What is the density of wet air with 75% relative humidity at 1atm and 300K? Given : vapour pressure of #H_2O# is 30 torr and average molar mass of air is 29g/mol?
#a)# #"1.174 g/L"#
#b)# #"1.156 g/L"#
#c)# #"1.178 g/L"#
#d)# #"1.143 g/L"#
1 Answer
It should be
The relative humidity
#phi = P_(H_2O)/(P_(H_2O)^"*") = 0.75# where
#P_(H_2O)# is the partial vapor pressure of water in the air and#"*"# indicates the substance in isolation.
In this case we have
#P_(H_2O)^"*" = "30 torr"#
#P_(H_2O) = 0.75 xx "30 torr" = "22.5 torr"#
at
Since we know the pressure of water vapor in the air is
#P_"dry air" = "760 torr" - "22.5 torr" = "737.5 torr"#
or
This variant on the ideal gas law, assuming air is an ideal gas, can be used to find its density:
#PM = DRT# where:
#P# is the pressure in#"atm"# of the ideal gas. Here we treat wet air as an ideal gas.#M# is the molar mass of a given component in the sample in#"g/mol"# ,#D# is the density of the ideal gas in#"g/L"# .#R = "0.082057 L"cdot"atm/mol"cdot"K"# is the universal gas constant.#T# is the temperature in#"K"# .
In a given volume of air at a given temperature and total pressure, the density for ideal gases is additive, i.e.
#D = D_1 + D_2#
Thus, the density of the wet air is given by:
#color(blue)(D_"wet air") = D_"dry air" + D_(H_2O)#
#= (P_"dry air"M_"air")/(RT) + (P_(H_2O)M_(H_2O))/(RT)#
#= (P_"dry air"M_"air" + P_(H_2O)M_(H_2O))/(RT)#
#= (0.970 cancel"atm" cdot "29 g/"cancel"mol" + 0.030 cancel"atm" cdot "18.015 g/"cancel"mol")/("0.082057 L"cdotcancel"atm""/"cancel"mol"cdotcancel"K" cdot 300 cancel"K")#
#=# #color(blue)("1.165 g/L")#
