A certain amount of N2 gas at 27 degree celsius is measured for Pressure in a gas monometer . The Pressure observed is 800 mm of Hg . If aqueous tension at this temperature is 20mm of Hg.Find the mass of gas if its volume is 2l.
Answers
Answer:
2.336g
Explanation:
P(dry) = P (moist) - aqueous tension
=800-20
=780
PV=nRT
(780/760) *2 = (m/28)*300*0.082
156/76= (m/28)*3*8.2
m= (156*28)/(3*8.2*76)
m=2.336g
The mass of nitrogen gas present is 2.3 g.
Given,
Temperature of nitrogen gas=27°C
Pressure observed through manometer=800 mm of Hg
Aqueous tension at this temperature=20 mm of Hg
Volume of the gas=2 L.
To find,
the mass of nitrogen gas present.
Solution:
The gas whose pressure is being measured with the help of manometer is a wet gas.
The pressure of a wet gas is given as:
P(wet gas)=P(dry gas)+Aqueous tension.
So the pressure of the dry gas will be given as:
P(dry gas)=P(wet gas)-Aqueous tension.
The pressure that the manometer measures is that of a wet nitrogen gas, thus the pressure of dry nitrogen gas will be:
P(dry nitrogen)=P(wet nitrogen)-Aqueous tension
P(dry nitrogen)=(800-20) mm of Hg
P(dry nitrogen)=780 mm of Hg.
Now this pressure of 780 mm of Hg needs to be converted into atm.
760 mm of Hg equals to 1 atm.
780 mm of Hg will be equal to=(1/760) x 780
780 mm of Hg will be equal to 1.02 atm.
Let the mass of the gas be m, so its moles will be equal to m/28.
Molar mass of nitrogen gas is 28 g/mol.
Temperature in degree celsius needs to be converted into Kelvin, T=273+27=300K.
According to ideal gas equation,
PV=nRT
1.02 x 2=(m/28) x 0.0821 x 300
2.04=(m/28) x 0.0821 x 300
(2.04 x 28)/0.0821 x 300=m
m=57.12/0.0821 x 300
m=57.12/24.63 g
m=2.3 g.
Hence, the mass of gas is 2.3 g.
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