# 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"#.

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#a)# #"55.3 torr"#

#b)# #"14.5 torr"#

#c)# #"100 torr"#

#d)# #"760 torr"#

##### 1 Answer

#"55.3 torr"# , approximately.

Well, look at a phase diagram for water:

At

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_2 = P_1"exp"(-(DeltabarH_(vap))/R[1/T_2 - 1/T_1])#

If we approximate that

#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

Or, converting, we get

(In fact, if we knew that the enthalpy of vaporization of water at