Answer to Question #169526 in Molecular Physics | Thermodynamics for Khushi

Question

Derive Clausius-Clapeyron equation. Explain why the boiling point of water

increases with increasing pressure.

Answer 1

According to Maxwell's relation

$(∂P/∂T)v = (∂S/∂V)T$

For a liquid vapour equilibrium, the change of pressure is independent on mass of gas hence the volume

So,

$(∂P/∂T)v = dP/dT$ _____[1]

the heat of vapourization independent on temperature so,

$(∂S/∂V)T = ∂Q/T∂V$

Or,$(∂S/∂V)T = Lm/{T(Vg - Vl)}$__[2]

Where,

$Lm$ = Molar heat of vapourization

$Vg$ and $Vl$ are molar volume of vapour and liquid respectively

From [1] and [2] we get

$dP/dT = Lm/{T(Vg - Vl)}$ ___[3]

Since, $Vg >> Vl$

so, $Vg - Vl ≈ Vg$

[3] turned out to be....

$dP/dT = Lm/T Vg$

From ideal gas equation, $PV = nRT$

Or, $V = nRT/P$

So, $dP/dT = P Lm/{nRT²}$

Or, $∫dP/P = (Lm/nR)∫(dT/T²)$

Integrating between limit, $P1$ to $P2$ and $T1$ to $T2$ we get....

Or, $ln (P2/P1) = (Lm/nR)[(1/T1)–(1/T2)]$

This is the required Clausius Clapeyron equation.

On increasing the pressure,

i.e when $P2 > P1$

$ln (P2/P1) = +ve$

$So, [(1/T1)–(1/T2)] = +ve$

$Or, (1/T1) > (1/T2)$

$Or, T2 > T1$

Hence on increasing the pressure the boiling point of water increase.