# What new pressure must be overcome for water to boil at 40^@ "C"? DeltabarH_(vap) = "40.65 kJ/mol" for water at 25^@ "C" and "1 atm".

## a) $\text{55.3 torr}$ b) $\text{14.5 torr}$ c) $\text{100 torr}$ d) $\text{760 torr}$

May 18, 2017

$\text{55.3 torr}$, approximately.

Well, look at a phase diagram for water: At ${40}^{\circ} \text{C}$, we move up to cross the liquid-vapor coexistence curve, and we hit the curve ($A C$) somewhere between $\text{0.0060 atm}$ and $\text{1.00 atm}$, closer to $\text{0.0060 atm}$.

We could actually calculate it using the Clausius-Clapeyron Equation:

$\boldsymbol{\ln \left({P}_{2} / {P}_{1}\right) = - \frac{\Delta {\overline{H}}_{v a p}}{R} \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]}$

where:

• ${P}_{i}$ is the pressure at which the phase transition occurs.
• $\Delta {\overline{H}}_{v a p}$ is the enthalpy of vaporization, at ${100}^{\circ} \text{C}$, which is $\text{40.65 kJ/mol}$.
• $R = \text{8.314472 J/mol"cdot"K}$ is the universal gas constant.
• ${T}_{i}$ is the temperature in $\text{K}$.

If we let ${P}_{1}$ be $\text{1 atm}$ and ${T}_{1}$ be $100 + 273.15 = \text{373.15 K}$, then ${T}_{2} = 40 + 273.15 = \text{313.15 K}$ and:

${P}_{2} = {P}_{1} \text{exp} \left(- \frac{\Delta {\overline{H}}_{v a p}}{R} \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]\right)$

If we approximate that $\Delta {\overline{H}}_{v a p}$ does not change significantly from ${100}^{\circ} \text{C}$ to ${40}^{\circ} \text{C}$, then:

P_2 ~~ ("1.00 atm")cdot"exp"(-("40.65 kJ"/"mol" xx (1000 "J")/("1 kJ"))/("8.314472 J/mol"cdot"K") [1/("40 + 273.15 K") - 1/("100 + 273.15 K")])

$=$ $\text{0.0812 atm}$,

which is indeed between $\text{0.0060 atm}$ and $\text{1.00 atm}$, while being much closer to the former than the latter.

Or, converting, we get $\text{61.74 torr}$. This is closest to $\textcolor{b l u e}{\text{55.3 torr}}$.

(In fact, if we knew that the enthalpy of vaporization of water at ${40}^{\circ} \text{C}$ was actually around $\text{43.29 kJ/mol}$ based on the reference value here of $\text{2403.25 kJ/kg}$ at ${41.53}^{\circ} \text{C}$, then we would have gotten $\text{52.45 torr}$, which is even closer.)