Question

If the external pressure on water is 422 mmHg, will water boil approximately at
100°C, 80°C, or 60°C (refer to figure 5.1 in our lab manual)?
om
100

Answer 1

The boiling point of water is required at a pressure of 422 mmHg.

The boiling point of water will be $80^{\circ}\text{C}$.

$T_0$ = Boiling point of water at sea level = $100^{\circ}\text{C}=373.15\ \text{K}$

$P_0$ = Pressure at sea level = 760 mmHg

R = Gas constant = 8.314 J/mol K

$\Delta H_v$ =  Heat of vaporization of water = 40660 J/mol

Boiling point is given by the Clausius-Clapeyron equation

$T_B=\left(\dfrac{1}{T_0}-\dfrac{R\ln\dfrac{P}{P_0}}{\Delta H_v}\right)^{-1}\\\Rightarrow T_B=\left(\dfrac{1}{373.15}-\dfrac{8.314\ln\dfrac{422}{760}}{40660}\right)^{-1}\\\Rightarrow T_B=357.12-273.15=83.97^{\circ}\text{C}$

The boiling point of water will be approximately $80^{\circ}\text{C}$.

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