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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. |
Answer» 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` `therefore (P_(1))/(P_(2))=(mu_(1))/(mu_(2))` where 1 and 2 refer to neon and oxygen respectively Given `(P_(1))/(P_(2))=(3)/(2) therefore (mu_(1))/(mu_(2))=(N_(2))/(N_(A))` where `N_(1)` and `N_(2)`are the number of molecules of 1 and 2 `therefore` The ratio of number of molecules `(N_(1))/(N_(2))=(mu_(1))/(mu_(2))=(3)/(2)` (ii) If `rho_(1)` and `rho_(2)` are mass densities of 1 and 2 respectively, we have `(rho_(1))/(rho_(2))=(m_(1)//V)/(m_(2)//V)=(m_(1))/(m_(2))` But we can also write `mu_(1)=(m_(1))/(M_(1))` and `mu_(2)=(m_(2))/(M_(2))` `therefore (m_(1))/(m_(2))=(mu_(1))/(mu_(2))xx(M_(1))/(M_(2))=(3)/(2)xx(20.2)/(32.0)=0.947` |
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