Engineering Chemistry S Datta

86 ENGINEERING CHEMISTRY
equilibrium. The drop in free energy of the liquid is –∆G 1
and corresponding gain in free energy
of the vapour is +∆G 2
. Since the system is in equilibrium,
as net ∆G = 0 ∴ ∆G 1
= ∆G 2
or,
V 1
∆P – S 1
∆T = V 2
∆P – S 2
∆T.
or,
∆P
∆T 2 − S 1 ∆S
=
V − V V V
2 1
2 − 1
Since,
∆S = q T
dP
∴
dT = ∆S q
=
V2 − V1
T(V2 − V 1)
where q is molar heat of transition from one phase to another.
Thus, d P
dT =
q
. This equation is the famous Clapeyron-Clausius equation.
T(V2 − V 1)
Application of Clapeyron-Clausius equation
1. Determination of latent heat.
We know
dP
dT =
q
.
T(V2 − V 1)
When two phases of liquid (Vol. V 1
) and vapor (Vol. V 2
) are in equilibrium, the volume
V 1
of the liquid can be neglected as it is very small compared to the volume of vapor V 2
.
or,
∴
Now, for ideal gas
∴
dP
dT =
q
TV 2
=
L
TV 2
PV = RT, i.e., V = RT
P
dP
dT = LP
RT 2
dP
P
= L . dT
RT 2
Assuming, latent heat is constant and on integrating,
ln P 2
P1
or, log P 2
P1
= L R
=
L
N
M
1
−
T
L
2.303R
1
T
1 2
L
N
M
O
Q
P or,
1 2
O
Q
T2 − T1P.
TT
(q = L = latent heat of vaporization)
log P P
2
L
N
M
L 1 1
= −
2.303R T T
1 1 2
Thus, from the vapour pressures P 1
and P 2
at different temperatures T 1
and T 2
we can
determine the molar heat of vaporization with the integrated Clapeyron-Clausius equation.
2. Effect of pressure on boiling point. With the help of the integrated Clapeyron-Clausius
equation we can determine the boiling point of a liquid at a different pressure when the boiling
point of it is known at one pressure along with the latent heat of vaporization.
O
Q
P