Question

(a) What is the vapor pressure of water at 20.0^◦C? (b) What percentage of atmospheric pressure does this correspond to? (c) What percent of 20.0^◦C air is water vapor if it has 100% relative humidity? (The density of dry air at 20.0^◦C is 1.20 kg/m³).

I got that the vapor pressure of the water will be 2.33x10³ Pa, but I am getting hung up on the formula for part b, which is where I can really use the extra help and explanation. Thanks!

Answer 1

b)

Divide the vapor pressure by atmospheric pressure:

2.33x10³ / (1.01*10⁵) *100 = 2.30%

c)

The density of water in this air is equal to the saturation vapor density of water at this temperature, taken as 1.72*10^-2 kg/m^3 Dividing by the density of dry air, we can get the percentage of water in the air:

1.72*10^-2 kg/m^3 /(1.20 kg/m^3) *100 = 1.43 %