Boiling point at reduced pressure

Worked example

A liquid boils at 389 K under 101.325 kPa and has ΔvapH = 36.9 kJ mol⁻¹. At what temperature does it boil when the pressure is 75 kPa?

1. 1. Clausius–Clapeyron equation: solve for T₂

   $\dfrac{1}{T_2} = \dfrac{1}{T_1} - \dfrac{R\ln(p_2/p_1)}{\Delta_\mathrm{vap}H}$

   = 379.0 K

Answer: T₂ = 379 K

Method

Clausius–Clapeyron: 1/T₂ = 1/T₁ − R ln(p₂/p₁)/ΔvapH. Lower pressure gives a lower boiling point.

Formulas used

• Clausius–Clapeyron equation

  $\ln\dfrac{p_2}{p_1} = -\dfrac{\Delta_\mathrm{vap}H}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$

More thermodynamics questions

• ΔG from ΔH and ΔS
• K from ΔG°
• ΔG° from K
• K from ΔH° and ΔS°
• van 't Hoff: K at a new temperature
• Moles of an ideal gas
• ΔH at another temperature (Kirchhoff)
• Entropy change on heating
• Hess's law from enthalpies of formation
• ΔH from a bomb-calorimeter ΔU
• Enthalpy of combustion from a calorimetry experiment
• Enthalpy of neutralisation
• ΔfH from enthalpies of combustion
• ΔH from mean bond enthalpies
• An unknown bond enthalpy from ΔH
• Born–Haber cycle: lattice enthalpy of NaCl
• Born–Haber cycle: lattice enthalpy of MgCl₂
• ΔS from standard entropies
• Temperature at which a reaction becomes feasible
• ΔG from enthalpies of formation and entropies
• Enthalpy of solution from a thermodynamic cycle
• Methanol synthesis: K from tabulated data