# What is the new vapor pressure for water that now boils at 90^@ "C" instead of 100^@ "C"?

## To what number do we need to lower down pressure in order for water to boil on 90 degrees? Latent heat of evaporation is 71 849 J/g. EDIT: A more accurate number would be 2264.78 J/g. - Truong-Son

Jan 1, 2017

I get $\text{0.6961 atm}$, or $\text{529.1 torr}$, or $7.054 \times {10}^{4}$ $\text{Pa}$.

This should make physical sense, because we physically see water boiling more easily (i.e. at lower temperatures) at higher altitudes, at which there is a lower atmospheric pressure.

Keep in mind that the Clausius-Clapeyron equation assumes the $\ln$ of the pressure varies linearly with reciprocal temperature in a small temperature range.

The actual number is more like ${P}_{2} = \text{0.6918 atm}$, or $\text{525.8 torr}$, which gives 0.62% error.

Or, from the CRC Handbook of Chemistry and Physics, ${P}_{2} = \text{0.6920 atm}$, or $\text{525.9208 torr}$, giving 0.59% error.

(David R. Lide, ed. (2005). CRC Handbook of Chemistry and Physics. Boca Raton, Florida: CRC Press. p. 6-8.)

You may want to reference this answer for a derivation of the Clausius Clapeyron equation, which is most effective for ideal gases:

$\frac{\mathrm{dl} n P}{\mathrm{dT}} = \frac{\Delta {\overline{H}}_{\text{vap}}}{R {T}_{b}^{2}}$

or the logarithmic form,

$\ln \left({P}_{2} / {P}_{1}\right) = - \frac{\Delta {\overline{H}}_{\text{vap}}}{R} \left[\frac{1}{T} _ \left(b 2\right) - \frac{1}{T} _ \left(b 1\right)\right]$,

where we assume $\Delta {\overline{H}}_{\text{vap}}$ varies negligibly with temperature, and:

• DeltabarH_"vap" = (DeltaH_"vap")/n is the molar enthalpy of vaporization in $\text{kJ/mol}$.
• $R = \text{0.008314472 kJ/mol"cdot"K}$ is the universal gas constant.
• ${T}_{b}$ is the boiling point.
• $P$ is the pressure at which the substance boils. ${P}_{2}$ is the new pressure, and ${P}_{1}$ is some reference pressure.

We know that the normal boiling point of water is ${100}^{\circ} \text{C}$, or $\text{373.15 K}$, at an atmospheric pressure of $\text{1 atm}$.

Your enthalpy of vaporization doesn't look right, though... the molar enthalpy of vaporization of water is $\Delta {\overline{H}}_{\text{vap" = "40.8 kJ/mol}}$, or $\text{2264.78 J/g}$. I'm going to use that number instead...

When we set ${P}_{1} = \text{1 atm}$, ${T}_{1} = {100}^{\circ} \text{C" = "373.15 K}$, and ${T}_{2} = {90}^{\circ} \text{C" = "363.15 K}$, we get:

ln(P_2/"1 atm") = -(40.8 cancel"kJ/mol")/(0.008314472 cancel("kJ/mol")cdot"K")[1/("363.15 K") - 1/("373.15 K")]

= -("4907.11 K")[1/("363.15 K") - 1/("373.15 K")]

$= - 0.3621$

$\implies \textcolor{b l u e}{{P}_{2}} = \left(\text{1 atm}\right) {e}^{- 0.3621}$

$=$ $\textcolor{b l u e}{\text{0.6961 atm}}$