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If atmospheric pressure at lake tahoe is 649 torr, what will be the boiling point of water there?

User Cavel
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The boiling point of water at lake Tahoe is 95.6°C.

The variation of boiling point of water with pressure is determined by using the Clausius- Clapeyron equation.


ln((P_1)/(P_2) )=((\Delta H_vap)/(R))((1)/(T_2) -(1)/(T_1) )

Here,

P₁ is the pressure at temperature T₁, P₂is the pressure at temperature T₂;ΔHvap is the enthalpy of vaporization of water,and R is the universal gas constant.

rewrite the expression for T₂.


T_2=((R*ln(P_1)/(P_2))/(\Delta H_vap) +(1)/(T_1) )^-^1

The boiling point of water at atmospheric pressure of 760 torr is 373 K.

Substitute 760 torr for P₁, 649 torr for P₂, 373 K for T₁, 8.314 J/K mol for R and 40660 J/mol for Δ H_vap and simplify.


T_2=((R*ln(P_1)/(P_2))/(\Delta H_vap) +(1)/(T_1) )^-^1\\ =(((8.314 J/K mol)(ln(760 torr)/(649 torr)))/(40660 J/mol) +(1)/(373 K) )^-^1\\ =368.56 K

Express the temperature in Celcius.


T_2=368.56 K -273 \\ =95.6^oC

User Evelyne
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