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

Answer 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_1V=mu_1 RT and P_2 V=mu_2 RT,ie.(P_1//P_2)=(mu_1//mu_2)`.Here 1 and 2 refer to neon and oxgen respectively.Since `(P_1//P_2)=(3//2)(given),(mu_1//mu_2)=3//2`. (i) by definitaion `mu _1=(N_1//N_A) and mu_2 =(N_2//N_A) where N_1 and N_2 ` are the number of molecules of 1 and 2 `N_A` is the Avogardo'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 thier molecular masses.(both `M_1 and M_2 " as well as" m_2 " and" M_2` should be expressed in the same untis. 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_2xx((M_1)/M_2)`
`=3//2xx20.2//32.0=0.947`