Question

Solve the problem using the Clayperon Clausius equation. The vapour pressure of liquid aluminum is $400\mathrm{mm}\mathrm{Hg}$ at $2.59\times {10}^3\mathrm{K}$. Assuming that its molar heat of vaporization is constant at $296\mathrm{kJ}{\mathrm{mol}}^{-1}$, the vapour pressure of liquid Aluminum is______ $\mathrm{mm}\mathrm{Hg}$ at a temperature of $2.57\times {10}^3\mathrm{K}$.

Answer 1

Calculating the vapour pressure of Aluminium

Step 1: Analyzing the given data

The vapour pressure of liquid aluminum at $2.59\times {10}^3\mathrm{K}=400\mathrm{mm}\mathrm{Hg}$

The molar heat of vaporization = $296\mathrm{kJ}{\mathrm{mol}}^{-1}$

The temperature at which the vapour pressure of Aluminum has to be calculated $=2.57\times {10}^3\mathrm{K}$

Step 2: Applying Clayperon Clausius equation

The Clausius-Clapeyron equation connects vapour pressure with the heat of vaporization.

It is represented as $\ln \left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)=\left(-\frac{∆{\mathrm{H}}_{\mathrm{Vap}}}{\mathrm{R}}\right)\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right)$

where ${\mathrm{P}}_1$ and ${\mathrm{P}}_2$ are the substance's vapoUr pressures at Kelvin temperatures ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$, respectively, ${\mathrm{H}}_{\mathrm{vap}}$ is the substance's molar heat of vaporization, and $\mathrm{R}$ is the ideal gas constant

Substituting $\mathrm{R}$ value of $8.314\times {10}^{-3}\mathrm{kJ}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$ and other given values, we get

$\ln \frac{{\mathrm{P}}_1}{400\mathrm{mm}\mathrm{Hg}}=\left(\frac{-296\mathrm{kJ}{\mathrm{mol}}^{-1}}{8.314\times {10}^{-3}\mathrm{kJ}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}}\right)\left(\frac{1}{2.57\times {10}^3\mathrm{K}}-\frac{1}{2.59\times {10}^3\mathrm{K}}\right)$

$\ln \frac{{\mathrm{P}}_1}{400\mathrm{mm}\mathrm{Hg}}=\left(-0.0356{\mathrm{K}}^{-1}\right)\left(\frac{1}{389.1}\right)-\left(\frac{1}{386.1}\right)$

$\mathrm{P}=4.8\mathrm{mm}\mathrm{Hg}$

Therefore, the vapour pressure of liquid Aluminum is $4.8\mathrm{mm}\mathrm{Hg}$ at a temperature of $2.57\times {10}^3\mathrm{K}$