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:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor: 1 atm = 760 mmHg)
[tex]P_2[/tex] = final pressure = 240. mmHg
[tex]\Delta H_{vap}[/tex] = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature or normal boiling point of water = [tex]100^oC=[100+273]K=373K[/tex]
[tex]T_2[/tex] = final temperature = ?
Putting values in above equation, we get:
[tex]\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[/tex]
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
[tex]T(K)=T(^oC)+273[/tex]
[tex]342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC[/tex]
Hence, the boiling point of water in Tibet is 69.9°C
