Question
Question: If the four tubes of a car are filled to the same pressure with \({N_2},{O_2},{H_2}\) and \(He\) sep...
If the four tubes of a car are filled to the same pressure with N2,O2,H2 and He separately then which one will be filled first:
A) N2
B) O2
C) H2
D) He
Solution
Refer to Graham's law of diffusion. According to the Graham’s law of diffusion, at constant temperature and pressure, diffusion rate of gases are inversely proportional to the square root of their density or hence molar mass.
Complete step by step solution:
Graham’s law gives us an empirical relationship that states that the rate of diffusion or effusion of two gases is inversely proportional to the square root of their molar masses.
Thus, we can write:
r2r1=M1M2
Where, r1 is the rate of diffusion of one gas, having molar mass M1 and r2 is the rate of diffusion of an another gas, having molar mass M2.
This relationship is derived from one of the postulates of the kinetic theory of gases which states that temperature in Kelvin is proportional to the kinetic energy of the gas molecules.
We can relate from the Graham’s law equation that, if a gas has higher molar mass, then its rate of diffusion or effusion will be lesser since rate of diffusion is inversely proportional to the molar mass.
Now, we are given gases N2,O2,H2 and He.
Molar mass of gas N2= 14 g/mol
Molar mass of gas O2 = 16 g/mol
Molar mass of gas H2= 2 g/mol
Molar mass of gas He = 4 g/mol
Thus, order of molar mass of gases:
H2<He<N2<O2
Hence, H2 has the least molar mass, thus it will have the highest rate of effusion or diffusion.
Therefore, if the four tubes of a car are filled to the same pressure with N2,O2,H2 and He separately then, H2will be filled first.
Hence, option C is correct.
Note: The rate at which a molecule diffuses or effuses is directly proportional to the speed at which it moves. Also, Graham’s law is a direct consequence of the fact which is given by the kinetic theory of gases that gas molecules at the same temperature possess the same average kinetic energy.