# 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?

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#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 **dry air** is found by subtraction:

#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")#