The ambient temperature is 85.0°F and the humidity of the surrounding air is reported to be 68.0%. Using the Clausius-Clapeyron equation and the boiling point of water as 100.0°C at 760 torr, calculate the vapor pressure (in torr) of water in the air. Use 40.7 kJ/mol as the ∆Hvap of water.

The vapor pressure of water in the air is 27.58 torr

Explanation:

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

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b