Question

Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperature of water. A particular pressure cooker has a safety valve that is set to vent steam if the pressure exceeds 3.4 atm. What is the approximate maximum temperature

Answer 1

Answer:

$\boxed{\text{139 \, ^{\circ} C}}$

Explanation:

The question is asking, "At what temperature does the vapour pressure of water equal 3.4 atm?"

To answer this question, we can use the Clausius-Clapeyron equation:

$\ln \left (\dfrac{p_{2}}{p_{1}} \right) = \dfrac{\Delta_{\text{vap}}H}{R} \left(\dfrac{1 }{ T_{1} } - \dfrac{1}{T_{2}} \right)$

Data:

p₁ = 1 atm;    T₁ = 373.15C

p₂ = 3.4atm; T₂ = ?

R  = 8.314 J·K⁻¹mol⁻¹

$\Delta_{\text{vap}}H = \text{39.67 kJ//mol}$

(The enthalpy of vaporization changes with temperature. Your value may differ from the one I chose.)

Calculation:

$\begin{array}{rcl}\ln \left (\dfrac{p_{2}}{p_{1}} \right)& = & \dfrac{\Delta_{\text{vap}}H}{R} \left( \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right)\\\\\ln \left (\dfrac{3.4}{1} \right)& = & \dfrac{39670}{8.314} \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\\ln3.4 & = & 4771 \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\1.224 & = & 12.78 - \dfrac{4771}{T_{2}}\\\\\dfrac{4771}{T_{2}} & = & 11.56\\\\\end{array}$

$T_{2} & = & \dfrac{4771}{11.56} = \text{412.6 K} = \textbf{139 \, ^{\circ} C}\\\\\text{The maximum temperature is } \boxed{\textbf{139 \, ^{\circ} C}}$