The normal boiling**point of water #=100\ ""^oC# or # 273 \ K#

Normal boiling point #=># atmospheric pressure #= 1.00 \ atm#

Note that, for boiling to take place, the atmospheric pressure must be equal to the pressure of water vapor. #(P_(atm) = P_(wv) = 1.00 \ atm)#

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

#P_1 \ # is the pressure of water vapor at temperature #T_1#

#P_1 = 1.00 \ atm " and " T_1 = 373\ K#

#P_2# is the pressure of water vapor at temperature #T_2#

#P_2= ?? \ atm " and " T_2 = 273\ K " "(0^oC)#

The enthalpy change of vaporization is known at #0^@ "C"# to be

#Delta H_(vap)= 43.9 (kJ)/ (mol.)#

(note that at #100^@ "C"#, we would have the more familiar #40.65 # #(kJ)/(mol)#.)

#ln(P_2/P_1) =( 4.39xx10^4 \ J*mol^-1) / (8.31 \ J * mol^-1*K^-1)[1/(373\ K) - 1/(273\ K)]#

#ln(P_2/P_1) =( 4.39xx10^4 \ cancel(J*)cancel(mol^-1))/ (8.31 \cancel( J * )cancel (mol^-1)*cancel(K^-1))[1/(373\ cancel(K))- 1/(273\ cancel(K))]#

#ln(P_2/P_1) = -5.19#

#ln(P_2/( 1.00 \ atm)) = -5.19#

#P_2 ~= 0.0056 \ atm#

At a pressure lower than #0.0056 \ atm \ (~= 567 Pa) " and " 273\ K\ #water will exist in the vapor phase.