Clausius–Clapeyron Relation

Every coexistence line in a phase diagram — the boiling curve, the melting curve, the sublimation curve — has a slope, and that slope is not free. It is fixed, exactly and without approximation, by two discontinuities that jump across the transition: the latent heat and the volume change. This is the content of the Clausius–Clapeyron relation,

\frac{\mathrm{d}p}{\mathrm{d}T} = \frac{L}{T\,\Delta v},

one of the most useful equations in all of thermodynamics. It tells you how a boiling point shifts with pressure, why a pressure cooker cooks faster, why food takes longer to cook on a mountain, and — through a famous sign — why ice can be melted by squeezing it. The derivation is disarmingly short. Two phases coexist only when their chemical potentials are equal, $\mu_1 = \mu_2$ . Demand that this equality persist as you slide along the coexistence line, differentiate using the Gibbs–Duhem relation, and the slope drops out. No equation of state, no model of the liquid, no approximation enters — the result is a rigorous consequence of equilibrium thermodynamics. This article derives the relation cleanly from equality of chemical potentials, integrates it (with two mild approximations) to the vapor-pressure curve $\ln p = -L/RT + \text{const}$ , and reads the physics off the numbers: the ~6% per kelvin rise of water's saturation pressure near room temperature, the 121 °C boiling point inside a 2-atm pressure cooker, and the negative melting slope of water — the anomaly, worth about −134 bar/K, that follows directly from ice being less dense than liquid water. Where a first-order line ends at a critical point, both $L$ and $\Delta v$ vanish and the relation must hand over to the Ehrenfest equations.

Key takeaways

The Clausius–Clapeyron relation gives the slope of any coexistence curve as $\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v) = \Delta s / \Delta v$ , tying it to the latent heat and volume jump.
It follows exactly from demanding that the two phases keep equal chemical potentials, $\mu_1 = \mu_2$ , as $(T,p)$ move along the boundary — no equation of state needed.
With the vapor treated as an ideal gas and the condensed volume neglected, it integrates to $\ln p = -L/RT + \text{const}$ : a straight line of $\ln p$ versus $1/T$ with slope $-L/R$ .
The sign of $\mathrm{d}p/\mathrm{d}T$ tracks the sign of $\Delta v$ : water's melting curve slopes negatively because ice is less dense than liquid water.
Melting curves are nearly vertical (large slope) because $\Delta v$ between two condensed phases is tiny; boiling curves are gentler because the vapor expands enormously.
KEY IDEA: equal chemical potentials, held along the curve, force the slope to be latent heat over $T\,\Delta v$ — geometry of the phase diagram encoded in two thermodynamic jumps.

The core insight

Coexistence means equal chemical potentials. Demanding that equality survive an infinitesimal step $(\mathrm{d}T, \mathrm{d}p)$ along the boundary, and using $\mathrm{d}\mu = -s\,\mathrm{d}T + v\,\mathrm{d}p$ for each phase, forces the slope to equal the ratio of the entropy jump to the volume jump — that is, latent heat over $T\,\Delta v$ .

Where this lives today

The Clausius–Clapeyron relation is a daily working tool across physics, chemistry, engineering, and Earth science. Chemists extract latent heats of vaporization and sublimation by plotting $\ln p$ against $1/T$ and reading the slope — a one-line experiment. Meteorologists live by the ~7% per kelvin rise of saturation vapor pressure: it sets how much moisture the atmosphere holds and underlies the expected intensification of heavy rainfall in a warming world (the "Clausius–Clapeyron scaling" of extreme precipitation). Engineers size refrigeration and steam cycles around the vapor-pressure curves of working fluids. The negative slope of water's melting line — the sign that follows directly from $\Delta v < 0$ — explains regelation and the anomalous behavior of ice under pressure, and its magnetic cousin $\mathrm{d}H_c/\mathrm{d}T = L/(T\,\Delta M)$ governs the phase boundary of type-I superconductors, where the field replaces pressure and the magnetization jump replaces $\Delta v$ . Wherever two phases meet, the same short derivation fixes the geometry of their boundary.

Origins · 1 minÉmile Clapeyron and how this was discovered

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Émile Clapeyron

1799–1864 · French engineer and physicist · A graduate of the École Polytechnique, Clapeyron gave Carnot's abstract cycle its analytic and graphical form, and in an 1834 memoir cast phase equilibrium into the coexistence-slope relation that Clausius later put on an entropic footing.

Benoît Paul Émile Clapeyron trained as a mining and railway engineer and spent years building railways in Russia, but his lasting mark is theoretical. His 1834 memoir Puissance motrice de la chaleur rescued Sadi Carnot's neglected 1824 work, translating it into the pressure–volume diagrams and analytic expressions that made thermodynamics usable — and in it he derived the slope of a coexistence curve in terms of the latent heat and volume change. Two decades later Rudolf Clausius, having introduced entropy, recast Clapeyron's result in the modern form $\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v)$ , and the relation has carried both their names ever since.

History

1824

— Sadi Carnot analyzes the motive power of heat, laying groundwork later built upon for phase equilibria

1834

— Émile Clapeyron derives the coexistence-curve slope in terms of latent heat and volume change

1850

— Rudolf Clausius introduces entropy and recasts the relation as

\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v)

1869

— Thomas Andrews maps the liquid–gas critical point of carbon dioxide, where the coexistence line terminates

1873

— van der Waals's equation of state supplies a model whose coexistence reproduces the Clausius–Clapeyron slope

The physical picture

Think of each phase as a free-energy sheet $\mu(T,p)$ floating over the pressure–temperature plane. The two sheets intersect along a curve: the coexistence line. To walk along that line you must move $(T,p)$ so as to stay on the intersection — which means the two sheets must descend at matching rates as you step. The rate of descent in the $T$ -direction is $-s$ per particle, and in the $p$ -direction is $+v$ per particle (that is the Gibbs–Duhem relation). The direction you are forced to walk is therefore set entirely by the ratios of $s$ and $v$ in the two phases, and that ratio is the Clausius–Clapeyron slope.

Figure

Two chemical-potential sheets μ₁(T,p) and μ₂(T,p) intersecting along the coexistence line; staying on the intersection forces the slope dp/dT = Δs/Δv

Physically: to boil a liquid you must pay latent heat $L$ to liberate each molecule, and the resulting vapor needs far more room, a large $\Delta v$ . Warming the system favors the high-entropy vapor; to keep the two phases balanced you must raise the pressure, and how steeply you must raise it is set by how much heat each conversion costs ( $L$ ) divided by how much room it needs ( $T\,\Delta v$ ). Because vaporization has a huge $\Delta v$ , the boiling curve is comparatively gentle; melting, with a tiny $\Delta v$ between two dense condensed phases, gives a nearly vertical line. The sign is a density statement. If the high-temperature phase is less dense ( $\Delta v > 0$ going low- $T$ →high- $T$ phase), the curve slopes upward. If it is denser, the slope is negative. Water–ice is the celebrated exception: the solid floats, $\Delta v < 0$ on melting, and the melting line tilts backward. This is the punchline plot below — a straight $\ln p$ -versus- $1/T$ vapor-pressure line whose slope gives the latent heat, with the anomalous negative-slope water melting branch called out.

Figure

Punchline: ln p versus 1/T. The vapor-pressure line is straight with slope −L/R (extract L from the slope). The water melting branch is the anomaly — a NEGATIVE dp/dT because ice floats

Near a critical point the two free-energy sheets become tangent rather than crossing transversally; the jumps $\Delta s$ and $\Delta v$ both close to zero, and the sharp slope formula gives way to the smooth physics of critical fluctuations. Beyond the critical point no distinction between phases survives, and there is no coexistence line at all.

The core equations

The relation itself packages three equivalent statements — slope equals entropy-jump over volume-jump, equals latent heat over $T\,\Delta v$ . It is the exact geometric content of a first-order boundary.

\frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta s}{\Delta v} = \frac{L}{T\,\Delta v}

where: $\mathrm{d}p/\mathrm{d}T$ is the slope of the coexistence curve in the pressure–temperature plane, $\Delta s = s_2 - s_1$ the entropy jump, $\Delta v = v_2 - v_1$ the volume jump (per particle or per mole), $L = T\,\Delta s$ the latent heat, and $T$ the coexistence temperature. Read physically: a large latent heat or a small volume change makes for a steep boundary. The whole derivation rests on the coexistence condition — the same one that anchors the latent-heat article. Equal $T$ and $p$ are necessary for thermal and mechanical equilibrium, but particles keep flowing until chemical potentials also match.

\mu_1(T,p) = \mu_2(T,p)

where: $\mu_i$ is the Gibbs free energy per particle of phase $i$ . This single equation, holding at every point of the boundary, is what we will differentiate. The tool that turns the coexistence condition into a slope is the Gibbs–Duhem relation, the per-particle differential of the Gibbs free energy.

\mathrm{d}\mu = -s\,\mathrm{d}T + v\,\mathrm{d}p

where: $s$ and $v$ are the entropy and volume per particle. It follows from $G = N\mu$ and $\mathrm{d}G = -S\,\mathrm{d}T + V\,\mathrm{d}p + \mu\,\mathrm{d}N$ ; recall `aq-sm-free-energies-enthalpy-gibbs`. The two coefficients $-s$ and $+v$ are exactly the slopes of the free-energy sheet that the physical picture invoked. With one mild approximation — a dilute ideal vapor and negligible condensed-phase volume — the slope equation becomes a simple ODE whose integral is the vapor-pressure curve.

\ln p = -\frac{L}{R T} + \text{const}, \qquad \frac{\mathrm{d}\ln p}{\mathrm{d}(1/T)} = -\frac{L}{R}

where: $R$ is the gas constant (use $k_B$ per particle) and $L$ the molar latent heat, treated as constant over the range. A plot of $\ln p$ against $1/T$ is a straight line whose slope $-L/R$ delivers the latent heat directly — the standard laboratory method.

Derivation AClausius–Clapeyron from equality of chemical potentials

This is the cleanest derivation in phase-transition thermodynamics, and it deserves to be seen assembled in full because everything downstream — the vapor-pressure law, the sign of the water melting line, the magnetic analog — rides on it. The strategy is a single idea executed carefully: two phases coexist only where their chemical potentials are equal, so if we insist on staying on the coexistence curve as we vary temperature and pressure, that equality must be preserved to first order. Differentiating it, with the Gibbs–Duhem relation supplying the differentials, forces the slope. No equation of state, no microscopic model, and no approximation enters — the result is exact.

Step 1. Impose the equilibrium condition at every point of the boundary: $\mu_1(T,p) = \mu_2(T,p)$ . Why this and not just equal $T,p$ : equal temperature and pressure ensure thermal and mechanical equilibrium, but particles keep converting until $\mu_1 = \mu_2$ stops net transfer. This selects the unique coexistence curve.

Step 2. Move an infinitesimal step $(\mathrm{d}T, \mathrm{d}p)$ along the curve. Since $\mu_1 = \mu_2$ holds both before and after the step, the changes must match: $\mathrm{d}\mu_1 = \mathrm{d}\mu_2$ . Why this is the crux: it converts a condition on values into a condition on differentials, which is where the slope lives.

Step 3. Insert the Gibbs–Duhem differential for each phase, $\mathrm{d}\mu_i = -s_i\,\mathrm{d}T + v_i\,\mathrm{d}p$ : $-s_1\,\mathrm{d}T + v_1\,\mathrm{d}p = -s_2\,\mathrm{d}T + v_2\,\mathrm{d}p$ . Why Gibbs–Duhem: it expresses how each phase's chemical potential responds to changes in the shared variables $T$ and $p$ , with coefficients that are exactly the measurable entropy and volume.

Step 4. Collect the $\mathrm{d}T$ and $\mathrm{d}p$ terms: $(v_2 - v_1)\,\mathrm{d}p = (s_2 - s_1)\,\mathrm{d}T$ , hence $\mathrm{d}p/\mathrm{d}T = (s_2 - s_1)/(v_2 - v_1) = \Delta s / \Delta v$ . Why the algebra is trivial yet the result is deep: the entire physics was already encoded in Step 1; the rest is bookkeeping.

Step 5. Substitute the latent heat $L = T\,\Delta s$ (from `aq-sm-latent-heat-phase-coexistence`): $\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v)$ . Why this form is the workhorse: $L$ and $\Delta v$ are directly measurable, and the sign of $\Delta v$ immediately fixes the sign of the slope — positive for a less-dense high- $T$ phase, negative for water's floating ice. Dimensionally $[L/(T\,\Delta v)] = \text{J}/(\text{K}\cdot\text{m}^3) = \text{Pa/K}$ , exactly the units of a slope in the $p$ – $T$ plane. Verified symbolically in verify_clausius-clapeyron-relation.py (Steps 3–5: $\mathrm{d}p/\mathrm{d}T = \Delta s/\Delta v = L/(T\,\Delta v)$ from $\mu_1=\mu_2$ and Gibbs–Duhem; dimensional check Pa/K).

Go deeper · Derivation B · 1 minIntegrating to the vapor-pressure curve

Derivation BIntegrating to the vapor-pressure curve

The goal now is to turn the exact-but-implicit slope formula into an explicit prediction for how vapor pressure grows with temperature — the curve that governs boiling points, humidity, and pressure cookers. This requires two physically transparent approximations valid far below the critical point: the vapor is dilute enough to treat as an ideal gas, and its volume so exceeds the liquid's that we can neglect the condensed phase entirely. With those in hand the slope equation becomes a separable first-order ODE that integrates in one line, yielding the celebrated straight-line relation between $\ln p$ and $1/T$ .

Step 1. Approximate the volume jump: $\Delta v = v_{\text{gas}} - v_{\text{liq}} \approx v_{\text{gas}}$ , and use the ideal-gas per-particle volume $v_{\text{gas}} = k_B T / p$ (or $RT/p$ per mole). Why valid: a vapor is typically $\sim 10^3$ times less dense than its liquid, so dropping $v_{\text{liq}}$ costs a fraction of a percent far from the critical point.

Step 2. Substitute into Clausius–Clapeyron: $\mathrm{d}p/\mathrm{d}T = L / (T \cdot RT/p) = L p / (R T^2)$ . Why this shape matters: the pressure now appears on the right, making the equation separable — a hallmark of exponential growth.

Step 3. Separate variables: $\mathrm{d}p/p = (L/R)\,\mathrm{d}T/T^2$ , assuming $L$ roughly constant over the integration interval. Why the constant- $L$ assumption: the latent heat varies only slowly with temperature away from the critical point, so treating it as constant is accurate over modest ranges.

Step 4. Integrate, using $\int \mathrm{d}T/T^2 = -1/T$ : $\ln(p/p_0) = -(L/R)(1/T - 1/T_0)$ , i.e. $p(T) = p_0\,\exp[-(L/R)(1/T - 1/T_0)]$ . Why this is powerful: vapor pressure rises faster than any power of $T$ , and a plot of $\ln p$ against $1/T$ is a straight line of slope $-L/R$ — the routine way to extract a latent heat from measured boiling data. Verified symbolically and numerically in verify_clausius-clapeyron-relation.py ( $\int L/(R T^2)\,\mathrm{d}T = -L/(RT)$ ; slope $\mathrm{d}\ln p/\mathrm{d}(1/T) = -L/R$ ; water ~6%/K near 300 K; pressure-cooker 2-atm boiling at 121 °C).

Limiting cases

Case	Result	Physical meaning
Vapor pressure (gas + condensed)	$\mathrm{d}p/\mathrm{d}T \approx Lp/(RT^2)$ , $p \propto e^{-L/RT}$	Neglect condensed volume, ideal vapor; valid far below $T_c$ where the vapor is dilute
Melting (two condensed phases)	$\mathrm{d}p/\mathrm{d}T = L_{\text{fus}}/(T\,\Delta v_{\text{fus}})$ , steep	$\Delta v_{\text{fus}}$ tiny ⟹ nearly vertical line; sign of slope follows sign of $\Delta v_{\text{fus}}$ (negative for water)
Approach to the critical point	$L \to 0$ , $\Delta v \to 0$ , ratio finite	Both jumps vanish but their ratio stays finite, giving a definite terminal slope
Continuous transition	$L = \Delta v = 0 \Rightarrow$ Ehrenfest: $\dfrac{\mathrm{d}p}{\mathrm{d}T} = \dfrac{\Delta C_p}{T V\,\Delta\alpha}$	With no first-derivative jumps, l'Hôpital gives ratios of second derivatives ( $\alpha$ , $\kappa_T$ , $C_p$ )

Common trapΔs and Δv are jumps between phases, not derivatives along the curve

The most common slip is to read $\Delta s$ and $\Delta v$ as changes along the coexistence line. They are not: they are the discontinuous jumps between the two phases, evaluated at the same point $(T,p)$ on the curve. A second trap is over-trusting the tidy $p \propto e^{-L/RT}$ form — it requires neglecting the liquid volume, treating the vapor as ideal, and holding $L$ constant, all of which fail as the critical point is approached (where $\Delta v$ shrinks and the vapor is no longer dilute). Finally, at a genuinely continuous transition $L = \Delta v = 0$ , so Clausius–Clapeyron gives the indeterminate $0/0$ and must be replaced by the Ehrenfest relations involving jumps in second derivatives.

Worked examples

Example 1 — The pressure cooker: boiling water at 2 atm. We predict the boiling temperature of water inside a pressure cooker held at 2 atm, using the integrated vapor-pressure curve.

Step 1. Reference point: at $p_0 = 1\ \text{atm}$ , water boils at $T_0 = 373.15\ \text{K}$ , with molar latent heat $L = 40\,660\ \text{J/mol}$ and $R = 8.314\ \text{J/mol}\,\cdot\,\text{K}$ .

Step 2. Invert $\ln(p/p_0) = -(L/R)(1/T - 1/T_0)$ for $T$ at $p = 2\ \text{atm}$ : $\dfrac{1}{T} = \dfrac{1}{T_0} - \dfrac{R}{L}\ln\!\dfrac{p}{p_0} = \dfrac{1}{373.15} - \dfrac{8.314}{40\,660}\ln 2$ .

Step 3. Evaluate: $R/L \cdot \ln 2 = (2.045\times 10^{-4})(0.6931) = 1.417\times 10^{-4}\ \text{K}^{-1}$ , so $1/T = 2.680\times 10^{-3} - 1.417\times 10^{-4} = 2.538\times 10^{-3}\ \text{K}^{-1}$ .

Step 4. Invert: $T = \mathbf{394.0\ \text{K} = 121\ ^\circ\text{C}}$ . Water in a 2-atm pressure cooker boils near 121 °C, roughly 21 degrees hotter than in the open — which is why pressure cookers cook so much faster. (Conversely, on a mountain where the pressure is lower, the boiling point drops and food cooks more slowly.)

Example 2 — The sign and steepness of the water melting line. We compute the slope of the ice–water melting curve and confirm it is negative and steep.

Step 1. At $T_m = 273.15\ \text{K}$ , the latent heat of fusion is $L_{\text{fus}} = 6010\ \text{J/mol}$ (positive: melting absorbs heat). We need the volume change.

Step 2. Densities at 0 °C: $\rho_{\text{ice}} = 916.7\ \text{kg/m}^3$ , $\rho_{\text{water}} = 999.8\ \text{kg/m}^3$ ; molar mass $M = 18.015\times 10^{-3}\ \text{kg/mol}$ . Molar volumes: $v_{\text{ice}} = M/\rho_{\text{ice}} = 1.965\times 10^{-5}\ \text{m}^3/\text{mol}$ , $v_{\text{water}} = M/\rho_{\text{water}} = 1.802\times 10^{-5}\ \text{m}^3/\text{mol}$ .

Step 3. The volume change on melting (ice → water) is $\Delta v_{\text{fus}} = v_{\text{water}} - v_{\text{ice}} = -1.633\times 10^{-6}\ \text{m}^3/\text{mol}$ — negative, because ice is less dense than liquid water.

Step 4. Slope: $\dfrac{\mathrm{d}p}{\mathrm{d}T} = \dfrac{L_{\text{fus}}}{T_m\,\Delta v_{\text{fus}}} = \dfrac{6010}{273.15 \times (-1.633\times 10^{-6})} = \mathbf{-1.35\times 10^{7}\ \text{Pa/K} \approx -135\ \text{bar/K}}$ . The slope is negative (raising pressure lowers the melting point) and steep (a 135-bar pressure change shifts the melting point by only 1 K), consistent with the nearly vertical, backward-tilting ice line in every water phase diagram. This is the thermodynamic root of pressure-melting under a skate blade.

Going further

For deeper treatments: Pathria & Beale, Statistical Mechanics, 3rd ed., §4.6; Kardar, Statistical Physics of Particles, Ch. 1–4 (phase coexistence); Reif, Fundamentals of Statistical and Thermal Physics, §8.5; Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 11; Landau & Lifshitz, Statistical Physics Part 1, §81–82; Callen, Thermodynamics and an Introduction to Thermostatistics, Ch. 9. Provenance: the slope formula and Gibbs–Duhem derivation are verifiable in Pathria & Beale §4.6, Callen Ch. 9, Reif §8.5, and Landau–Lifshitz Part 1 §81–82. Numbers verified independently (verify_clausius-clapeyron-relation.py): pressure-cooker boiling at 394.0 K (121 °C) at 2 atm; water melting slope $-135\ \text{bar/K}$ from $\Delta v_{\text{fus}} = -1.63\ \text{cm}^3/\text{mol}$ ; saturation-pressure scaling $\approx 6\%$ /K near 300 K. No nonstandard values are asserted. Prerequisites: First-Order Phase Transitions · Latent Heat and Phase Coexistence · Order Parameter. Leads to: Continuous Phase Transitions · Critical Exponents · Landau Theory of Phase Transitions.

Is the Clausius–Clapeyron relation exact or approximate?

The core relation $\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v)$ is exact — it follows purely from equality of chemical potentials along the coexistence curve and the Gibbs–Duhem relation, with no model of the substance. What is approximate is the integrated vapor-pressure form $p \propto e^{-L/RT}$ , which additionally assumes an ideal vapor, negligible condensed volume, and constant $L$ . Keep the two separate: the slope law is rigorous thermodynamics; the exponential curve is a useful special case.

Why does the boiling point rise with pressure but the melting point of water fall?

Both follow from $\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v)$ with $L > 0$ ; only the sign of $\Delta v$ differs. For boiling, the gas is far less dense than the liquid, so $\Delta v > 0$ and the slope is positive: more pressure ⟹ higher boiling point (the pressure cooker). For water melting, ice is less dense than liquid water, so $\Delta v < 0$ and the slope is negative: more pressure ⟹ lower melting point. Water is anomalous precisely because its solid floats; most substances have a positive melting slope.

How do I get a latent heat out of vapor-pressure data?

Plot the natural logarithm of the measured vapor pressure against $1/T$ . The integrated relation says $\ln p = -L/(RT) + \text{const}$ , so the data fall on a straight line with slope $-L/R$ . Multiply the slope by $-R$ and you have the molar latent heat of vaporization (or of sublimation, if you are on the solid–vapor line). This is the standard calorimetry-free route to a latent heat and is accurate as long as the temperature range is modest enough that $L$ is roughly constant.

What replaces Clausius–Clapeyron at a second-order transition?

At a continuous transition the entropy and volume are continuous ( $L = 0$ , $\Delta v = 0$ ), so the slope formula degenerates into $0/0$ . The resolution, due to Ehrenfest, is to apply l'Hôpital's rule to the coexistence condition, which produces slope relations built from the second derivatives that do jump — heat capacity, thermal expansivity, and compressibility. The result is the pair of Ehrenfest equations, e.g. $\mathrm{d}p/\mathrm{d}T = \Delta C_p / (T V\,\Delta\alpha)$ . (In real continuous transitions these second derivatives often diverge rather than jump, so the Ehrenfest picture is itself only a first approximation — see the critical-exponents article.)

Does the relation apply to magnetic and superconducting transitions?

Yes, with the conjugate variables swapped. For a magnetic system the pressure $p$ is replaced by the field $H$ and the volume $v$ by the magnetization $M$ , giving $\mathrm{d}H_c/\mathrm{d}T = L/(T\,\Delta M)$ for the critical field of a type-I superconductor. The derivation is identical: equality of Gibbs free energies of the normal and superconducting phases along the boundary. The Clausius–Clapeyron structure is generic to any first-order transition once you identify the right pair of conjugate thermodynamic variables.

Why is the sublimation curve steeper than the vaporization curve at the triple point?

Because the sublimation latent heat is the sum of the fusion and vaporization latent heats, $L_{\text{sub}} = L_{\text{fus}} + L_{\text{vap}}$ (energy to go solid→gas equals solid→liquid plus liquid→gas). With $\Delta v \approx v_{\text{gas}}$ nearly the same for both curves near the triple point, the slope $\mathrm{d}p/\mathrm{d}T = L/(T\,\Delta v)$ is larger for sublimation. This is why, at the triple point, the sublimation line rises more steeply than the boiling line — a small but robust consequence of the additivity of latent heats.

Does the relation predict the boiling point drop at altitude?

Directly. Atmospheric pressure falls roughly exponentially with altitude, and the integrated Clausius–Clapeyron curve then lowers the boiling temperature accordingly. At the pressure atop a 5000-m peak (about 0.5 atm), inverting $\ln(p/p_0) = -(L/R)(1/T - 1/T_0)$ gives a boiling point near 83 °C — cool enough that cooking times lengthen noticeably. The same equation that raises the boiling point in a pressure cooker lowers it on a mountain.