5 Thermodynamics & Kinetics-Clausius-Clapeyron Equation

5.60 Spring 2007 Lecture #19 page 1



Clausius-Clapeyron Equation

Let’s revisit solid-gas & liquid-gas equilibria. We can make an
approximation:

V gas
>> V solid
, V liquid
∆Vsubl , ∆Vvap ≈ V gas




We can ignore the molar volume of the condensed phase compared to
the gas.

Taking the Clapeyron equation (exact), e.g. for solid-gas eq. and using
the approximation above:

dp ∆Ssubl ∆H subl ∆H subl
= = ≈
dT ∆Vsubl T ∆Vsubl TV gas


RT
Assuming an ideal gas, V gas
=
p


dp p ∆H subl dp p d ln p ∆H subl
⇒ = = =
dT RT 2 dT dT RT 2

This is the Clausius-Clapeyron Equation
for liq-gas, replace ∆Hsub with ∆H vap

dp p ∆Hvap dp p d ln p ∆Hvap
i.e. = = =
dT RT 2 dT dT RT 2



The Clausius-Clapeyron equation relates the temperature dependence
of the vapor pressure of a liquid or a solid to ∆H vap or
∆Hsub (respectively).

, 5.60 Spring 2007 Lecture #19 page 2



We can make another approximation: Assuming ∆Hsubl
independent of T,

p2 dp ∆Hsubl T2 1 p2 ∆Hsubl ⎛ 1 1 ⎞ ∆Hsubl ⎛T2 −T1 ⎞
∫p = ∫T T dT ln =− ⎜ − ⎟= ⎜ ⎟
1 p R 1
2
p1 R ⎝T2 T1 ⎠ R ⎝ TT 1 2 ⎠




This is the Integrated Clausius-Clapeyron Equation
(for liq-gas, replace ∆Hsub with ∆H vap )

p2 ∆Hvap ⎛ 1 1 ⎞ ∆H vap ⎛T2 −T1 ⎞
i.e. ln =− ⎜ − ⎟= ⎜ ⎟
p1 R ⎝T2 T1 ⎠ R ⎝ TT 1 2 ⎠




In practice this is how you determine vapor pressure over a liquid or
solid as a function of T.

Clausius-Clapeyron problems have the two following forms:

1. You know (T1,p1) and (T2,p2) for s-g or ℓ-g coexistence and want
to know ∆Hsub or ∆H vap
2. You know (T1,p1) and ∆Hsub or ∆H vap for s-g or ℓ-g coexistence and
want to know (T2,p2) (coexistence).

This allows you, for example, to calculate that the boiling point in
Denver is 97°C.