# If given a boiling point, how do you find vapor pressure?

Jun 29, 2017

Experimentally... or the normal boiling point and enthalpy of vaporization are already well-known and you wish to find the vapor pressure at a different boiling point.

If you were asked to find the vapor pressure needed to boil water at a new temperature of ${T}_{b , 2} = {120.00}^{\circ} \text{C}$, given that the molar enthalpy of vaporization at ${100.00}^{\circ} \text{C}$ is $\text{40.67 kJ/mol}$, then you would use the Clausius-Clapeyron equation.

This describes the variation of boiling point with vapor pressure:

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

where:

• ${P}_{v a p , i}$ is the vapor pressure for a chosen state $i$ in units of, say, $\text{atm}$.
• $\Delta {\overline{H}}_{v a p}$ is the molar enthalpy of vaporization, assumed to vary little between the two boiling points.
• ${T}_{b , i}$ is the boiling point for a chosen state $i$ in units of $\text{K}$.
• $R$ is the universal gas constant in $\text{J/mol"cdot"K}$.

Thus, to find the new vapor pressure needed for water to boil at ${120}^{\circ} \text{C}$, we first recognize that

${T}_{b , 1} = {100.00}^{\circ} \text{C}$ at $\text{1.00 atm}$ is the normal boiling point of water.

That gives us all the necessary info.

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

where $\text{exp} \left(x\right) = {e}^{x}$ is written for readability.

If we assign

${T}_{b , 1} = 100.00 + \text{273.15 K}$,

${T}_{b , 2} = 120.00 + \text{273.15 K}$,

${P}_{v a p , 1} = \text{1.00 atm}$,

then the new vapor pressure needed to boil at ${120.00}^{\circ} \text{C}$ is therefore:

$\textcolor{b l u e}{{P}_{v a p , 2}} = {P}_{v a p , 1} \times \text{exp} \left\{- \frac{\Delta {\overline{H}}_{v a p}}{R} \left[\frac{1}{T} _ \left(b , 2\right) - \frac{1}{T} _ \left(b , 1\right)\right]\right\}$

= ("1.00 atm") xx "exp"{-("40.67 kJ/mol")/("0.008314472 kJ/mol"cdot"K")[1/(120.00 + "273.15 K") - 1/(100.00 + "273.15 K")]}

= ("1.00 atm") xx "exp"{-"4891.47 K"[-"0.0001363 K"^(-1)]}

$= \left(\text{1.00 atm}\right) \times {e}^{0.6668}$

$=$ $\textcolor{b l u e}{\text{1.95 atm}}$

This says that water at lower altitudes boils at higher temperatures, and vice versa.

Jul 13, 2017

The vapor pressure is the atmospheric or system pressure over the boiling liquid.

#### Explanation:

For accuracy, a boiling point always should reference the ambient (or system) pressure. So, given a boiling point (temperature) and the pressure, you know that the vapor pressure of the compound is the same as the system pressure.

This is why "pressure cookers" cook food faster. The higher pressure in the cooker means that the water boils at a high temperature, in order to create the vapor pressure of water that matches it.

Similarly, "boiling" foods at high elevations (lower vapor pressure) takes longer because the "boiling point" temperature of water is lower, as it does not need to generate as much water vapor pressure to "boil".

Working a boiling point pressure back to a liquid vapor pressure at cooler (ambient) temperatures is another matter.