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A vessel contains two non reactive gases : neon (monatomic) 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 Ne=20.2 u, molecular mass of O_(2) = 32.0 u. |
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Answer» Solution :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 referto 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` |
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