Clausius-Clapeyron Vapor Pressure Calculator
Solve the Clausius-Clapeyron equation to find a substance's enthalpy of vaporization from two vapor-pressure readings, predict its vapor pressure at a new temperature, or find the temperature — including the normal boiling point — where it reaches a target pressure, with full step-by-step working.
Pick what you want to solve for, then enter the values you already know.
Enter two vapor-pressure/temperature readings for the same substance — they can come from a data table or two lab measurements.
Result
Enthalpy of vaporization
≈ 8.1789 kcal/mol
293.15 K
T1 in Kelvin
373.15 K
T2 in Kelvin
127.65 °C
Predicted normal boiling point
8.314 J/(mol·K)
Gas constant used
Step-by-Step Clausius-Clapeyron Calculation
Here's exactly how this answer was calculated, one step at a time.
Step 1: Convert both temperatures to Kelvin
T1 = 293.15 K; T2 = 373.15 KStep 2: Take the natural log of the pressure ratio
ln(P2 ÷ P1) = ln(355.1 ÷ 17.5) = 3.0102Step 3: Apply the two-point Clausius-Clapeyron equation
R is the universal gas constant, 8.314 J/(mol·K), in the form the Clausius-Clapeyron equation uses.
ΔHvap = R × ln(P2/P1) ÷ (1/T1 − 1/T2) = 8.314 × 3.0102 ÷ (0.0034112 − 0.0026799)Step 4: Result
ΔHvap ≈ 34.2207 kJ/mol (8.1789 kcal/mol)Step 5: Bonus: predicted normal boiling point (P = 1 atm)
Using point 1 as the reference and the ΔHvap just calculated, this is the temperature at which the substance's vapor pressure would reach standard atmospheric pressure.
Tb ≈ 127.65 °C
Clausius-Clapeyron Vapor Pressure Calculator: Three Tools in One
This free calculator solves the Clausius-Clapeyron equation from any direction you need it. Give it two vapor-pressure readings at two different temperatures and it works out the substance's enthalpy of vaporization (ΔHvap) directly, no rearranging required. Give it a known ΔHvap and one reference point and it predicts the vapor pressure at any other temperature, or — flip it around — solves for the exact temperature at which the substance reaches a target vapor pressure, including its normal boiling point at 1 atmosphere.
Every calculation shows its full step-by-step working, so whether you're a general or physical chemistry student checking a homework problem, or an engineer sketching a quick estimate for a distillation, refrigeration, or vacuum process, you can see exactly how each number was produced and copy the full solution for your notes.
What Is the Clausius-Clapeyron Equation?
The Clausius-Clapeyron equation captures a simple physical fact: every pure liquid's vapor pressure rises as temperature rises, and it does so in a very specific, predictable way — roughly exponentially, not linearly. Named after Benoît Paul Émile Clapeyron and Rudolf Clausius, the two-point form most commonly used in chemistry classes is ln(P2/P1) = −(ΔHvap/R) × (1/T2 − 1/T1), where R is the universal gas constant, 8.314 J/(mol·K).
The core idea is that raising the temperature gives more liquid molecules enough kinetic energy to overcome the intermolecular forces holding them in the liquid phase and escape into the vapor. How steeply vapor pressure climbs with temperature depends directly on ΔHvap — the amount of energy it takes to vaporize one mole of the substance — which is exactly why that quantity sits at the center of the equation.
Finding ΔHvap From Two Data Points
One of the most practical uses of this equation is running it in reverse: if you already have two vapor-pressure measurements for a substance at two different temperatures, you can solve directly for ΔHvap without ever needing a calorimeter. This is exactly what this calculator's first mode does — plug in P1 at T1 and P2 at T2, and it rearranges the equation to ΔHvap = R × ln(P2/P1) ÷ (1/T1 − 1/T2).
This approach is extremely common in a lab setting: a student or researcher measures a liquid's vapor pressure at just two convenient temperatures (often room temperature and boiling point) and back-calculates ΔHvap, rather than running a full calorimetry experiment. Because ΔHvap is assumed constant across the temperature range in this simplified equation, the two points you choose should ideally sit reasonably close together for the most accurate result.
Predicting Vapor Pressure at a New Temperature
The second mode flips the problem around: given a known ΔHvap and one reference vapor-pressure/temperature pair, it predicts the vapor pressure at any other temperature. This is the classic textbook application — a data table gives a substance's ΔHvap and its vapor pressure at one standard temperature, and the question asks what the vapor pressure will be at some new temperature.
This is genuinely useful outside the classroom too. Anyone working with volatile solvents, refrigerants, or fuels in a lab, a plant, or a piece of process equipment often needs to know how much a vapor pressure will shift with a temperature change — for solvent storage and handling safety, refrigerant cycle design, or estimating evaporation rates at a different ambient temperature than a reference data sheet provides.
Finding the Temperature — and the Normal Boiling Point
The third mode solves for the missing temperature instead: given ΔHvap, a reference point, and a target pressure, it finds exactly what temperature produces that pressure. Set the target pressure to standard atmospheric pressure (1 atm ≈ 760 mmHg ≈ 101.325 kPa) and this becomes a direct calculation of the substance's normal boiling point — the temperature at which its vapor pressure finally matches the surrounding atmospheric pressure and it can boil freely, not just evaporate from the surface.
This same logic is exactly how altitude affects boiling: atmospheric pressure drops as you climb higher, so the vapor pressure needed to reach a full boil is lower, and water boils at a temperature below 100°C on a mountaintop. Swap in the local atmospheric pressure as your target and this calculator answers that question directly too.
Why Kelvin, and Why the Ideal Gas Assumption Matters
The equation only works correctly with absolute temperature (Kelvin), because it's derived by substituting the ideal gas law into the exact Clapeyron equation — and the ideal gas law is built entirely on absolute temperature. This calculator handles that conversion automatically whenever you work in Celsius, so you never have to remember to add 273.15 yourself.
The ideal-gas substitution is also where the equation's core approximation lives: it assumes the vapor above the liquid behaves like an ideal gas, and that the liquid's own molar volume is negligible compared to the vapor's. Both assumptions are excellent well below a substance's critical temperature and pressure, but start to break down as conditions approach the critical point, where liquid and vapor properties converge.
Real-World Uses of the Clausius-Clapeyron Equation
Beyond the chemistry classroom, this equation shows up constantly in engineering and everyday science. Refrigeration and air-conditioning engineers use it to characterize refrigerants and predict how their vapor pressure — and therefore cooling capacity — shifts across an operating temperature range. Meteorologists use a closely related version to describe how the atmosphere's saturation water-vapor pressure rises with temperature, which underpins humidity calculations and cloud formation models.
Vacuum technology and freeze-drying (lyophilization) both depend on knowing precisely what pressure needs to be maintained to keep a substance evaporating (or subliming) at a chosen low temperature, and distillation process design uses it to estimate how a component's volatility shifts across a column's temperature profile — all standard, everyday applications of the same equation this calculator solves.
Common Mistakes When Using This Equation
The most frequent error is mixing temperature units — plugging Celsius directly into 1/T terms instead of converting to Kelvin first, which silently produces a wildly wrong answer without throwing any obvious error. A close second is sign confusion in the exponent: because vapor pressure always increases with temperature, a correctly worked calculation should always show ΔHvap coming out positive when P2 (the higher-temperature pressure) is greater than P1.
It's also easy to forget that ΔHvap is only assumed constant over the range being used — stretching a calculation across an extremely wide temperature span, or across a phase change like melting that this equation doesn't cover, will produce a noticeably less accurate answer than using two points that sit reasonably close to the conditions you actually care about.
Clausius-Clapeyron Calculator: Quick Reference and Disclaimer
Quick formula: ln(P2/P1) = −(ΔHvap/R) × (1/T2 − 1/T1), with R = 8.314 J/(mol·K), ΔHvap in J/mol, and both temperatures in Kelvin. Rearranged forms: ΔHvap = R × ln(P2/P1) ÷ (1/T1 − 1/T2); P2 = P1 × exp[−(ΔHvap/R) × (1/T2 − 1/T1)]; 1/T2 = 1/T1 − (R/ΔHvap) × ln(P2/P1).
This calculator is intended for educational and general reference use, including general and physical chemistry coursework and preliminary engineering estimates. It assumes ideal-gas vapor behavior and a temperature-independent ΔHvap, and does not replace measured vapor-pressure data, an Antoine-equation fit, or professional process engineering software for real industrial design work.
Frequently Asked Questions
What is the Clausius-Clapeyron equation?
The Clausius-Clapeyron equation describes how a pure substance's vapor pressure changes with temperature along its liquid-vapor (or solid-vapor) equilibrium line. In its most common two-point form it's written as ln(P2/P1) = −(ΔHvap/R) × (1/T2 − 1/T1), where P1 and P2 are vapor pressures at absolute temperatures T1 and T2, ΔHvap is the molar enthalpy of vaporization, and R is the universal gas constant. It comes from combining the Clapeyron equation with the ideal gas law and assuming ΔHvap barely changes over the temperature range in question.
What units does ΔHvap need to be in?
This calculator expects ΔHvap in kJ/mol and internally converts it to J/mol to match R = 8.314 J/(mol·K). If a textbook or data table gives you ΔHvap in kcal/mol, multiply by 4.184 first; if it's given per gram rather than per mole, multiply by the substance's molar mass to get J/mol or kJ/mol before entering it.
Why do I need to convert temperature to Kelvin?
The Clausius-Clapeyron equation is derived from the ideal gas law, which only holds when temperature is measured on an absolute scale where zero really means zero molecular motion. Using Celsius or Fahrenheit directly would make the 1/T terms meaningless. This calculator converts any Celsius input to Kelvin automatically before running the math.
Can I use this calculator to predict a normal boiling point?
Yes. A substance's normal boiling point is simply the temperature at which its vapor pressure equals standard atmospheric pressure (1 atm, roughly 760 mmHg or 101.325 kPa). In the 'Find the Temperature' mode, set the target pressure P2 to 1 atm's equivalent in whatever unit you're working in, and the calculator solves directly for that boiling point. The 'Find ΔHvap' mode also shows this as a bonus result automatically.
How accurate is the Clausius-Clapeyron equation?
It's an approximation, not an exact law. The full Clapeyron equation is exact, but getting to the simpler Clausius-Clapeyron form requires two assumptions: that the vapor behaves as an ideal gas, and that ΔHvap stays roughly constant over the temperature range used. Both assumptions hold up well for modest temperature ranges well below a substance's critical point, which is why this equation is so widely used, but it drifts from real measured data as temperatures get closer to the critical point or span a very wide range.
What's the difference between this and Raoult's Law?
Raoult's Law describes how a solvent's vapor pressure drops when something else is dissolved in it, at a single fixed temperature. The Clausius-Clapeyron equation describes how a pure substance's own vapor pressure changes as temperature changes. They answer two completely different questions — concentration effects versus temperature effects — and are often used together: Clausius-Clapeyron gives you a pure solvent's vapor pressure at whatever temperature you're working at, and Raoult's Law then tells you how a solution made from that solvent compares.
Where does the equation come from?
It starts from the exact Clapeyron equation, dP/dT = ΔH / (TΔV), which relates the slope of a phase boundary to the enthalpy and volume change of the phase transition. For a liquid-to-vapor transition, the vapor's molar volume is vastly larger than the liquid's, so ΔV is approximated as just the vapor's volume, and that volume is replaced using the ideal gas law, V = RT/P. Substituting and integrating between two temperature-pressure points gives the two-point logarithmic form this calculator uses.
Can this be used for sublimation (solid to vapor) too?
Yes, the same mathematical form applies to any phase change into a vapor, including sublimation — you'd just use the enthalpy of sublimation (ΔHsub) in place of ΔHvap. Since ΔHsub is always larger than ΔHvap (it includes the energy of melting as well as vaporizing), don't mix the two: use vapor-pressure data measured for the same phase transition you're solving for.