Question

The temperature inside a pressure cooker is 115⁰C. Calculate the vapor pressure of water inside the pressure cooker. What would be the temperature inside the pressure cooker if the vapor pressure of water was 3.50atm?

Short answer

The vapor pressure of water inside the pressure cooker is 1.41atm. The temperature inside the pressure cooker is 140⁰C when the vapor pressure of water was 3.50atm.

Step-by-step solution

Determine the vapor pressure of water inside the pressure cooker

Using theClausius-Clapeyron formula:

$\ln \frac{p_1}{p_2}=\frac{\Delta H_{vap}}{R}\times \left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Where, $\Delta H_{vap}=40790J/mol$

$$\begin{gathered} T_2=273+115 \\ =383K \\ {T}_1=273+110 \\ =373Kl \end{gathered}$$

$$\begin{gathered} p_1=1atm \\ R=8.314J/Kmol \end{gathered}$$

Substituting the values in equation (i):

$$\begin{gathered} \ln \frac{1}{p_1}=\frac{49790}{8.314}\times \left(\frac{1}{383}-\frac{1}{373}\right) \\ \ln 1-\ln p_2=-0.3434 \\ \ln p_2=0.3434 \\ {p}_2=1.41atm \end{gathered}$$

So, the vapor pressureis 1.41atm.

Determine the temperature of water inside the pressure cooker if the vapor pressure of water is 3.50atm.

Here, $\Delta H_{vap}=40790J/mol$

$$\begin{gathered} T_1=273+110=373K \\ {p}_1=1atm \\ {p}_2=3.5atm \\ R=8.314J/Kmol \end{gathered}$$

Substituting the values in equation (i):

$$\begin{gathered} \ln \frac{1}{3.5}=\frac{49790}{8.314}\times \left(\frac{1}{T_2}-\frac{1}{373}\right) \\ -1.2428=4906.18K^{-1}\times \frac{1}{T_2}-13.153 \\ {T}_2=\left(413-273\right)°C \\ {T}_2=140°C \end{gathered}$$

So, thetemperature is 140⁰C.