Question

2 A mountaineer climbs Mount Everest and wishes to make a strong cup of tea. He boils his kettle, but the final drink tastes lousy because the water boiled at too low of a temperature, because the pressure at the top of the mountain is 0.4 bar. Given that the A = 50 and the normal boiling point of water as 373 K. calculate the temperature the water boils at the top of the mountain.

Answer 1

2.

From Clausius-Clapeyron equation,

$ln\frac{P_{2}}{P_{1}} = - \frac{\Delta _{vap}H^{\ominus }}{R}\left ( \frac{1}{T_{2}} - \frac{1}{T_{1}}\right )$

where,

P₁ = atmospheric pressure for normal boiling point = 1 bar

P₂ = pressure at the top of the mountain = 0.4 bar

T₁ = normal boiling point of water = 373 K

T₂ = temperature at which water boils on top of mountain

$\Delta _{vap}H^\ominus$= molar heat of vaporization = 50 kJ/mol = 50000 J/mol     ($\because$1 kJ = 1000 J)

R = gas constant = 8.314 J/mol.K

Now, substituting these values in the above equation to get,

$ln\frac{0.4}{1} = - \frac{50000}{8.314}\left ( \frac{1}{T_{2}} - \frac{1}{373}\right )$

or, - 0.916 = - 6014(1/T₂ - 1/373)

or, (1/T₂ - 1/373) = 1.52 x 10^-4

or, 1/T₂ = 0.00283

or, T₂ = 353 K

Hence, the temperature at which water boils at the top of the mountain = 353 K