What is the density of wet air with 75 relative humidity?
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It should be 1.165 g/L. There is a typo in the given answers...
The relative humidity ϕ of an air-water mixture is given by:
ϕ=PH2OP*H2O=0.75
where PH2O is the partial vapor pressure of water in the air and * indicates the substance in isolation.
In this case we have
P*H2O=30 torr
PH2O=0.75×30 torr=22.5 torr
at Pwet 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:
Pdry 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⋅atm/mol⋅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=D1+D2
Thus, the density of the wet air is given by:
Dwet air=Ddry air+DH2O
=Pdry airMairRT+PH2OMH2ORT
=Pdry airMair+PH2OMH2ORT
=0.970atm⋅29 g/mol+0.030atm⋅18.015 g/mol0.082057 L⋅atm/mol⋅K⋅300K
= 1.165 g/L
The relative humidity ϕ of an air-water mixture is given by:
ϕ=PH2OP*H2O=0.75
where PH2O is the partial vapor pressure of water in the air and * indicates the substance in isolation.
In this case we have
P*H2O=30 torr
PH2O=0.75×30 torr=22.5 torr
at Pwet 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:
Pdry 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⋅atm/mol⋅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=D1+D2
Thus, the density of the wet air is given by:
Dwet air=Ddry air+DH2O
=Pdry airMairRT+PH2OMH2ORT
=Pdry airMair+PH2OMH2ORT
=0.970atm⋅29 g/mol+0.030atm⋅18.015 g/mol0.082057 L⋅atm/mol⋅K⋅300K
= 1.165 g/L
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