What is the density of wet air with 75% relative humidity at 1 atm and 300 K? Given : vapour pressure of `H_(2)O` is 30 torr and average molar mass of air is 29 g `mol^(_1)` .
A
1.614 g/L
B
0.96 g/L
C
1.06 g/L
D
1.164 g/L
Text Solution
Verified by Experts
The correct Answer is:
d
(d) % Relative humidity `=("Partial pressure of "H_(2)O)/("Vapour pressure of " H_(2)O)xx100` `75=(P_(H_(2)O))/(30)xx100` `P_(H_(2)O)=22.5"torr,"` % of `H_(2)O` vapour in air `=((22.5))/(760)xx100=2.96` molar mass of wet air `=(29xx97.04+2.96xx18)/(100)` `=(2814.16+53.28)/(100)=28.67` density of wet air `=(PM)/(RT)=(1xx28.67)/(0.0821xx300)` `=1.164g//L`
Topper's Solved these Questions
GASEOUS STATE
NARENDRA AWASTHI|Exercise Level 1 (Q.31 To Q.60)|1 Videos
GASEOUS STATE
NARENDRA AWASTHI|Exercise Level 1 (Q.121 To Q.150)|1 Videos
What is the density of wet air with 40% relative humidity at 1 atm and 284.5 K. [Given vapour pressure of H_(2)O=47.5 torr, Average molar mass of dry air = 29 g mol^(-1) ]
Density of air (in g/L) at 307°C and 2 atm pressure is (Average molar mass of air is 29 g mol-1)
821mLN_(2) (g) was collected over liquid water at 300 K and 1 atm.If vapour pressure of H_(2)O is 30 torr then moles of N_(2) (g) in moist gas mixture is :
Find the mass of water vapour per cubic metre of air at temperature 300K and rekative humidity 50%. The saturation vapour pressure at 300K is 3.6kPa and the gas constant R=8.3JK^(-1)mol^(-1) .
The masspercentage in composition of dry air at sea level contains contains approximately 75.5% of N_(2) . If the total atmosphere pressure is 1 atm then what will be the partial pressure of nitrogen? [Given : average molecular mass of air = 29 ]
How much water vapour is contained in a cubic room of 4m along an edge if the relative humidity 1s 50% water at 27^(@)C The vapour pressure of water at 27^(@)C is 26.7 torr (The relative humidity expresses the partial pressure of water as a per cent of water vapour pressure) .
