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.
