Statistical Mechanics

2. The Clausius-Clapeyron equation relates the latent heat (heat of
transformation) of vaporization or condensation to the rate of
change of vapour pressure with temperature.
In the case of a solid-liquid transformation, it relates the latent heat of
fusion or solidification to the rate of change of melting point with
pressure.
Clausius-clapeyron Equation

3. We consider an infinitely small reversible
Carnot cycle whose isotherms are at
temperature T-dT.
These two isotherms represent two states
of a two-phase system.
Let the system absorbs the amount of
heat Q1 in transition from phase 2.
During this transition the volume of system changes from V2 to V1 at a pressure P.
In the reverse process, the volume changes from V1to V2 at a pressure P-dP, and the
system liberates the amount of heat Q2.
since the heat Q1 is utilised in the phase transition, Q1=L, where L is the latent heat
of transition.

4. Carnot Cycle for derivation of the Clausius-Clapeyron equation
The work done in this cycle will be
W = P(V1 – V2) + (P – dP) (V2 – V1)
= (V1 – V2) dP →(1)
Then the efficiency of the cycle is
ƞ = W/Q1 = (V1 – V2) dP / L → (2)
On the other hand, the efficiency of the Carnot cycle is given by
ƞ = 1 – (T2/T1) = 1 – ((T–dT)/T) = dT/T →(3)
Comparing (2) and (3), we obtain
→(4)
This equation is known as the Clausius-Clapeyron equation.
dP/dT = L / T(V1 – V2)

5. Explain melting, vaporization and sublimation.
In case of the liquid – vapour phase transition, the volume change ΔV =
Vg – V1is always positive and hence dP/dT > 0
i.e. the boiling point increases with increase of pressure.
If the vapour pressure is low, Vg>> V1
dP/dT = Lv/TVg →(5)
where Lv is the latent heat of vaporization. Regarding the vapour as an
ideal gas, Vg= RT/P and then
dP/P = (Lv/R) (dT/T2) →(6)

6. On integrating, we get
In [P(T) / Pc(Tc)] = (Lv/R)((1/Tc) – (1/T)
or
P(T) = Pc(Tc) exp [(Lv/R)((1/Tc) – (1/T))] →(7)
Where Pc and Tc are the critical pressure and critical temperature,
respectively, corresponding to the critical point C.
 For the solid- liquid phase transition, (4) can be represented as
→(8)
Where Lm is the latent heat of melting, Vg and Vs are the specific volumes of
the liquid and solid phase, respectively, on the melting line.
dP/dT = Lm/T (V1 – V2)

7. In this case, V1and V2 are usually close to each other and two cases are
possible:
The volume change ΔV is negative, then dP/dT < 0 and the melting point
decreases with increase of pressure.
In case of the solid- vapour phase transition, (4) takes the form
→(9)
Where Ls is the latent heat of sublimation, Vg and Vs are the specific
volume of vapour and solid phase, respectively, on the sublimation line. In
this case Vg>>Vs and Vg = RT/P.
Review of thermodynamics
→(10)
on integration, we get
→(11)
dP/dT = Ls/T(Vg – Vs)
dP/dT = (Ls/R)(dT/T2)
InP = - (Ls/RT) + A

8. Vapour pressure of solids are usually measure over only a small range of
temperature. Within this range ,(11) can be expressed in the form
→(12)
Where A and B are constant of a given system.
The triple point satisfies the condition
→(13)
 The temperature and pressure corresponding to the triple point are
donated by Tr and Pr for water, they are Tr = 0.010C and Pr = 6Pa.
Consequently under normal condition and atmospheric pressure equilibrium
of all phase water cannot be observed at the triple point.
InP = - (B/T) + A
gs= g1 = gg

9. We have briefly revised the salient feature of equilibrium thermodynamics
which is based on laws of thermodynamics; from which all the
thermodynamics (macroscopic) properties can be deduced.
However thermodynamics gives no information at all regarding the
explanation of thermodynamics properties of the system on the basis of a
molecular theory.
It is statistical mechanics that provides a molecular theory of the
thermodynamic properties of a macroscopic system.