Clausius-Clapeyron calculator
Vapor pressure, boiling point or enthalpy of vaporization from two points.
What is the Clausius-Clapeyron equation?
It links a liquid’s vapor pressure to temperature. In the two-point form from OpenStax Chemistry 2e: ln(P₂ ÷ P₁) = (ΔHvap ÷ R)(1/T₁ − 1/T₂), with temperatures in kelvins, R = 8.314 J/(mol·K) and ΔHvap the enthalpy of vaporization. Given one known point (T₁, P₁) and two of T₂, P₂ and ΔHvap, it finds the third.
Example: boiling point at a lower pressure
OpenStax asks for benzene’s boiling point in Denver, where the air pressure is 83.4 kPa. Benzene boils at 80.1 °C at 101.3 kPa and ΔHvap = 30.8 kJ/mol, so 1/T₂ = 1/353.25 − 8.314 × ln(83.4 ÷ 101.3) ÷ 30,800, giving T₂ ≈ 346.8 K or about 73.7 °C. The book, rounding T₁ to 353.3 K, reports 73.8 °C.
How do I find the enthalpy of vaporization?
Measure the vapor pressure at two temperatures. OpenStax: isooctane has 10.0 kPa at 34.0 °C and 100.0 kPa at 98.8 °C, which gives ΔHvap ≈ 33.8 kJ/mol. For ethanol, 5.95 kPa at 20.0 °C and 53.3 kPa at 63.5 °C give about 41.4 kJ/mol.
Does the pressure unit matter?
No — only the ratio P₂ ÷ P₁ enters, so any unit works as long as both pressures use it. Temperatures must be converted to kelvins; the calculator does that for °F and °C.
How accurate is it?
The equation assumes ΔHvap stays constant, which is only approximately true, so the answer is an estimate that gets worse over wide temperature ranges. For acetone (boiling point 56.5 °C, ΔHvap 31.3 kJ/mol) it predicts about 30.3 kPa at 25.0 °C; OpenStax gives 30.1 kPa for the same problem.
Why does water boil at a lower temperature on a mountain?
Water boils when its vapor pressure equals the air pressure. At high elevation the air pressure is lower, so a lower temperature is enough. With ΔHvap = 40.65 kJ/mol, boiling at 101.325 kPa occurring at 100 °C, the calculator gives a boiling point of about 89.8 °C at 70.1 kPa, which is close to the pressure at 3,000 m in the standard atmosphere.
Why does the enthalpy I get differ from the table value?
The equation assumes ΔHvap is constant, but it changes with temperature. Using the vapor pressure of water at 20 °C (about 2.34 kPa) and at 100 °C (101.325 kPa) gives ΔHvap ≈ 42.8 kJ/mol, compared with 40.65 kJ/mol at the boiling point. The two-point form gives an average over the range.
How the equation works
Vapor pressure rises steeply with temperature, so the natural logarithm of pressure is nearly a straight line against 1/T. The equation is that line: its slope is −ΔHvap ÷ R. It is derived by treating the vapor as an ideal gas, neglecting the volume of the liquid and assuming that ΔHvap does not change with temperature. With two known points on the line you can find the third value in the two-point form. The calculator accepts T₁ and P₁ plus any two of T₂, P₂ and ΔHvap, converts the temperatures to kelvins and uses R = 8.314 J/(mol·K).
Common mistakes
- Using Celsius or Fahrenheit. The formula needs absolute temperature in kelvins. Adding 273.15 to °C is part of the conversion the calculator does for you.
- Mixing kJ and J. R is 8.314 J/(mol·K), so ΔHvap has to be in joules per mole. An enthalpy of 40.65 kJ/mol is 40,650 J/mol.
- Using the equation across a phase change. It describes liquid–vapor equilibrium. Between solid and vapor, the enthalpy of sublimation applies, and the equation changes its meaning.