Question

A vessel contains two non-reactive gases neon (monatomic) and oxygen (diatomic). A ratio of their partial pressures is 3:2. The ratio of number of molecules is:
A. $\dfrac{3}{2}$
B. $\dfrac{2}{3}$
C. $\dfrac{1}{3}$
D. $\dfrac{1}{2}$

Answer 1

The volume and the temperature of the two gases would be the same as they are in the same container. Ideal gas equations can be applied to find the required ratio.

Complete answer:
The two gases are non-reactive and would exert pressure on the walls of the container individually equal to their partial pressures as if the other gas were not present. So, the volume and temperature would remain same for the two gases
Application of the ideal gas equation to the two gases separately:
Let ${P_1}$ be the partial pressure of Neon and ${P_2}$ be the partial pressure of Oxygen.
Then, For Neon we can write: ${P_1}V = {n_1}RT$--------(i)
For Oxygen we can write: ${P_2}V = {n_2}RT$ ---------(ii)
Where, V and T are the volume and temperature of the container in which two gases are contained.
Dividing (i) by (ii)
$\dfrac{{{P_1}V}}{{{P_2}V}} = \dfrac{{{n_1}RT}}{{{n_2}RT}}$
$\Rightarrow \dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{n_1}}}{{{n_2}}}$
But, $\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{3}{2}$
Therefore, $\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{{P_1}}}{{{P_2}}} = \dfrac{3}{2}$
That is, the ratio of number of moles of neon to oxygen is $3:2$ and the same would be the ratio of the number of molecules as Avogadro’s number will be the common factor when moles are converted to the number of molecules.

Hence the correct answer is (A).

Note: Since the comparison has to be made between the number of molecules of the two gases, the total pressure of the system is not needed. The separate ideal gas equations must be written and then divided to get the ratio of moles in terms of ratio of partial pressures. The ratio of the number of moles is the ratio of the number of molecules as the number of moles are converted to the number of molecules by multiplying with the common factor which is Avogadro’s number.