Clausius–Clapeyron Calculator
Estimate the vapor pressure of a liquid at any temperature from one known pressure–temperature pair and the enthalpy of vaporization, using the integrated Clausius–Clapeyron equation. Enter the known conditions, the target temperature, and ΔHvap.
Estimate Vapor Pressure at a New Temperature
Clausius–Clapeyron Formula
Where:
- \(P_1\) — vapor pressure at the known temperature (any pressure unit)
- \(P_2\) — vapor pressure at the target temperature (same unit as \(P_1\))
- \(T_1\) — known temperature (kelvin)
- \(T_2\) — target temperature (kelvin)
- \(\Delta H_{vap}\) — enthalpy of vaporization (J/mol)
- \(R\) — gas constant, \(8.314\ \text{J/(mol·K)}\)
This is the integrated form of the Clausius–Clapeyron equation. It assumes the vapor behaves as an ideal gas and that the enthalpy of vaporization is constant over the temperature range, and it predicts how vapor pressure changes exponentially with temperature.
Clausius–Clapeyron – Calculation Example
Example: water boils at 100 °C at 760 mmHg (1 atm). Estimate the vapor pressure of water at 25 °C, using \(\Delta H_{vap} = 40.7\ \text{kJ/mol}\).
Convert temperatures to kelvin: \(T_1 = 373.15\ \text{K}\), \(T_2 = 298.15\ \text{K}\), and \(\Delta H_{vap} = 40\,700\ \text{J/mol}\).
\[ \ln\!\left(\frac{P_2}{760}\right) = -\frac{40\,700}{8.314}\left(\frac{1}{298.15} - \frac{1}{373.15}\right) = -4894.9 \times 0.000674 = -3.300 \]
\[ P_2 = 760 \times e^{-3.300} \]
Result: \(P_2 \approx 28.04\ \text{mmHg}\), which is about 3.74 kPa or 0.0369 atm. (The measured value is 23.76 mmHg — the difference comes from ΔHvap not being perfectly constant; see the notes below.)
How to Use the Calculator
- Enter the Known Vapor Pressure (P₁) of the liquid and select its unit (mmHg, kPa, atm, Pa, or bar). A boiling point at a known pressure works well here.
- Enter the Known Temperature (T₁) at which that pressure was measured, in °C, K, or °F.
- Enter the Target Temperature (T₂) at which you want to estimate the vapor pressure, in any of the same temperature units.
- Enter the Enthalpy of Vaporization (ΔHvap) and choose kJ/mol, J/mol, or kcal/mol. Reference tables for common liquids list this value.
- Read the results: the estimated vapor pressure at T₂ is shown in your chosen pressure unit, plus conversions to kPa, atm, and mmHg.
What Is the Clausius–Clapeyron Equation?
The Clausius–Clapeyron equation describes the relationship between the vapor pressure of a liquid and temperature. Vapor pressure rises steeply with temperature because more molecules have enough energy to escape the liquid phase. The equation quantifies this: a plot of \(\ln P\) against \(1/T\) is a straight line with slope \(-\Delta H_{vap}/R\). Liquids with a high enthalpy of vaporization, such as water, show a steeper temperature dependence than volatile liquids like acetone or ether.
The integrated form used by this calculator needs only one reference point — typically a boiling point — to estimate vapor pressure at any other temperature. This is the standard back-of-the-envelope method in chemistry and chemical engineering for interpolation, since direct measurements are not always available at the temperature of interest.
Interpreting the result is straightforward: if \(T_2 T_1\), it rises. For example, the calculation above shows why water evaporates at room temperature even though it boils at 100 °C — its vapor pressure at 25 °C is still about 0.037 atm. Vapor pressure also determines boiling behavior: a liquid boils when its vapor pressure equals the surrounding pressure, which is why water boils below 100 °C at altitude.
If you work with gas-phase problems generally, you may also find our Combined Gas Law calculator and Dalton's Law calculator useful. The same exponential temperature dependence appears in kinetics — see the Arrhenius Equation calculator.
Assumptions, Limitations, and Practical Tips
- Constant ΔHvap. The enthalpy of vaporization actually decreases with temperature and reaches zero at the critical point. Over ranges of 20–50 K the error is usually small, but over hundreds of kelvin it grows — the water example above overestimates by about 18% because ΔHvap falls from 40.7 kJ/mol at 100 °C to about 44 kJ/mol near 25 °C in reality. Use a ΔHvap value measured near the midpoint of your temperature range for better accuracy.
- Ideal gas vapor. The derivation neglects the molar volume of the liquid and assumes the vapor obeys the ideal gas law. This is an excellent approximation at low pressures (below a few atmospheres).
- Only for vapor–liquid equilibrium. The equation applies to a pure substance. It does not describe solutions, where Raoult's law or activity coefficients are needed, and it should not be extrapolated above the critical temperature.
- Choose a reliable reference point. A normal boiling point (1 atm) from a handbook is a good anchor; so is any tabulated vapor pressure value.
- Watch the units. Temperature must be in kelvin in the formula itself; the calculator converts °C and °F for you. Keep ΔHvap and R consistent — the calculator handles the conversion to J/mol internally.
For quick reference, enthalpies of vaporization at the normal boiling point include water (40.7 kJ/mol), ethanol (38.6 kJ/mol), methanol (35.2 kJ/mol), acetone (29.1 kJ/mol), and benzene (30.8 kJ/mol).
Frequently Asked Questions
How do I calculate vapor pressure using the Clausius–Clapeyron equation by hand?
Take the natural log form \( \ln(P_2/P_1) = -\Delta H_{vap}/R \times (1/T_2 - 1/T_1) \), plug in your known pressure, both temperatures in kelvin, and ΔHvap in J/mol. Compute the right-hand side, exponentiate: \( P_2 = P_1 \, e^{\text{exponent}} \). The most common hand-calculation error is forgetting to convert Celsius to kelvin — always add 273.15 before taking reciprocals.
Can I use the Clausius–Clapeyron equation to find a boiling point instead of a pressure?
Yes. Rearrange so the unknown temperature is isolated: \( 1/T_2 = 1/T_1 + (R/\Delta H_{vap})\ln(P_1/P_2) \). Set \( P_2 \) equal to the external pressure (e.g. 700 mmHg at altitude), solve for \( T_2 \), then convert kelvin back to °C. This is the standard way to estimate boiling points at high elevation or in vacuum distillation setups.
Why is my calculated vapor pressure different from the tabulated value?
The integrated equation assumes ΔHvap is constant, but it actually drops as temperature rises. Over a wide range (like 373 K down to 298 K for water) the error can reach 15–20%. Use a ΔHvap measured near the midpoint of your temperature range, or switch to the Antoine equation, which is fitted to real vapor pressure data and is more accurate across wide ranges.
What is the difference between the Clausius–Clapeyron and Antoine equations?
Both predict vapor pressure versus temperature. Clausius–Clapeyron is derived from thermodynamics and needs one reference point plus ΔHvap; it works for any liquid but assumes constant ΔHvap. The Antoine equation, \( \log_{10} P = A - B/(C + T) \), is purely empirical with substance-specific constants — more accurate for a given liquid within its fitted range, but useless for substances without published constants.
Can I use this equation for mixtures or solutions like ethanol–water?
No, not directly. The Clausius–Clapeyron equation applies only to pure substances. For solutions you need Raoult's law (ideal mixtures) plus activity coefficients for real ones, and the vapor above a mixture has a different composition than the liquid. For each pure component of a mixture, however, you can still use this calculator to estimate its individual vapor pressure at the mixture's temperature.
What's a common mistake when entering ΔHvap units?
Mixing up kJ/mol and J/mol. The formula requires J/mol because R is in J/(mol·K), so 40.7 kJ/mol must become 40,700 J/mol. Also, older tables often list values in kcal/mol (1 kcal = 4184 J). The calculator converts all of these automatically — just make sure the dropdown matches your source's units, otherwise the result can be off by a factor of 1000.
Does vapor pressure always increase with temperature?
Yes, for a stable liquid, vapor pressure rises exponentially with temperature and keeps climbing until it reaches the critical point, where liquid and vapor become identical and ΔHvap falls to zero. This is why the calculator should not be used near or above the critical temperature. Below the freezing point, note the result describes the hypothetical supercooled liquid rather than the solid.