Assuming that dry air contains 79% `N_(2)` and 21% `O_(2)` by volume, calculate the density of dry air at if it has a relative humidity of 40%. The vapour pressure of water at `25^(@)C` is 23.76 mm.

Assuming that dry air contains 79% `N_(2)` and 21% `O_(2)` by volume,

1.	Assuming that dry air contains 79% `N_(2)` and 21% `O_(2)` by volume, calculate the density of dry air at if it has a relative humidity of 40%. The vapour pressure of water at `25^(@)C` is 23.76 mm.
Answer» Calculation of density of dry air. Density of air means the mass of air per litre. first let us convert 1 L `(1000 cm^(3))` of air at `25^(@)C` 1 atm pressure to the volume at S.T.P. `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `:. " " (1 atmxx1000" cm"^(3))/(298" K")=(1 atm xxV_(2))/(273 K) " or " V_(2)=916.1" cm"^(3)` As air contains 79%`N_(2)` and 21% `O_(2)`, `:. " "` Volume of `N_(2)` at S.T.P. in the air `=(79)/(100)xx916.1cm^(3)=723.72" cm"^(3)` Volume of `O_(2)` at S.T.P. in the air `=(21)/(100)xx916.1" cm"^(3)=193.38" cm"^(3)` Mass of `723.72" cm"^(3)` of `N_(2)` at S.T.P. `=(28)/(22400xx723.72=0.905" g"` Mass of `193.38" cm"^(3)` of `O_(2)` at S.T.P. `=(32)/(22400)xx923.38=0.276" g"` `:. ` Total mass of 1 L of air at `25^(@)C` and 1 atm `=0.905+0.276" g"=1.181" g"` Hence, density of dry air `=1.181" g "L^(-1)` Calculation of density of moist air % Relative humidity`=("Partial pressure of "H_(2)O "vapours")/("Vapour pressure of "H_(2)O "at the same temp")xx100` `:. " "` Partial pressure of `H_(2)O` vapours`=(40xx23.76)/(100) =9.5" mm"` As total pressure =1 atm=760 mm, `:. " "` Partial pressure of `N_(2)=(79)/(100)xx750.5" mm "=592.9" mm "` and Partial pressure of `O_(2)=750.5-529.9=157.6" mm"` Applying gas equation PV=nRT, i.e., `n=(PV)/(RT)` No. of moles of `H_(2)O` in the air `(n_(H_(2)O))=((9.5//760)"atm "xx1L))/(0.0821" L atm "K^(-1)mol^(-1)xx298" K ")=5.1xx10^(-4)` No. of moles of `N_(2)` in the air `(n_(N_(2)))=((592.9//760)xx1)/(0.0821xx298)=3.19xx10^(-2)` No. of moles of `O_(2)` in the air `(n_(O_(2))=((157.6//760)xx1)/(0.0821xx298)=8.48xx10^(-2)` Total mass of 1 L of air `=5.1xx10^(-4)xx18+3.19xx10^(-2)xx28+8.48xx10^(-3)xx32` `=0.00918+0.8932+0.02714=1.174" g"` Hence, density of moist air `=1.174" g "L^(-1)` Note. Density of moist dry air can also be calculated by this method.

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Assuming that dry air contains 79% `N_(2)` and 21% `O_(2)` by volume, calculate the density of dry air at if it has a relative humidity of 40%. The vapour pressure of water at `25^(@)C` is 23.76 mm.

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