<u>Answer:</u> The vapor pressure of water in the air is 27.58 torr
<u>Explanation:</u>
We are given:
Temperature of water in air = 85.0°F
Converting the temperature from degree Fahrenheit to degree Celsius is:
where,
= temperature in Fahrenheit
= temperature in centigrade
So,
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
where,
= initial pressure which is the pressure at normal boiling point = 760 torr
= final pressure = ?
= Enthalpy of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature =
= final temperature =
Putting values in above equation, we get:
We are given:
68.0 % of water in the air
This means that 68 grams of water is present in the air
Mass of air = 100 - 68 = 32 g
To calculate the number of moles, we use the equation:
Molar mass of water = 18 g/mol
Average molar mass of air = 29 g/mol
Mole fraction of a substance is given by:
To calculate the mole fraction of substance, we use the equation given by Raoult's law, which is:
where,
= vapor pressure of water = ?
= total pressure = 35.72 torr
= mole fraction of water = 0.774
Putting values in above equation, we get:
Hence, the vapor pressure of water in the air is 27.58 torr