Question

g Tibet (altitude above sea level is 29,028 ft) has an atmospheric pressure of 240. mm Hg. Calculate the boiling point of water in Tibet. The heat of vaporization for water is 40.7 kJ/mol.

Answer 1

Answer: The boiling point of water in Tibet is 69.9°C

Explanation:

To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg      (Conversion factor:  1 atm = 760 mmHg)

$P_2$ = final pressure = 240. mmHg

$\Delta H_{vap}$ = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature or normal boiling point of water = $100^oC=[100+273]K=373K$

$T_2$ = final temperature = ?

Putting values in above equation, we get:

$\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC$

Hence, the boiling point of water in Tibet is 69.9°C