Question

In Breckenridge, Colorado, the typical atmospheric pressure is 520. torr. What is the boiling point of water (${\mathbf{\Delta H}}_{\mathrm{vap}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{7}\mathbf{kJ}\mathbf{/}\mathbf{mol}$ ) in Breckenridge? What pressure would have to be applied to steam at to condense the steam to liquid water?

Short answer

In Breckenridge, Colorado, the typical atmospheric pressure is 194 atm.

Step-by-step solution

Calculation of the boiling point

The boiling point of a liquid can be determined by using the formula:

$\begin{array}{c}\ln \left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)=\frac{{\mathrm{\Delta H}}_{\mathrm{vap}}}{\mathrm{R}}\left(\frac{\mathrm{I}}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right)\\ \mathrm{or}\\ \ln \left(\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}\right)=\frac{{\mathrm{\Delta H}}_{\mathrm{vap}}}{\mathrm{R}}\left(\frac{\mathrm{I}}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)\end{array}$

$\begin{array}{c}\ln \left(\frac{520\mathrm{torr}}{760\mathrm{torr}}\right)=\frac{40.7\times {10}^3\mathrm{J}/\mathrm{mol}}{8.3145\mathrm{J}/\mathrm{K}.\mathrm{mol}}\left(\frac{1}{373\mathrm{K}}-\frac{1}{{\mathrm{T}}_2}\right)\\ -7.75\times {10}^{-5}=\left(\frac{1}{373\mathrm{K}}-\frac{1}{{\mathrm{T}}_2}\right)\\ \frac{1}{{\mathrm{T}}_2}=2.76\times {10}^{-3}\\ {\mathrm{T}}_2=362\mathrm{K}\mathrm{or}89°\mathrm{C}\end{array}$

Pressure would have to be applied to steam at  to condense the steam to liquid water

$\begin{array}{c}\ln \left(\frac{{\mathrm{P}}_2}{1.00}\right)=\frac{40.0\times {10}^3\mathrm{J}/\mathrm{mol}}{8.3145\mathrm{J}/\mathrm{K}.\mathrm{mol}}\left(\frac{1}{373\mathrm{K}}-\frac{1}{623\mathrm{K}}\right)\\ {\mathrm{P}}_2={\mathrm{e}}^{5.27}\\ {\mathrm{P}}_2=194\mathrm{atm}\end{array}$

The pressure needed to apply is 194 atm.