Question

Under identical conditions of temperature and pressure the ratio of rates of effusion of ${{O}_{2}}\text{ and C}{{\text{O}}_{2}}$ gases are given by-
(A) $\dfrac{rate\text{ of effusion of oxygen}}{rate\text{ of effusion of C}{{\text{O}}_{2}}}=0.87$
(B) $\dfrac{rate\text{ of effusion of oxygen}}{rate\text{ of effusion of C}{{\text{O}}_{2}}}=1.17$
(C) $\dfrac{rate\text{ of effusion of oxygen}}{rate\text{ of effusion of C}{{\text{O}}_{2}}}=8.7$
(D) $\dfrac{rate\text{ of effusion of oxygen}}{rate\text{ of effusion of C}{{\text{O}}_{2}}}=0.117$

Answer 1

Hint: To solve this, firstly find out the molecular weights of carbon dioxide and oxygen. You can use Graham’s law of effusion to solve this. The law says that the effusion of a gas is inversely proportional to the square root of the molecular weight of the particles in the gas.

Complete step by step solution:
Before answering the question, let us what effusion means.
In simple words, we can say that effusion is the process by which a gas escapes a container through an opening. It is basically the movement of the gas particles through a small hole.
The rate of effusion of a gas is explained to us by Graham’s law of effusion. The law gives us a relation between the rate of effusion and the molecular weights of the particles in the gas.
The law states that- The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of particles in it. We can write it as-
\[Rate\text{ }of\text{ }effusion\propto \sqrt{\dfrac{1}{Molecular\text{ }weight}}\]
For two gases, we can write law relation as-
\[\dfrac{Rat{{e}_{1}}}{Rat{{e}_{2}}}=\sqrt{\dfrac{M.{{W}_{2}}}{M.{{W}_{1}}}}\]
Here we have used 1 and 2 to denote the first and the second gas. M.W is the molecular weight of the corresponding gases and rate is the rate of effusion.
Now, let us use this relation to find the rate of effusion of oxygen and carbon dioxide at similar temperature and pressure.
For that, we have to calculate the molecular weights of oxygen and carbon dioxide.
For oxygen, molecular weight = 32.
For carbon dioxide molecular weight = 12 + 32 = 44.
Let us consider that the first gas is oxygen and the second gas is carbon dioxide.
Therefore, we can write that-
\[\dfrac{Rat{{e}_{1}}}{Rat{{e}_{2}}}=\sqrt{\dfrac{44}{32}}=1.17\]
From the above calculation we can understand that the rate of effusion of oxygen upon the rate of effusion of carbon dioxide is 1.17.

Therefore, the correct answer is the option [B] $\dfrac{rate\text{ of effusion of oxygen}}{rate\text{ of effusion of C}{{\text{O}}_{2}}}=1.17$.

Note: The same law is also valid for diffusion and is called Graham’s law of diffusion. The rate of diffusion of any gas is inversely proportional to the square root of the molecular weight of the particle in the gas.Also, in the same condition of temperature and pressure, the molar mass is proportional to the mass density i.e. the rate of diffusion of different gases is inversely proportional to the square root of their mass densities-
\[r\propto \dfrac{1}{\sqrt{d}}\]