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

$$\begin{gathered} \frac{dP}{dT}=\frac{L}{T·\Delta V} \\ \Rightarrow \frac{dP}{dT}=\frac{L}{T·V_g} \\ =\frac{L}{T\times \frac{RT}{P}} \\ =\frac{P·L}{R{T}^2} \\ \therefore \frac{dP}{P}=\frac{L}{R}·\frac{dT}{T^2} \end{gathered}$$

02

Calculation

Integrating both sides

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

Exponentiating both sides

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

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!