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 ΔHvap is independent of temperature, integration of the above equation can be performed
If the vapor pressure P1 is known at boiling point temperature T1, this equation can be used to estimate the boiling point temperature T2 at another pressure P2.
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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 ΔHvap of water at 373.15K is 40657 Jmol-1 and the gas constant R is 8.3145 JK-1mol-1.
Use the integrated Clausius-Clapeyron equation to solve for T2
The SI units are:
T = K
R = JK-1mol-1
ΔH = Jmol-1
The units of P1 and P2 are immaterial, so long as they are the same.
Note that P1 and P2 are vapor pressures. Since T1 and T2 refer to boiling point temperatures, the vapor pressures P1 and P2 will be the same as the externally applied pressures.
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.