What is the new vapor pressure for water that now boils at #90^@ "C"# instead of #100^@ "C"#?

To what number do we need to lower down pressure in order for water to boil on 90 degrees?
Latent heat of evaporation is 71 849 J/g.

EDIT: A more accurate number would be 2264.78 J/g.

- Truong-Son

1 Answer
Jan 1, 2017

I get #"0.6961 atm"#, or #"529.1 torr"#, or #7.054xx10^4# #"Pa"#.

This should make physical sense, because we physically see water boiling more easily (i.e. at lower temperatures) at higher altitudes, at which there is a lower atmospheric pressure.

Keep in mind that the Clausius-Clapeyron equation assumes the #ln# of the pressure varies linearly with reciprocal temperature in a small temperature range.

The actual number is more like #P_2 = "0.6918 atm"#, or #"525.8 torr"#, which gives #0.62%# error.

Or, from the CRC Handbook of Chemistry and Physics, #P_2 = "0.6920 atm"#, or #"525.9208 torr"#, giving #0.59%# error.

(David R. Lide, ed. (2005). CRC Handbook of Chemistry and Physics. Boca Raton, Florida: CRC Press. p. 6-8.)


You may want to reference this answer for a derivation of the Clausius Clapeyron equation, which is most effective for ideal gases:

#(dlnP)/(dT) = (DeltabarH_"vap")/(RT_b^2)#

or the logarithmic form,

#ln(P_2/P_1) = -(DeltabarH_"vap")/(R)[1/T_(b2) - 1/T_(b1)]#,

where we assume #DeltabarH_"vap"# varies negligibly with temperature, and:

  • #DeltabarH_"vap" = (DeltaH_"vap")/n# is the molar enthalpy of vaporization in #"kJ/mol"#.
  • #R = "0.008314472 kJ/mol"cdot"K"# is the universal gas constant.
  • #T_b# is the boiling point.
  • #P# is the pressure at which the substance boils. #P_2# is the new pressure, and #P_1# is some reference pressure.

We know that the normal boiling point of water is #100^@ "C"#, or #"373.15 K"#, at an atmospheric pressure of #"1 atm"#.

Your enthalpy of vaporization doesn't look right, though... the molar enthalpy of vaporization of water is #DeltabarH_"vap" = "40.8 kJ/mol"#, or #"2264.78 J/g"#. I'm going to use that number instead...

When we set #P_1 = "1 atm"#, #T_1 = 100^@ "C" = "373.15 K"#, and #T_2 = 90^@ "C" = "363.15 K"#, we get:

#ln(P_2/"1 atm") = -(40.8 cancel"kJ/mol")/(0.008314472 cancel("kJ/mol")cdot"K")[1/("363.15 K") - 1/("373.15 K")]#

#= -("4907.11 K")[1/("363.15 K") - 1/("373.15 K")]#

#= -0.3621#

#=> color(blue)(P_2) = ("1 atm")e^(-0.3621)#

#=# #color(blue)("0.6961 atm")#