Question

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

Answer 1

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.)

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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")`