At 20'c , the vapour pressure of pure water is 17.5 torr, how many mol of water molecules are present per litre of air at 20'c, when relative humidity is 45 %
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20 degree centigrade = 293 K
The saturation vapour pressure is:
es=Aexp[−BT]es=Aexp[−BT]
where A = 2.53x10^11 Pa and B = 5420 K and T is in Kelvin
Plugging these values, es=2341 Paes=2341 Pa
The Vapour pressure is:
e=es∗RH100e=es∗RH100
So, we have e = 1053.48 Pa
The specific humidity q is given by the relation:
q=0.622e/pq=0.622e/p
Where P is the atmospheric pressure.
Assuming P = 10^5 Pa, then we have q = 6.553 grams of water vapour per kg of air.
The density of air is given by the relation:
ρ=P/[RT]ρ=P/[RT]
where R is 287 J/KgK. Plugging these values, we have ρ=1.1892kg/m3ρ=1.1892kg/m3
The amount of Water per m^3 of air is :
W = density x q
So, we have W = 7.793 grams of water vapour per m^3 of air.
Since, 1000 L = 1 m^3, in one litre of air, we have 7.793 milli grams of water vapour
The saturation vapour pressure is:
es=Aexp[−BT]es=Aexp[−BT]
where A = 2.53x10^11 Pa and B = 5420 K and T is in Kelvin
Plugging these values, es=2341 Paes=2341 Pa
The Vapour pressure is:
e=es∗RH100e=es∗RH100
So, we have e = 1053.48 Pa
The specific humidity q is given by the relation:
q=0.622e/pq=0.622e/p
Where P is the atmospheric pressure.
Assuming P = 10^5 Pa, then we have q = 6.553 grams of water vapour per kg of air.
The density of air is given by the relation:
ρ=P/[RT]ρ=P/[RT]
where R is 287 J/KgK. Plugging these values, we have ρ=1.1892kg/m3ρ=1.1892kg/m3
The amount of Water per m^3 of air is :
W = density x q
So, we have W = 7.793 grams of water vapour per m^3 of air.
Since, 1000 L = 1 m^3, in one litre of air, we have 7.793 milli grams of water vapour
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