# At approximately what pressure will water be a vapor a 0°C?

Jun 4, 2016

At a pressure lower than $0.0056 \setminus a t m$

#### Explanation:

The normal boiling**point of water =100\ ""^oC or $273 \setminus K$

Normal boiling point $\implies$ atmospheric pressure $= 1.00 \setminus a t m$

Note that, for boiling to take place, the atmospheric pressure must be equal to the pressure of water vapor. $\left({P}_{a t m} = {P}_{w v} = 1.00 \setminus a t m\right)$

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

${P}_{1} \setminus$ is the pressure of water vapor at temperature ${T}_{1}$

${P}_{1} = 1.00 \setminus a t m \text{ and } {T}_{1} = 373 \setminus K$

${P}_{2}$ is the pressure of water vapor at temperature ${T}_{2}$

P_2= ?? \ atm " and " T_2 = 273\ K " "(0^oC)

The enthalpy change of vaporization is known at ${0}^{\circ} \text{C}$ to be

$\Delta {H}_{v a p} = 43.9 \frac{k J}{m o l .}$

(note that at ${100}^{\circ} \text{C}$, we would have the more familiar $40.65$ $\frac{k J}{m o l}$.)

$\ln \left({P}_{2} / {P}_{1}\right) = \frac{4.39 \times {10}^{4} \setminus J \cdot m o {l}^{-} 1}{8.31 \setminus J \cdot m o {l}^{-} 1 \cdot {K}^{-} 1} \left[\frac{1}{373 \setminus K} - \frac{1}{273 \setminus K}\right]$

$\ln \left({P}_{2} / {P}_{1}\right) = \frac{4.39 \times {10}^{4} \setminus \cancel{J \cdot} \cancel{m o {l}^{-} 1}}{8.31 \setminus \cancel{J \cdot} \cancel{m o {l}^{-} 1} \cdot \cancel{{K}^{-} 1}} \left[\frac{1}{373 \setminus \cancel{K}} - \frac{1}{273 \setminus \cancel{K}}\right]$

$\ln \left({P}_{2} / {P}_{1}\right) = - 5.19$

$\ln \left({P}_{2} / \left(1.00 \setminus a t m\right)\right) = - 5.19$

${P}_{2} \cong 0.0056 \setminus a t m$

At a pressure lower than $0.0056 \setminus a t m \setminus \left(\cong 567 P a\right) \text{ and } 273 \setminus K \setminus$water will exist in the vapor phase.