Clausius Clapeyron Calculator
Compute clausius clapeyron using validated scientific equations. See step-by-step derivations, unit analysis, and reference values.
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)\nln(P2/101.325) = 4890.34 * (0.002680 - 0.002544)\nln(P2/101.325) = 4890.34 * 0.000136 = 0.6651\nP2 = 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)\n-0.7065 = 4890.34 * (1/373.15 - 1/T2)\n1/T2 = 1/373.15 + 0.7065/4890.34 = 0.002680 + 0.000144 = 0.002824\nT2 = 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.
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.
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