1.947844
atm
197.365
kPa
1.947844
0.947844
atm
20
°C
0.666723
1.947844
atm
197.365
kPa
1.947844
0.947844
atm
20
°C
0.666723
The Clausius-Clapeyron Equation Calculator determines the vapor pressure of a substance at a new temperature given a known vapor pressure at a reference temperature and the enthalpy of vaporization. This fundamental thermodynamic relationship is essential for predicting how vapor pressure changes with temperature without needing detailed Antoine constants for every substance.
The Clausius-Clapeyron equation is one of the most important equations in physical chemistry and chemical engineering. It connects the macroscopic observable (vapor pressure) with the molecular property (enthalpy of vaporization), providing deep insight into phase equilibria.
The integrated form of the Clausius-Clapeyron equation is:
$$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$
Solving for \(P_2\):
$$P_2 = P_1 \cdot \exp\left[-\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\right]$$
where:
This equation assumes that \(\Delta H_{vap}\) is constant over the temperature range (a good approximation for small ranges) and that the vapor behaves as an ideal gas. The derivation starts from the Clapeyron equation \(dP/dT = \Delta H_{vap}/(T \Delta V)\) and applies the ideal gas approximation for the molar volume of the vapor.
The final vapor pressure (P₂) shows how much the vapor pressure changes when temperature changes from T₁ to T₂. If T₂ > T₁, P₂ > P₁ (vapor pressure increases). The pressure ratio indicates the factor by which the vapor pressure changes. A substance with a high enthalpy of vaporization will show a steeper pressure-temperature curve, meaning its vapor pressure is more sensitive to temperature changes.
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Starting from water's boiling point (1 atm at 100°C), the vapor pressure nearly doubles to about 1.99 atm at 120°C. This is the principle behind pressure cookers — the sealed vessel reaches higher pressures and temperatures.
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Cooling ethanol from its boiling point (78.37°C) to 50°C reduces the vapor pressure to about 0.30 atm (30.5 kPa), showing that ethanol still has significant vapor pressure well below its boiling point.
The Clausius-Clapeyron equation is a fundamental thermodynamic relation that describes how the vapor pressure of a substance changes with temperature. In its integrated form: ln(P₂/P₁) = -ΔHvap/R × (1/T₂ - 1/T₁). It is derived from the equality of chemical potentials at phase equilibrium.
The integrated Clausius-Clapeyron equation assumes: (1) the enthalpy of vaporization is constant over the temperature range, (2) the vapor behaves as an ideal gas, and (3) the molar volume of the liquid is negligible compared to the vapor. These assumptions are reasonable for moderate temperatures far from the critical point.
The enthalpy of vaporization (ΔHvap) is the energy required to convert one mole of liquid to gas at the boiling point. Common values: water (40,660 J/mol), ethanol (38,560 J/mol), benzene (30,720 J/mol), mercury (59,110 J/mol). Higher values indicate stronger intermolecular forces.
Yes! You can rearrange the equation to solve for T₂ given P₂: 1/T₂ = 1/T₁ - (R/ΔHvap)×ln(P₂/P₁). This is exactly how the Boiling Point at Altitude calculator works — it finds the temperature at which vapor pressure equals the reduced atmospheric pressure.
For temperature ranges of 20-30°C around the boiling point, accuracy is typically within 1-5%. Accuracy decreases for larger temperature ranges because ΔHvap actually varies with temperature. Near the critical point, the equation becomes unreliable because the ideal gas assumption fails.
The Clausius-Clapeyron equation involves reciprocals of temperature (1/T), which requires an absolute temperature scale. Using Celsius would give incorrect results because 0°C is not true zero. The Kelvin scale starts at absolute zero, making 1/T physically meaningful as a measure of thermal energy.
Plot ln(P) versus 1/T for experimental vapor pressure data. The Clausius-Clapeyron equation predicts this should be a straight line with slope equal to -ΔHvap/R. Therefore, ΔHvap = -slope × R. This is a standard experiment in physical chemistry courses.
The Clapeyron equation (dP/dT = ΔH/(TΔV)) is exact and applies to any phase transition. The Clausius-Clapeyron equation applies specific approximations (ideal gas, negligible liquid volume) to the liquid-vapor transition, yielding the integrated ln(P) vs 1/T form that is easier to use.
Yes, ΔHvap generally decreases as temperature increases, reaching zero at the critical temperature where the distinction between liquid and vapor phases disappears. For water, ΔHvap is about 45,050 J/mol at 0°C and decreases to 40,660 J/mol at 100°C and 2,260 J/mol near the critical point (374°C).
Yes, the same equation applies to the solid-vapor equilibrium (sublimation). Simply use the enthalpy of sublimation (ΔHsub) instead of ΔHvap. Since sublimation involves both melting and vaporization, ΔHsub = ΔHfus + ΔHvap and is larger than ΔHvap alone.
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