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

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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")#