Question

3. At sea level, the atmospheric pressure is 101.3 kPa. Atop Mount Everest, the atmospheric pressure is 33.7 kPa. Considering the normal boiling point of water (100 °C) and its heat of vaporization (40.7 kJ/mol), at what temperature will water boil atop Mount Everest? m

Answer 1

Answer : The temperature at top mount everest will be 344 K

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = atmospheric pressure at at sea level = 101.3 kPa

$P_2$ = atmospheric pressure at top mount everest = 33.7 kPa

$T_1$ = normal boiling point of water = $100^oC=273+100=373K$

$T_2$ = temperature at top mount everest = ?

$\Delta H_{vap}$ = heat of vaporization = 40.7 kJ/mole = 40700 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{33.7}{101.3})=\frac{40700J/mole}{8.314J/K.mole}\times (\frac{1}{373}-\frac{1}{T_2})$

$T_2=344.141K\approx 344K$

Hence, the temperature at top mount everest will be 344 K