Question #e4ab2

This looks like a job for the Clausius - Clapeyron equation, which looks like this

#color(blue)(ul(color(black)(ln(P_1/P_2) = (DeltaH_"vap")/R * (1/T_2 - 1/T_1))))#

Here

• #P_1# is the vapor pressure of the liquid at #T_1#
• #P_2# is the vapor pressure of the liquid at #T_2#
• #R# is the universal gas constant, equal to #"8.3145 J mol"^(-1)"K"^(-1)#
• #DeltaH_"vap"# is the enthalpy of vaporization of the liquid, equal to #"40,660 J mol"^(-1)# for water (see here)

Now, you know that at an unknown atmospheric pressure #P_2#, water boils at

#T_2 = 110^@"C" + 273.15 = "383.15 K"#

Moreover, you know that at normal pressure, i.e. at #P_1 = "1 atm"#, water has a normal boiling point of

#T_1 = 100^@"C" + 273.15 = "373.15 K"#

As you know, a liquid boils when its vapor pressure is equal to the atmospheric pressure, so you can use the two atmospheric pressures in the Clausius - Clapeyron equation as #P_1# and #P_2#.

So, plug in your values into the equation to get

#ln("1 atm"/P_2) = ("40,660" color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("mol"^(-1)))))/(8.3145 color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("mol"^(-1)))) color(red)(cancel(color(black)("K"^(-1))))) * (1/(383.15 color(red)(cancel(color(black)("K")))) - 1/(273.15color(red)(cancel(color(black)("K")))))#

#ln("1 atm"/P_2) = -0.342041625#

This will be equivalent to

#e^ln("1 atm"/P_2) = e^(-0.342041625)#

which will get you

#"1 atm"/P_2 = 0.71032#

You can thus say that

#P_2 = "1 atm"/0.71032 = color(darkgreen)(ul(color(black)("1.4 atm")))#

Now, does this result make sense?

As atmospheric pressure increases, the molecules of water will need more energy in order to escape from the liquid, which implies that the temperature of the water will be higher at the new boiling point.

So in your case, a higher boiling temperature implies a higher a higher value for the vapor pressure of water at its boiling point, i.e. a higher atmospheric pressure.