What new pressure must be overcome for water to boil at #40^@ "C"#? #DeltabarH_(vap) = "40.65 kJ/mol"# for water at #25^@ "C"# and #"1 atm"#.

#"55.3 torr"#, approximately.

Well, look at a phase diagram for water:

At #40^@ "C"#, we move up to cross the liquid-vapor coexistence curve, and we hit the curve (#AC#) somewhere between #"0.0060 atm"# and #"1.00 atm"#, closer to #"0.0060 atm"#.

We could actually calculate it using the Clausius-Clapeyron Equation:

#bb(ln(P_2/P_1) = -(DeltabarH_(vap))/R[1/T_2 - 1/T_1])#

where:

• #P_i# is the pressure at which the phase transition occurs.
• #DeltabarH_(vap)# is the enthalpy of vaporization, at #100^@ "C"#, which is #"40.65 kJ/mol"#.
• #R = "8.314472 J/mol"cdot"K"# is the universal gas constant.
• #T_i# is the temperature in #"K"#.

If we let #P_1# be #"1 atm"# and #T_1# be #100 + 273.15 = "373.15 K"#, then #T_2 = 40+273.15 = "313.15 K"# and:

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

If we approximate that #DeltabarH_(vap)# does not change significantly from #100^@ "C"# to #40^@ "C"#, then:

#P_2 ~~ ("1.00 atm")cdot"exp"(-("40.65 kJ"/"mol" xx (1000 "J")/("1 kJ"))/("8.314472 J/mol"cdot"K") [1/("40 + 273.15 K") - 1/("100 + 273.15 K")])#

#=# #"0.0812 atm"#,

which is indeed between #"0.0060 atm"# and #"1.00 atm"#, while being much closer to the former than the latter.

Or, converting, we get #"61.74 torr"#. This is closest to #color(blue)("55.3 torr")#.

(In fact, if we knew that the enthalpy of vaporization of water at #40^@ "C"# was actually around #"43.29 kJ/mol"# based on the reference value here of #"2403.25 kJ/kg"# at #41.53^@ "C"#, then we would have gotten #"52.45 torr"#, which is even closer.)