A 30 litre sample of moist air at 50c and 1 atm has 75% relative humidity. Then what will be the final volume if it is compressed isothermally until 100% relative humidity. (given : vapour pressure of water at 50c is 0.2 atm)
Answers
Step-by-step explanation:
It should be 1.165 g/L.
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The relative humidity ϕ of an air-water compound is given by:
ϕ=PH2OP*H2O=0.75
where PH2O is the incomplete vapour pressure of the liquid in the space and * means the object in obscurity.
In this instance, we have
P*H2O=30 torr
PH2O=0.75×30 torr=22.5 torr
at P wet air=1 atm=760 torr and T=300 K.
As we understand the pressure of water vapour in the air is 22.5 torr, the vapour pressure of the dry air is determined by reduction:
P dry air=760 torr−22.5 torr=737.5 torr
or 0.970 atm. And therefore, the vapour pressure of the liquid in the air mixture is 0.030 atm.
This modification on the ideal gas law, allowing air is an absolute gas, can be used to find its density:
PM=DRT
where:
P is the pressure in the atmosphere
M is the molar mass
D is the density
R=0.082057 L⋅atm/mol⋅K is the universal gas constant.
T is the temperature in K.
In a provided amount 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:
D wet air=Ddry air+DH2O
=P dry air M air RT+PH2O M H2O RT
=P dry air M air+P H2O M H2O RT
=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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