Clausius Clapeyron Calculator
Calculate vapor pressure at different temperatures using the Clausius-Clapeyron equation. Find P2, T2, or enthalpy of vaporization from known phase
Clausius Clapeyron Calculator
Calculator
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Formula: ln(P2/P1) = (deltaH_vap / R) * (1/T1 - 1/T2)
Worked example โ Vapor pressure at 120 C: 197.05 kPa (almost double atmospheric pressure)
Formula
ln(P2/P1) = (deltaH_vap / R) * (1/T1 - 1/T2)
Where P1 and P2 are vapor pressures at temperatures T1 and T2 (in Kelvin), deltaH_vap is the molar enthalpy of vaporization (J/mol), and R is the universal gas constant (8.314 J/mol-K). The equation relates how vapor pressure changes with temperature during a phase transition.
Worked Examples
Example 1: Vapor Pressure of Water at 120 C
Problem:Water boils at 100 C (373.15 K) at 101.325 kPa. What is its vapor pressure at 120 C (393.15 K)? deltaH_vap = 40,660 J/mol.
Solution:ln(P2/101.325) = (40660/8.314) * (1/373.15 - 1/393.15) ln(P2/101.325) = 4890.34 * (0.002680 - 0.002544) ln(P2/101.325) = 4890.34 * 0.000136 = 0.6651 P2 = 101.325 * e^0.6651 = 101.325 * 1.9448 = 197.05 kPa
Result:Vapor pressure at 120 C: 197.05 kPa (almost double atmospheric pressure)
Example 2: Boiling Point at Reduced Pressure
Problem:At what temperature does water boil under a vacuum of 50 kPa? Normal BP = 373.15 K, deltaH = 40,660 J/mol.
Solution:ln(50/101.325) = (40660/8.314) * (1/373.15 - 1/T2) -0.7065 = 4890.34 * (1/373.15 - 1/T2) 1/T2 = 1/373.15 + 0.7065/4890.34 = 0.002680 + 0.000144 = 0.002824 T2 = 354.14 K = 80.99 C
Result:Water boils at 81.0 C under 50 kPa vacuum
Frequently Asked Questions
What is the Clausius-Clapeyron equation?
The Clausius-Clapeyron equation describes the relationship between vapor pressure and temperature for a substance undergoing a phase transition, such as liquid to gas. The equation states that ln(P2/P1) = (deltaH_vap/R) * (1/T1 - 1/T2), where P1 and P2 are vapor pressures at temperatures T1 and T2 (in Kelvin), deltaH_vap is the molar enthalpy of vaporization, and R is the universal gas constant. This equation is derived from thermodynamic principles and assumes that the enthalpy of vaporization remains constant over the temperature range considered. It is one of the most important equations in physical chemistry for understanding phase equilibria.
What are the assumptions of the Clausius-Clapeyron equation?
The Clausius-Clapeyron equation relies on several simplifying assumptions. First, it assumes the enthalpy of vaporization (deltaH) is constant and does not change with temperature, which is approximately true over small temperature ranges but breaks down over large ones. Second, it assumes the vapor behaves as an ideal gas, following PV = nRT. Third, it assumes the molar volume of the liquid is negligible compared to the molar volume of the gas. These assumptions make the equation most accurate near the normal boiling point and for moderate temperature ranges. For high-precision work or near the critical point, the more general Clapeyron equation (dP/dT = deltaH / T*deltaV) should be used instead.
How is this equation used in meteorology?
In meteorology, the Clausius-Clapeyron equation is fundamental to understanding atmospheric moisture and weather patterns. It predicts that the saturation vapor pressure of water increases approximately exponentially with temperature, roughly doubling for every 10 degrees Celsius increase. This means warmer air can hold significantly more water vapor than cooler air. Climate scientists use this relationship to predict that for every 1 degree Celsius of global warming, the atmosphere can hold about 7 percent more moisture, leading to more intense precipitation events. The equation also explains why dew forms on cool mornings and why clouds form when moist air rises and cools.
Can the Clausius-Clapeyron equation be used for solid-liquid transitions?
Yes, the general Clapeyron equation applies to any first-order phase transition, including solid-liquid (melting), solid-gas (sublimation), and liquid-gas (vaporization). For the solid-liquid transition, the equation becomes dP/dT = deltaH_fus / (T * deltaV_fus), where deltaH_fus is the enthalpy of fusion and deltaV_fus is the volume change upon melting. However, the simplified logarithmic form (Clausius-Clapeyron) is specifically derived for transitions involving a gas phase where volume changes are large. For melting, where volume changes are small and relatively constant, the simplified linear form dT/dP = T*deltaV/deltaH is often more appropriate and useful.
How do the sublimation and vaporisation curves meet at the triple point?
They meet at the single temperature and pressure where solid, liquid and vapour coexist, which for water is 273.16 K and 611.657 Pa. Below that point the solid-vapour boundary applies and the enthalpy entered here must be the enthalpy of sublimation, which by Hess's law is the enthalpy of fusion plus the enthalpy of vaporisation. Because that is the larger quantity, the sublimation curve rises more steeply than the vaporisation curve where the two join. Carrying a single enthalpy value straight across the triple point therefore smooths over a real kink and quietly biases the answer.
How do I extract an enthalpy of vaporisation from a plot of ln P against 1/T?
Rearranged, the relation is linear: ln P = -(deltaH_vap/R)(1/T) + C. Measure vapour pressure at several temperatures, plot ln P against 1/T in reciprocal kelvin, and the best-fit line has slope -deltaH_vap/R. Multiplying that slope by -8.314 J/mol/K recovers the enthalpy, so a slope of -4,890 K gives 4,890 x 8.314 = 40,660 J/mol, the accepted figure for water. Visible curvature in the plot is the warning that deltaH_vap is drifting over the range you measured.
What is Trouton's rule and how does it sanity-check an enthalpy of vaporisation?
Trouton's rule observes that for normal, non-associating liquids the entropy of vaporisation at the normal boiling point, deltaH_vap divided by Tb, clusters near 88 J/mol/K. Benzene obeys it closely: 30,800 / 353.25 = 87 J/mol/K. Water does not, at 40,660 / 373.15 = 109 J/mol/K, because hydrogen bonding leaves the liquid unusually ordered and expensive to disrupt. Applied to a fitted enthalpy, a ratio far below about 80 J/mol/K usually means the data, the temperature units or the pressure units are wrong.
Why must the temperatures be entered in kelvin rather than Celsius?
Temperature enters as 1/T1 minus 1/T2, as reciprocals of absolute temperature rather than as an interval, so a Celsius value changes the size of every term and at exactly 0 degrees Celsius divides by zero. The gas constant 8.314 J/mol/K is defined on the absolute scale as well. Convert with T(K) = T(C) + 273.15 first, and keep the enthalpy in J/mol so it matches R: entering 40.66 instead of 40,660 shrinks the exponent a thousandfold and returns a P2 almost identical to P1.
How does this relationship explain vacuum distillation and freeze drying?
Read backwards, the curve says that lowering the external pressure lowers the temperature at which a liquid's vapour pressure matches it, so the liquid boils cooler. That is the whole basis of the rotary evaporator and vacuum distillation, where a heat-sensitive compound that would decompose at its atmospheric boiling point can be distilled tens of degrees lower. The sublimation branch governs freeze drying, with ice held below the triple point so water leaves the solid directly as vapour. A pressure cooker runs the same argument upward.
What is the difference between vapour pressure and boiling point?
Vapour pressure is the pressure exerted by vapour in equilibrium with its liquid at a stated temperature. Boiling occurs at the temperature where that vapour pressure rises to equal the pressure pressing on the liquid surface. The two are the same liquid-vapour coexistence curve read along different axes, which is exactly why Clausius Clapeyron Calculator can solve either for P2 at a chosen T2 or for T2 at a chosen P2. The normal boiling point is defined where vapour pressure reaches 101.325 kPa, which for water is 373.15 K.
References
Background & Theory
History
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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