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 ((P_2)/(P_1))=(\Delta H_(vap))/(R)* ((1)/(T_1)-(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 ((33.7)/(101.3))=(40700J/mole)/(8.314J/K.mole)* ((1)/(373)-(1)/(T_2))`

`T_2=344.141K\approx 344K`

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