Two tanks of equal volume contain equal masses of oxygen and nitrogen at 410 K. find the ratio of (i) the number of molecules of the gases (ii) the presesure exerted by the gases in the two tanks .

Explanation

Data `: m_(o)=m_(N)` (equal masses of the two gases),equal volumne and the same temperature , `M_(O_(2)) = 32 "g/mol" M_(N_(2)) = 28` g/ mol
`n_(o) and n_(N)` the respective number of moles, and `N_(A)` the Avogardro number.
Then `n_(0) = N_(o)/N_(A) and n_(N) = N_(N)/(N_(A))`
`m_(o) = n_(o) . M_(o)_(2) = N_(0)/N_(A) . M_(0_(2)) and M_(n) =n_(N) = N_(N)/N_(A).M_(N_(2))`
(i) since ` m_(0) = n_(0), N_(0)M_(0) = N_(N) M_(N)`
`N_(o)/N_(N) = (M_(n_(2)))/(M_(o_(2))) = 28/32 = 7/8`
This gives the ratio of the number or oxygen molecules to that of nitrogen.
(ii) Using the ideal gas equation , pV = nRT
` P_(o) =n_(o) (RT)/V and p_(n) = n_(N) = (RT)/V`
Since the gases occupy the same volume V and are at the same temperature T.
` p_(o)/p_(N) = n_(o)/n_(N) = M_(N_(2))/M_(o_(2)) " " therefore n_(o)/ M_(o_(2)) = n_(N) . M_(N_(2)) ` From Eq (2) ]
` 28/32 = 7/8 `
This gives the corresponding pressure ratio.