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
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
I get
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
The actual number is more like
Or, from the CRC Handbook of Chemistry and Physics,
(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
Your enthalpy of vaporization doesn't look right, though... the molar enthalpy of vaporization of water is
When we set
#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")#