Question

assume that dry air contains 79%N2 and 21%O2 by volume calculate the density of dry air and moist air at 25°c and 1atm pressure when relative humidity is 60% The total vapour pressure of water at 25°c is 23.76mm Hg​

Answer 1

The density of moist air is 1.16 g/L  

Explanation:

As given the relative humidity is 60 % i.e the partial pressure of water at which it is equal to the vapour pressure of water.

Hence the partial pressure of water in vapour at this condition will be pv =  

                            = 0.6 × 23.76 mm of Hg = 14.256 mm of Hg = 0.01875 atm

As the atmospheric pressure = 760 mm of Hg  

The rest of the pressure is the pressure of the dry air pd =  

                                              = 760 -  14.256 mm of Hg  

                                              = 745.744 mm of Hg  = 0.981 atm  

Molar mass of dry air = 28.94 g/ mol

Molar mass of water vapour = 18 g / mol  

Now, putting these values in formula :  

​ρhumid air = ρdMd+ ρvMvRT

= 0.981×28.94 + 0.01875×18 0.0821×298                

= 28.7124.77                

= 1.16 g / L

So, the density of moist air is 1.16 g/L  

Answer 2

$\huge \tt 1.16 \: g \: {l}^{ - 1}$