Chapter 15, Problem 15.82P

Interpretation Introduction

Interpretation:

The vapor pressure of water at $5°\text{C, 25}°\text{C, 50}°\text{C, 95}°\text{C}$ has to be determined.

Concept Introduction:

If $P_1$ and $P_2$ are the vapor pressures at two temperatures $T_1$ and $T_2$ then they are related by Clausius-Clapeyron Equation as follows:

  $ln\left(\frac{P_1}{P_2}\right)=\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

If enthalpy of vaporization and vapor pressure at one temperature is known then this equation can be used to estimate vapor pressure at other temperature.

Answer to Problem 15.82P

The vapor pressure of water at $5°\text{C, 25}°\text{C, 50}°\text{C, 95}°\text{C}$ is $\text{5}\text{.9925}\text{torr}$, $\text{2}1.4656\text{torr}$, $\text{84}\text{.7257}\text{torr}$ and $626.871\text{torr}$, respectively.

Explanation of Solution

Conversion factor for temperature in $°\text{C}$ to $\text{K}$ is as follows:

  $\text{temperature}\left(\text{K}\right)=\text{temperature}\left(°\text{C}\right)+273.15$        (1)

Substitute $5°\text{C}$ for temperature $\left(°\text{C}\right)$ in equation (1) to calculate $\text{temperature}\left(\text{K}\right)$.

  $\begin{array}{c}\text{temperature}\left(\text{K}\right)=5+273.15\\ =278.15\text{K}\end{array}$

Substitute $25°\text{C}$ for temperature $\left(°\text{C}\right)$ in equation (1) to calculate $\text{temperature}\left(\text{K}\right)$.

  $\begin{array}{c}\text{temperature}\left(\text{K}\right)=25+273.15\\ =298.15\text{K}\end{array}$

Substitute $50°\text{C}$ for temperature $\left(°\text{C}\right)$ in equation (1) to calculate $\text{temperature}\left(\text{K}\right)$.

  $\begin{array}{c}\text{temperature}\left(\text{K}\right)=50+273.15\\ =323.15\text{K}\end{array}$

Substitute $95°\text{C}$ for temperature $\left(°\text{C}\right)$ in equation (1) to calculate $\text{temperature}\left(\text{K}\right)$.

  $\begin{array}{c}\text{temperature}\left(\text{K}\right)=95+273.15\\ =368.15\text{K}\end{array}$

Water generally boils at $373.15\text{K}$ and $760\text{torr}$ pressure.

The formula to calculate pressure at a particular temperature is as follows:

  $\ln \left(\frac{P_2}{P_1}\right)=-\left(\frac{\Delta H_{\text{Vap}}}{R}\right)\left(\frac{T_1-T_2}{T_1{T}_2}\right)$        (2)

Here,

$P_1$, $P_2$ is initial and final pressure.

$\Delta H_{\text{Vap}}$ is molar enthalpy of vaporization.

$R$ is gas constant

$T_1$, $T_2$ is initial and final temperature.

At $5°\text{C}$,

Substitute $760\text{torr}$ for $P_1$, $43.99\text{kJ}\cdot {\text{mol}}^{-1}$ for $\Delta H_{\text{Vap}}$, $0.008314\text{kJ/mol}$ for R, $278.15\text{K}$ for $T_2$ and $373.15\text{K}$ for $T_1$ in equation (2) to calculate pressure at $5°\text{C}$.

  $\begin{array}{c}\ln \left(\frac{P_2}{760\text{torr}}\right)=-\left(\frac{43.99\text{kJ}\cdot {\text{mol}}^{-1}}{0.008314\text{kJ/mol}}\right)\left(\frac{\left(373.15\text{K}\right)-\left(278.15\text{K}\right)}{\left(373.15\text{K}\right)\left(278.15\text{K}\right)}\right)\\ =-4.8428\end{array}$        (3)

Rearrange equation (3) further to determine pressure at $5°\text{C}$.

  $\begin{array}{c}{\text{P}}_{\text{2}}={\text{e}}^{-4.8428}\left(760\text{torr}\right)\\ =\text{5}\text{.9925}\text{torr}\end{array}$

At $25°\text{C}$,

Substitute $760\text{torr}$ for $P_1$, $43.99\text{kJ}\cdot {\text{mol}}^{-1}$ for $\Delta H_{\text{Vap}}$, $0.008314\text{kJ/mol}$ for R, $298.15\text{K}$ for $T_2$ and $373.15\text{K}$ for $T_1$ in equation (2) to calculate pressure at $25°\text{C}$.

  $\begin{array}{c}\ln \left(\frac{P_2}{760\text{torr}}\right)=-\left(\frac{43.99\text{kJ}\cdot {\text{mol}}^{-1}}{0.008314\text{kJ/mol}}\right)\left(\frac{\left(373.15\text{K}\right)-\left(298.15\text{K}\right)}{\left(373.15\text{K}\right)\left(298.15\text{K}\right)}\right)\\ =-3.5668\end{array}$        (4)

Rearrange equation (4) further to determine pressure at $25°\text{C}$.

  $\begin{array}{c}{\text{P}}_{\text{2}}={\text{e}}^{-3.5668}\left(760\text{torr}\right)\\ =\text{2}1.4656\text{torr}\end{array}$

At $50°\text{C}$,

Substitute $760\text{torr}$ for $P_1$, $43.99\text{kJ}\cdot {\text{mol}}^{-1}$ for $\Delta H_{\text{Vap}}$, $0.008314\text{kJ/mol}$ for R, $323.15\text{K}$ for $T_2$ and $373.15\text{K}$ for $T_1$ in equation (2) to calculate pressure at $50°\text{C}$.

  $\begin{array}{c}\ln \left(\frac{P_2}{760\text{torr}}\right)=-\left(\frac{43.99\text{kJ}\cdot {\text{mol}}^{-1}}{0.008314\text{kJ/mol}}\right)\left(\frac{\left(373.15\text{K}\right)-\left(323.15\text{K}\right)}{\left(373.15\text{K}\right)\left(323.15\text{K}\right)}\right)\\ =-2.1939\end{array}$        (5)

Rearrange equation (5) further to determine pressure at $50°\text{C}$.

  $\begin{array}{c}{\text{P}}_{\text{2}}={\text{e}}^{-2.1939}\left(760\text{torr}\right)\\ =\text{84}\text{.7257}\text{torr}\end{array}$

At $95°\text{C}$,

Substitute $760\text{torr}$ for $P_1$, $43.99\text{kJ}\cdot {\text{mol}}^{-1}$ for $\Delta H_{\text{Vap}}$, $0.008314\text{kJ/mol}$ for R, $368.15\text{K}$ for $T_2$ and $373.15\text{K}$ for $T_1$ in equation (2) to calculate pressure at $95°\text{C}$.

  $\begin{array}{c}\ln \left(\frac{P_2}{760\text{torr}}\right)=-\left(\frac{43.99\text{kJ}\cdot {\text{mol}}^{-1}}{0.008314\text{kJ/mol}}\right)\left(\frac{\left(373.15\text{K}\right)-\left(368.15\text{K}\right)}{\left(373.15\text{K}\right)\left(368.15\text{K}\right)}\right)\\ =-0.192578\end{array}$        (6)

Rearrange equation (6) further to determine pressure at $95°\text{C}$.

  $\begin{array}{c}{\text{P}}_{\text{2}}={\text{e}}^{-0.192578}\left(760\text{torr}\right)\\ =626.871\text{torr}\end{array}$

Conclusion

The vapor pressure of water at $5°\text{C, 25}°\text{C, 50}°\text{C, 95}°\text{C}$ is $\text{5}\text{.9925}\text{torr}$, $\text{2}1.4656\text{torr}$, $\text{84}\text{.7257}\text{torr}$ and $626.871\text{torr}$, respectively. The slight discrepancy in values from the experimental values reported is due to percentage error in calculations.