Chapter 11, Problem 158AP

Interpretation Introduction

Interpretation:

The atmospheric pressure on the surface of mars is to be calculated.

Concept introduction:

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron equation as follows:

$\log \left(\frac{P_1}{P_2}\right)=\frac{\Delta {\text{H}}_{vap}}{\text{R}}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Here, $\Delta {\text{H}}_{vap}$ is the molar heat of vaporisation, $\text{R}$ is the gas constant, $P_1\&P_2$ are the vapor pressures, and $T_1\&T_2$ are the temperatures.

Answer to Problem 158AP

Solution: $\text{8}\text{.3}\times {\text{10}}^{-3}\text{atm}$.

Explanation of Solution

Given information: The vapor pressure of atmosphere on earth is $P_1,1\text{ atm}$.

Temperatures of dry ice are $T_1=195\text{K,}{T}_2=150\text{K}$.

Molar heat of sublimation, $\Delta {\text{H}}_{\text{sub}}=25.9\text{ kJ}/\text{mol}$.

Gas constant, $\text{R}=8.314\text{ J}/\text{K}.\text{mol}$.

First, convert heat from kilo joule per mole to joule per mole, the expression is as follows:

$1\text{ kJ}/\text{mol = }{10}^3\text{ J}/\text{mol}$

Convert the heat in joule per mole as follows:

$\begin{array}{c}1\text{ kJ}/\text{mol = }{10}^3\text{ J}/\text{mol}\\ \text{25}\text{.9 kJ}/\text{mol}=25.9\times {10}^3\text{ J}/\text{mol}\\ =2.59\times {10}^4\text{ J}/\text{mol}\end{array}$

Now, calculate $P_2$ as follows:

$\log \left(\frac{P_1}{P_2}\right)=\frac{\Delta {\text{H}}_{\text{sub}}}{\text{R}}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Substitute $1\text{ atm}$ for $P_1$, $2.59\times {10}^4\text{ J}/\text{mol}$ for $\Delta {\text{H}}_{\text{sub}}$, $8.314\text{ J/K}.\text{mol}$ for $\text{R}$, and $195\text{ K}$ for $T_1$ and $150\text{ K}$ for $T_2$ as follows:

$\ln \left(\frac{\text{1 atm}}{P_2}\right)=\frac{2.59\times {10}^4\text{ J}/\text{mol}}{8.314\text{ J}/\text{K}.\text{mol}}\left(\frac{1}{\text{150 K}}-\frac{1}{\text{195 K}}\right)$

$\begin{array}{c}\ln \left(\frac{\text{1 atm}}{P_2}\right)=\frac{2.59\times {10}^4\text{ J}/\text{mol}}{8.314\text{ J}/\text{K}.\text{mol}}\left(\frac{1}{\text{150 K}}-\frac{1}{\text{195 K}}\right)\\ =3115.23\left(\frac{195\text{ K}-150\text{ K}}{195\text{ K}\times 150\text{ K}}\right)\\ =4.79\end{array}$

Taking antilogarithm on both sides as follows:

$\begin{array}{c}\ln \left(\frac{\text{1 atm}}{P_2}\right)=4.79\\ \frac{\text{1 atm}}{P_2}=\exp \left(4.79\right)\\ =120.3\end{array}$

Rearrange the above expression to obtain the vapor pressure as follows:

$\begin{array}{c}{P}_2=\frac{\text{1 atm}}{120.3}\\ =8.3\times {10}^{-3}\text{atm}\end{array}$

Conclusion

The atmospheric pressure at the surface of mars is $8.3\times {10}^{-3}\text{atm}$.