{kJ/mol}$ — physically sensible. ✓ Cross-check the sign logic: temperature up + pressure up ⇒ positive $L$, correct for evaporation (Latent Heat). ✓

---

Recall Which form for which cell?

Is a vapour involved? ::: Yes → integrated log form ($\ln P_2/P_1=-\frac{L}{R}(1/T_2-1/T_1)$); No (solid–liquid) → exact form $dP/dT=L/(T\Delta v)$. What decides the SIGN of the melting-curve slope? ::: The sign of $\Delta v=v_{liquid}-v_{solid}$; negative for water (ice less dense) gives a negative slope. What happens to $dP/dT$ as $\Delta v\to 0$ near a critical point? ::: It diverges (curve turns vertical); exactly at the critical point $L=0$ and $\Delta v=0$ give $0/0$ and the line ends. In the "doubling pressure" exam trick, what info do you actually need? ::: Only the pressure ratio $P_2/P_1$ (here $=2$), not the absolute pressures.

See also Gibbs Free Energy, Entropy and the Second Law, First Law of Thermodynamics and Maxwell Relations for where the exact form itself comes from.