Question

A 5.00g sample of liquid H2O is placed in a sealed 5.00L vessel at a temperature of 45°C. The vapor pressure of H2O at 45°C is 0.0103 atm.

A.After some time will liquid water be present?

B. At this point, what is the pressure of the container?

Answer 1

(A.) Vapor pressure of water = 0.0103 atm

volume of container = 5.00 L = volume of water vapor

Temperature of water = 45 ^oC = 318 K

According to ideal gas law,

moles of gas = (Pressure of gas * volume of gas) / (R * temperature of gas)

where R = gas constant = 0.0821 L-atm/mol-K

moles of water vapor = (Pressure of water vapor * volume of water vapor) / (R * temperature of water vapor)

moles of water vapor = [(0.0103 atm) * (5.00 L)] / [(0.0821 L-atm/mol-K) * (318 K)]

moles of water vapor = 1.97 x 10^-3 mol

mass of water vapor = (moles of water vapor) * (molar mass of water)

mass of water vapor = (1.97 x 10^-3 mol) * (18.0 g/mol)

mass of water vapor = 0.0355 g

Since mass of water vapor (0.0355 g) is less than mass of water placed intially (5.00 g), therefore, after some time, liquid water will be present.

(B.) At this point, pressure of container = vapor pressure of water at 45 ^oC = 0.0103 atm