Question

What's the boiling point water (in celsius) at the top of Mt Everest (atmospheric pressure =0.54 atm)? I've tried about everything and I keep getting wrong answers. At 25C Substance delta H (kj/mol) delta S (j/mol*k) delta G (kj/mol) H2O (l) -285.8 70.0 -237.1 H2O (g) -241.8 188.8 -228.6

Find Tboiling point in celsius.

Answer 1

In vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Change negative signs

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

we will ned dHVap,

dHVap = Hvapor - Hliquid = -241.8 - (-285.8) = 44 kJ/mol = 44000 J/mol

substitute

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

ln(0.54/1) = 44000/8.314*(1/(100+273)-1/(T + 273))

solve for T

ln(0.54/1)* 8.314 / 44000 - 1/(100+273) = -1/(T + 273)

-0.00279739 = -1/(T + 273)

T = 1/0.00279739 - 273

T = 84.4760 ºC