Clausius-Clapeyron Calculator
The Clausius-Clapeyron equation predicts how the vapor pressure of a liquid changes with temperature. Given a known vapor pressure at one temperature and the molar enthalpy of vaporization, you can find the vapor pressure at any other temperature. This calculator uses the two-point form, assuming a constant enthalpy of vaporization and ideal-gas vapor behavior. Enter your first pressure and temperature, the target temperature, and the enthalpy of vaporization. Temperatures must be in Kelvin and enthalpy in joules per mole. The molar gas constant R is fixed at the exact 2019 SI value.
Clausius-Clapeyron formula
ln(P2 / P1) = -(dHvap / R) * (1/T2 - 1/T1)
R = 8.314462618 J/(mol K)
P2 = P1 * exp( -(dHvap / R) * (1/T2 - 1/T1) )
T must be in Kelvin; dHvap in J/mol
P2 comes out in the same pressure unit you used for P1, because the equation works on the ratio P2/P1. Cooling the liquid (T2 below T1) lowers vapor pressure; heating it raises vapor pressure.
Using the equation correctly
- Convert Celsius to Kelvin by adding 273.15 before entering temperatures.
- The default values model water near its normal boiling point: P1 = 101,325 Pa at 373.15 K, with dHvap of 40,660 J/mol.
- Enthalpy of vaporization is assumed constant across the temperature span; this is the main source of error.
- The vapor is assumed ideal, valid well below the critical temperature.
- Look up dHvap for your substance in a reference such as the NIST Chemistry WebBook.
Clausius-Clapeyron: frequently asked questions
What is the Clausius-Clapeyron equation?
The two-point Clausius-Clapeyron equation relates two vapor pressures (P1, P2) to two absolute temperatures (T1, T2) and the molar enthalpy of vaporization: ln(P2/P1) = -(dHvap/R) times (1/T2 - 1/T1). It assumes the enthalpy of vaporization is constant over the temperature range and that the vapor behaves as an ideal gas.
What value of R should I use?
Use the molar gas constant R = 8.314462618 J/(mol K), the exact value adopted in the 2019 SI revision and published by NIST. Make sure your enthalpy of vaporization is in joules per mole (J/mol) to match. If your enthalpy is in kJ/mol, multiply by 1,000 first.
Why must temperatures be in Kelvin?
The equation uses 1/T terms derived from thermodynamics, which require an absolute temperature scale. Kelvin starts at absolute zero, so it is the correct scale. To convert from Celsius, add 273.15. Using Celsius or Fahrenheit directly produces wildly incorrect results.
What does this calculator solve for?
Enter any pressure pair and temperature pair plus the molar enthalpy of vaporization to find P2, or rearrange to verify enthalpy. This page solves for the second vapor pressure P2 given P1, T1, T2, and dHvap, the most common textbook arrangement.
How accurate is the two-point form?
It is an approximation. The enthalpy of vaporization actually decreases slightly as temperature rises toward the critical point, so the constant-enthalpy assumption introduces error over wide temperature ranges. Over a span of a few tens of degrees near normal boiling points, the two-point form is usually accurate to a few percent.
Official sources
Reviewed by the CalculatorHub team, edited by James Graham, 17 June 2026. See our methodology.