Chapter 16, Problem 86E

Interpretation Introduction

Interpretation: The boiling point of water in Breckenridge and the pressure that has to be applied to stem at ${350}^{0}C$ to condense the steam to liquid water should be determined.

Concept Introduction :

The Clausius-Clapeyron equation is represented as follows:

  $\ln \left(\frac{P_{vap1}}{P_{vap2}}\right)=\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

  $P_{vap1}\text{ and }{P}_{vap2}\text{ = vapor pressures}$

  $T_1{\text{ and T}}_2=\text{ temperatures }$

  $\Delta H_{vap}=\text{ Enthalpy of vaporization}$

  $R=\text{ universal gas constant}$

Answer to Problem 86E

Boiling point of water in Breckenridge= ${90}^{0}C$

The pressure that has to be applied to stem at ${350}^{0}C$ to condense the steam to liquid water= $278.6atm$

Explanation of Solution

Given information :

Typical atmospheric pressure- $520torr$

  $\Delta H_{vap}=40.7kjmo{l}^{-1}$

Temperature- ${350}^{0}C$

Under normal condition;

Temperature- $T_1=100+273=373K$

Pressure- $P_1=760torr$

  $P_2=520torr$

  $\begin{array}{c}\ln \left(\frac{P_{vap1}}{P_{vap2}}\right)=\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\\ \ln \left(\frac{760}{520}\right)=\frac{40700Jmo{l}^{-1}}{8.314J{K}^{-1}mo{l}^{-1}}\left(\frac{1}{T_2}-\frac{1}{373K}\right)\\ 0.379=4895.357K\times \left(\frac{1}{T_2}-\frac{1}{373K}\right)\\ \frac{1}{T_2}=363K\\ T_2={90}^{0}C\end{array}$

Boiling point of water in Breckenridge= ${90}^{0}C$

The pressure that has to be applied to stem at ${350}^{0}C$ to condense the steam to liquid water; P₃ Temperature- $T_2=363K$

Temperature- $T_3=350+273K=623K$

Pressure- $P_1=760torr$

  $\begin{array}{c}\ln \left(\frac{P_{vap1}}{P_{vap3}}\right)=\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_3}-\frac{1}{T_2}\right)\\ \ln \left(\frac{760}{P_3}\right)=\frac{40700Jmo{l}^{-1}}{8.314J{K}^{-1}mo{l}^{-1}}\left(\frac{1}{623K}-\frac{1}{363K}\right)\\ P_3=211783.2094torr\\ =278.6atm\end{array}$

Conclusion

Boiling point of water in Breckenridge= ${90}^{0}C$

The pressure that has to be applied to stem at ${350}^{0}C$ to condense the steam to liquid water $278.6atm$