# If at "71226 Pa" of pressure, ethoxyethane boils at 25^@ "C", then what vapor pressure is needed to boil at 78^@ "C" if across the temperature range we can assume that DeltaH_"vap" = "29.1 kJ/mol"?

Mar 4, 2017

$\text{418897 Pa}$, or $\text{4.13 atm}$.

Recall the Clausius-Clapeyron equation (at least the name):

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

where:

• $P$ is the vapor pressure of the liquid at the particular temperature $T$.
• $\Delta {H}_{\text{vap}}$ is the enthalpy for the vaporization process.
• $R = \text{8.314472 J/mol"cdot"K}$ is the universal gas constant.

Basically, this equation describes how to determine vapor pressure at a new temperature. You were given the following info:

• $\Delta {H}_{\text{vap" = "29.1 kJ/mol}}$
• ${P}_{1} = \text{71226 Pa}$ at ${T}_{1} = {25.0}^{\circ} \text{C}$
• P_2 = ??? at ${T}_{2} = {78.0}^{\circ} \text{ C}$

The $\Delta {H}_{\text{vap}}$ was written as $\Delta {H}_{\text{vap}}^{\circ}$ because ${T}_{1}$ happened to be the same temperature as defined for standard thermodynamic temperature and pressure (${25}^{\circ} \text{C}$ and $\text{1 atm}$, or sometimes $\text{1 bar}$ in some newer texts).

Anyways, to calculate the new vapor pressure, we arbitrarily chose ${P}_{1}$ for the given pressure and are to calculate ${P}_{2}$.

P_2/P_1 = "exp"[-(DeltaH_"vap")/(R)[1/T_2 - 1/T_1]]

P_2 = P_1"exp"[-(DeltaH_"vap")/(R)[1/T_2 - 1/T_1]]

where "exp"("stuff") is the function ${e}^{\text{stuff}}$ to keep the equation from looking too small.

So, plug stuff in to get:

color(blue)(P_2) = ("71226 Pa")"exp"[-(29.1 cancel"kJ/mol")/(0.008314472 cancel("kJ/mol")cdotcancel"K")[1/(78.0 + 273.15 cancel"K") - 1/(25.0 + 273.15 cancel"K")]]

$=$ $\textcolor{b l u e}{\text{418897 Pa}}$

That means the vapor pressure increased at a higher temperature.

This should make sense because a higher temperature implies a higher average kinetic energy, so the molecules at the surface of the solution can escape the solution more easily, increasing the vapor pressure above the solution.

In $\text{torr}$ this would be $\text{3141.99 torr}$, which is about $\text{4.13 atm}$, so this is rather unusual. But it make sense because the boiling point of ethoxyethane is about ${34}^{\circ} \text{C}$, and we're well past that.