Question

Estimate the boiling point of water in Leadville, Colorado, elevation 3170 m. To do this, use the barometric formula relating pressure and altitude: P=P0×10−Mgh/(2.303RT) (where P = pressure in atm; P0 = 1 atm; g = acceleration due to gravity; molar mass of air, M= 0.02896 kg mol−1 ; R = 8.3145 J mol−1 K−1; and T is the Kelvin temperature). Assume the air temperature is 10.0 ∘C and that ΔHvap= 41 kJ mol−1 H2O.

Answer 1

First we get the pressure:

P = 1 atm * 10^(-(0.02896 kg/mol * 9.81m/s² * 3170 m)/(2.303 * 8.3145 J/molK * 283K))

Units cancel out as J/mol over J/mol, so:

P = 1 atm * 0.682 = 0.682 atm

We now have a point of comparison, which is the normal boiling point at sea level (1 atm), and we will get the new point with the pressure obtained, using clausius clapeyron:

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

ln (1atm / 0.682 atm) = 41 kJ/mol / 0.0083145 kJ/mol * (1/T2 - 1/373.15)

0.382726 = 4931.14 * (1/T₂ - 0.00268)

7.7614 x 10^-5 = 1/T₂ - 0.00268

0.0027576 = 1/T₂

T₂ = 362.63 K = 89.48 ºC