Question

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"`

Answer 1

It should be `"1.165 g/L"`. There is a typo in the given answers...

---

The relative humidity `phi` of an air-water mixture is given by:

`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 `P_("wet air") = "1 atm" = "760 torr"` and `T = "300 K"`.

Since we know the pressure of water vapor in the air is `"22.5 torr"`, the vapor pressure of the dry air is found by subtraction:

`P_"dry air" = "760 torr" - "22.5 torr" = "737.5 torr"`

or `"0.970 atm"`. And thus, the vapor pressure of water in the air mixture is `"0.030 atm"`.

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")`