A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m3. Find the pressure of the mixture.
Answers
Therefore total pressure = 2,250 pa
Explanation:
Atomic mass: Atomic mass of a chemical component is the mass of chemical component of 1 mole.
Atomic mass of O₂ is = 32 gram / mole
It means the weight of 1 mole is 32 gram.
Therefore the number of mole in 1.6 gram of O₂ is gram
Atomic mass of N₂ is 28 gram.
It means the weight of 1 mole is 28 gram.
Therefore the number of mole in 1.6 gram of N₂ is = 0.1 gram.
A ideal gas always flow P V=nRT
P = pressure
V = volume= 0.166 m³
n = no. of molecules
R= gas constant= 8.3 J/mole
T = temperature = 300 k
The partial pressure of O₂ is
=750 pa
The partial pressure of N₂ is
=1550 pa
Therefore total pressure = (750+1550)pa=2,250 pa
The Pressure of the oxygen and nitrogen mixture is 2250 Pa
Explanation:
Given data in the question
V = 0.166
T = 300 K
Mass of Oxygen ( ) = 1.60 g
Molecular weight of Oxygen = 32 g
Mass of Nitrogen () = 2.80 g
Molecular weight = 28 g
Partial pressure is given by
Partial pressure is given by
Maximum pressure is the number of partial loads.
The pressure of the mixture which contains 1.60 g of Oxygen and 2.80 g of Nitrogen with 300 K temperature is 2250 Pa in a vessel of volume 0.166