A vessel contains two non-reactive gases neon (monoatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. Estimate the ratio of
(i) number of molecules, and
(ii) mass density of neon and oxygen in the vessel.
Atomic mass of neon = 20.2 u, and molecular mass of oxygen = 32.0 u.

Explanation

Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have `P_(1) V = mu_(1), RT` and `P_(2)V= mu_(2)RT`, i.e. `(P_(1)//P_(2)) = (mu_(1) // mu_(2))`. Here 1 and 2 refer to neon and oxygen respectively. Since `(P_(1)//P_(2)) = (3//2)` (given), `(mu_(1) // mu_(2))= 3//2`.
(i) By definition `mu_(1) = (N_(1)//N_(A))` and `u_(2) = (N_(2)//N_(A))` where `N_(1)` and `N_(2)` are the number of molecules of 1 and 2, and `N_(A)` is the Avogadro.s number. Therefore, `(N_(1)//N_(2)) = (mu_(1) // mu_(2)) = 3//2`.
(ii) We can also write `mu_(1) = (m_(1)//M_(1))` and `mu_(2) = (m_(2)//M_(2))` where `m_(1)` and `m_(2)` are the masses of 1 and 2, and `M_(1)` and `M_(2)` are their molecular masses. (Both `m_(1)` and `M_(1)`, as well as `m_(2)` and `M_(2)` should be expressed in the same units). If `rho_(1)` and `rho_(2)` are the mass densities of 1 and 2 respectively, we have
`(rho_(1))/(rho_(2))= (m_(1)//V)/(m_(2)//V)= (m_(1))/(m_(2))= (mu_(1) )/ (mu_(2)) xx (M_(1)/(M_(2)))`
`= (3)/(2) xx (20.2)/(32.0)= 0.947`