Our pressure cooker was a Prestige Skyline from the 1950s. The regulator is sticking up at the back.

A familiar object in the kitchen of my youth was our pressure cooker. It cooked the vegetables in a fraction of the time, saving significant amounts of energy and therefore cost.

Our pressure cooker was a stainless steel device equipped with a regulator that maintained an internal pressure of 2 atmospheres. I was thinking back to it the other day, and started wondering about the temperature of the superheated water inside the container. This can be calculated using the Clausius-Clapeyron equation

which approximates the slope of the liquid-vapor coexistence curve at vapor pressure P and boiling point temperature T. Using the mathematical identity

the Clausius-Clapeyron equation can also be written in the form

If we further assume that ΔH_{vap} is independent of temperature, integration of the above equation can be performed

If the vapor pressure P_{1} is known at boiling point temperature T_{1}, this equation can be used to estimate the boiling point temperature T_{2} at another pressure P_{2}.

**– – – –**

*Worked Example*

Water boils at 373.15K at a pressure of 1 atmosphere. At what temperature will water boil in a pressure cooker operating at a pressure of 2 atmospheres? The enthalpy of vaporization ΔH_{vap} of water at 373.15K is 40657 Jmol^{-1} and the gas constant R is 8.3145 JK^{-1}mol^{-1}.

__Strategy__

Use the integrated Clausius-Clapeyron equation to solve for T_{2}

The SI units are:

T = K

R = JK^{-1}mol^{-1}

ΔH = Jmol^{-1}

The units of P_{1} and P_{2} are immaterial, so long as they are the same.

Note that P_{1} and P_{2} are **vapor pressures**. Since T_{1} and T_{2} refer to boiling point temperatures, the vapor pressures P_{1} and P_{2} will be the same as the externally applied pressures.

__Calculation__

Water in a pressure cooker operating at a pressure of two atmospheres boils at 394K, or 121°C. This explains why vegetables cook so fast in these devices.

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