Q2(b). Integrated Clausius–Clapeyron equation from the Clapeyron equation

Reading: the two phases are in phase equilibrium (the paper says chemical equilibrium); the derivation is for the vapour pressure of a liquid (or solid).

Clapeyron equation: where it comes from

1. At equilibrium between phases 1 and 2 the molar Gibbs energies are equal, G₁=G₂. Along the coexistence curve they stay equal, so dG₁=dG₂.
2. dG=V dP-S dT for each phase: V₁dP-S₁dT=V₂dP-S₂dT ⇒ (dP)/(dT)=(S₂-S₁)/(V₂-V₁)
3. The transition is reversible at T, so Δ S=Δ H/T, which gives the starting expression (dP)/(dT)=(Δ H)/(T(V₂-V₁)). It is exact for any two-phase equilibrium.

Clausius' approximations (liquid ⇌ vapour)

1. The molar volume of the vapour is far larger than that of the liquid (water at 100°C: about 30 000 cm³ against 19 cm³), so V₂-V₁≈ V_g.
2. The vapour behaves as an ideal gas: V_g=RT/P. Then (dP)/(dT)=ΔᵥₐₚH PRT² (dln P)/(dT)=ΔᵥₐₚHRT² This is the differential Clausius–Clapeyron equation.

Integration

1. Take ΔᵥₐₚH as independent of temperature over the range, separate the variables and integrate: ∫ dln P=ΔᵥₐₚHR∫dTT² ln P=-ΔᵥₐₚHRT+C A plot of ln P against 1/T is a straight line of slope -ΔᵥₐₚH/R; C is a constant of integration.
2. Between limits (P₁,T₁) and (P₂,T₂): 4pt] ln(P₂)/(P₁)=-ΔᵥₐₚHR((1)/(T₂)-(1)/(T₁))=ΔᵥₐₚHR (T₂-T₁)/(T₁T₂) (integrated form)
3. The same equation holds for sublimation with Δ_(sub)H. It fails near the critical point, where Vₗ is not negligible and ΔᵥₐₚH falls to zero.

Application. water, ΔᵥₐₚH=40.7 kJ mol⁻¹, boils at 100°C under 1 atm. In a pressure cooker at 120°C: ln P₂=(40 700)/(8.314)((1)/(373.15)-(1)/(393.15)), P₂=1.95 atm. In the hills, at 0.70 atm, water boils at 90.1°C.