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Problem 5.35. The Clausius-Clapeyron relation 5.47 is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and $∆$V depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take $∆$V to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:

P= (constant) x e^-L/RT

This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.

Step by step solution

01

Given information

Using the ideal gas law,

$$\begin{gathered} P{V}_g=RT \\ {V}_g=\frac{RT}{P} \end{gathered}$$

Using Clausius Clapeyron Equation

$$\begin{gathered} \frac{dP}{dT}=\frac{L}{T·\mathrm{Δ}V} \\ ⇒\frac{dP}{dT}=\frac{L}{T·V_g} \\ =\frac{L}{T×\frac{RT}{P}} \\ =\frac{P·L}{R{T}^2} \\ ∴\frac{dP}{P}=\frac{L}{R}·\frac{dT}{T^2} \end{gathered}$$

02

Calculation

Integrating both sides

$$\begin{gathered} ∫\frac{dP}{P}=\frac{L}{R}·∫\frac{dT}{T^2} \\ \ln P=\frac{-L}{R}·\frac{1}{T}+\text{constant} \end{gathered}$$

Exponentiating both sides

$P=e^{\left(\frac{-L}{RT}+\text{const}\right)}$
role="math" localid="1646902017006" $⇒P=\text{constant}×e^{\frac{-L}{RT}}$

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