Peter Atkins • Julio de Paula Atkins’ Physical Chemistry Eighth Edition Chapter 4 – Lecture 2 Physical Transformations of Pure Substances Copyright © 2006 by Peter Atkins and Julio de Paula

The effect of applied pressure on vapor pressure Pressure applied to condensed phase squeezes molecules out of the liquid Papplied ∝ Pvapor

Fig 4.11 Two methods of applying pressure to a condensed phase

The effect of applied pressure on vapor pressure Pressure applied to condensed phase squeezes molecules out of the liquid Assumptions: Pressurizing gas does not change properties of the liquid Gas solvation does not occur Papplied ∝ Pvapor where P* = vapor pressure of liquid at 1 atm ΔP = applied pressure

Water is subjected to an applied pressure of 10 bar. At 25 °C, density = 0.997 g/cm3, ∴ Vm = 18.1 cm3 mol-1. Since VmΔP/RT << 1, Or an increase of 0.73 %

Fig 4.12 P vs T plot for a system of two phases When pressure is applied to a system in a two-phase equilibrium, the equilibrium is disturbed System can regain equilibrium by changing temperature i.e., Clapeyron eqn

Location of Phase Boundaries For two phases α and β in equilibrium: μα(P,T) = μβ(P,T) Phase boundaries most simply discussed in terms of their slopes, dP/dT The solid-liquid boundary: which approximates to:

Fig 4.13 Typical solid-liquid phase boundary (P2, T) ● ●(P1,T)

Location of Phase Boundaries The liquid-vapor boundary with PV = RT becomes and finally: Clausius-Clapeyron eqn

Fig 4.14 Typical liquid-vapor phase boundary (T, P2) ● (T1, P) ● ●(T2, P) ●(T, P1)

Clausius-Clapeyron Equation Molar heat of vaporization (DHvap) ≡ the energy required to vaporize 1.00 mole of a liquid ln P = − DHvap RT + C Clausius-Clapeyron Equation P = (equilibrium) vapor pressure T = temperature (K) R = gas constant (8.314 J/K•mol) m = −DHvap/R ln P 1/T

Clausius-Clapeyron Equation ln P1 = − DHvap RT1 + C ln P2 = − DHvap RT2 + C m = −DHvap/R ln P 1/T

Fig 4.15 Typical triple point ΔHsub > ΔHvap