What is the new vapor pressure for water that now boils at $90^{\circ} C$ $90^@ "C"$ instead of $100^{\circ} C$ $100^@ "C"$ ?

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What is the new vapor pressure for water that now boils at $90^{\circ} C$ $90^@ "C"$ instead of $100^{\circ} C$ $100^@ "C"$ ?

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

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Answer 1 · 2017-01-01T22:39:25

I get $0.6961 atm$ $"0.6961 atm"$ , or $529.1 torr$ $"529.1 torr"$ , or $7.054 \times 10^{4}$ $7.054xx10^4$ $Pa$ $"Pa"$ .

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 $ln$ $ln$ of the pressure varies linearly with reciprocal temperature in a small temperature range.

The actual number is more like $P_{2} = 0.6918 atm$ $P_2 = "0.6918 atm"$ , or $525.8 torr$ $"525.8 torr"$ , which gives $0.62 %$ $0.62%$ error.

Or, from the CRC Handbook of Chemistry and Physics, $P_{2} = 0.6920 atm$ $P_2 = "0.6920 atm"$ , or $525.9208 torr$ $"525.9208 torr"$ , giving $0.59 %$ $0.59%$ 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:

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

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

$= -0.3621$

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

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

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What is the new vapor pressure for water that now boils at $90^{\circ} C$ $90^@ "C"$ instead of $100^{\circ} C$ $100^@ "C"$ ?

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

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What is the new vapor pressure for water that now boils at 90^@ "C"90∘C90^@ "C" instead of 100^@ "C"100∘C100^@ "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

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Impact of this question

What is the new vapor pressure for water that now boils at $90^{\circ} C$ $90^@ "C"$ instead of $100^{\circ} C$ $100^@ "C"$ ?

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