Question

Estimate the boiling point at the top of Mount Everest. Assume it is 8800m above sea level and assume air is an ideal gas and the temperature of the atmosphere is 20 C. In reality, the temperature drops with increasing elevation, how would this affect the answer?

Answer 1

boil of water on Mount Everest is easier than sea level. At the mean sea level, the pressure is the highest whereas as the altitude decreases the pressure. Mount Everest, being the highest peak has the least pressure on the Earth. At the sea level when we heat water there is much atmospheric pressure acting on the molecules of the water. Hence the heat of the molecules cannot overcome the pressure easily. Hence it takes time for water to boil. Whereas on mount Everest, the pressure is too low and the temperature is around 72-degree celsius is enough for molecules to overcome the atmospheric pressure.

ln(p₂/p₁) = (∆Hv/R)∙(1/T₁ - 1/T₂)

You know
p₁ = 23.8 mm of Hg
T₁ = 25°C = 298.15K

Water boils at the temperature at which vapor pressure is equal to the ambient pressure. Ambient pressure on top of the Everest is 30% of standard atmospheric pressure, i.e.
p₂ = 0.3 X 760 mm of Hg = 228 mm of Hg

Solve equation above for T₂ and plug in the values:
ln(p₂/p₁) = (∆Hv/R)∙(1/T₁ - 1/T₂)

1/T1 - 1/T2 = ln (P2/P1)/ (∆Hv/R)

T₂ = 1/[ 1/T₁ - (R/∆Hv)∙ln(p₂/p₁) ]
= 1/[ 1/298.15K - (8.314472J/molK/40790J/mol)∙ln(228/23.8) ]
= 345.61K
= 72.46 °C