# 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"#?

##### 1 Answer

#"418897 Pa"# , or#"4.13 atm"# .

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

#ln(P_2/P_1) = -(DeltaH_"vap")/(R)[1/T_2 - 1/T_1]# where:

#P# is the vapor pressure of the liquid at the particular temperature#T# .#DeltaH_"vap"# is the enthalpy for the vaporization process.#R = "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:

#DeltaH_"vap" = "29.1 kJ/mol"# #P_1 = "71226 Pa"# at#T_1 = 25.0^@ "C"# #P_2 = ???# at#T_2 = 78.0^@" C"#

The

Anyways, to calculate the new vapor pressure, we arbitrarily chose

#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

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")]]#

#=# #color(blue)("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