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vlabodo [156]
3 years ago
11

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

to be 296.0 mm Hg. At what temperature (in °C) will water boil at base camp ? The vapor pressure of water at 373 K is 760.0 mm Hg. (ΔH°vap for H2O = 40.7 kJ/mol and R = 8.314 J/mol K)
Physics
1 answer:
Mamont248 [21]3 years ago
4 0

Answer:

74.86°C

Explanation:

P₂ = Vapour pressure of water at sea level = 760 mmHg

P₁ = Pressure at base camp = 296 mmHg

T₂ = Temperature of water = 373 K

ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol

R = Gas constant = 8.314 J/mol K

From Claussius Clapeyron equation

ln\frac{P_2}{P_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\\\Rightarrow ln\frac{760}{296}=\frac{40700}{8.314}\left(\frac{1}{T_1}-\frac{1}{373}\right)\\\Rightarrow ln\frac{760}{296}\times \frac{8.314}{40700}+\frac{1}{373}=\frac{1}{T_1}\\\Rightarrow 0.0028735=\frac{1}{T_1}\\\Rightarrow T_1=347.996\ K

T₁ = 347.996 K = 74.86°C

∴Water will boil at 74.86°C

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