Chapter 16, Problem 128AE

Interpretation Introduction

Interpretation:

The temperature inside the pressure cooker when the vapor pressure of water is 3.50 atm should be predicted.

Concept Introduction:

The ClausiusClapeyron equation is given as follows:

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

Where,

• $P_2$ is the vapour pressure at $T_2$
• $P_1$ is the vapour pressure at $T_1$
• $\Delta H_{vap}$ is the enthalpy of vaporization
• $R$ is the Gas constant.

Answer to Problem 128AE

The temperature inside the pressure cooker when the vapor pressure of water is 3.50 atmis $412.4\text{K}$ .

Explanation of Solution

Given information:

• Temperature inside the pressure cooker is $\left(T_2\right)$
• $115°\text{C}=\left(115+273\right)\text{K}=388\text{K}$

Vapor pressure of water inside the pressure cooker is $\text{3}\text{.50}\text{atm}$

The pressure at normal boiling point $\left(P_1\right)$ , normal boiling of water at 1 atm $\left(T_1\right)$ and $\Delta H_{vap}$ are taken as 1 atm, 373 K and 40,790 J/mol respectively.

The ClausiusClapeyron equation is given as follows:

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

Substitute the given values in above formula.

  $\begin{array}{c}\text{ln}\left(\frac{P_2}{1\text{atm}}\right)=\frac{-40790\text{J}{\text{mol}}^{-1}}{8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\left(\frac{1}{388\text{K}}-\frac{1}{373\text{K}}\right)\\ \text{ln}{P}_2=\left(-4895.35\text{K}\right)\left(-1.03645\times {10}^{-4}\text{K}\right)\\ P_2=e^{0.51}\\ =1.67\text{atm}\end{array}$

Therefore, the value of $P_2$ is $1.67\text{atm}$ .

Substitute the value of $P_2=3.50\text{atm}$ , $T_1=373\text{K}$ and $P_1=1\text{atm}$ in ClausiusClapeyron equation.

  $\begin{array}{c}\text{ln}\left(\frac{3.50\text{atm}}{1\text{atm}}\right)=\frac{-40790\text{J}{\text{mol}}^{-1}}{8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}}\left(\frac{1}{T_2}-\frac{1}{373\text{K}}\right)\\ 1.2527=\left(-4895.35\text{K}\right)\left(\frac{1}{T_2}-\frac{1}{373\text{K}}\right)\\ \frac{1}{T_2}-\frac{1}{373\text{K}}=\frac{1.2527}{-4895.35\text{K}}\\ \frac{1}{T_2}=\frac{1.2527}{-4895.35\text{K}}+\frac{1}{373\text{K}}\\ \frac{1}{T_2}=0.002425{\text{K}}^{-1}\\ T_2=412.4\text{K}\end{array}$

Therefore, the temperature inside the pressure cooker when the vapor pressure of water is 3.50 atmis $412.4\text{K}$ .

Conclusion

The temperature inside the pressure cooker when the vapor pressure of water is 3.50 atmis $412.4\text{K}$ .