The pressure in a bulb dropped from `2000` to `1500 mm Hg` in `47 min` when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight `79` in the molar ratio of `1:1` at a total pressure of `4000 mm` of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of `74 min`.

Explanation

Now at `P prop n ` ( moles ) , we define the rate of diffusion as the drope in the pressure per second. First we try to find the rate of diffusion of the gas B.
The rate of diffusion of `O_(2) = R_(0) = ( 2000 -1500) // 47 =10.638 m m // `min.
Assuming that gas B was present alone in the bulb. Let the rate of diffusion of `B =R_(B)`
From Graham's Law of diffusion,we have `:`
`( R_(B))/(R_(O)) = sqrt((M_(O_(2)))/( M_(B)))= sqrt((32)/( 79)) = 0.636`
`rArr R_(B) = 10.638 xx 0.636 =6.77m m //` min
Now the bulb contains mixture of `O_(2)` and B in the mole ration of `1:1` at total pressure of 4000 mm Hg.
`rArr P_(O_(2)) = P_(B) = 2000` mm of Hg
As the pressure and temperature conditions are same for both gases in the second case ( same bulb ) , so the rate of diffusion will remain same in the second case also.
Let `X_(0)` and `X_(B)` be the final pressure in the bulb after leakage for 74 minutes.
`R_(O) = ( 2000 -X_(0))/( 74) = 10.638`
`rArr X_(0) = 1212. 78`
`R_(B)=( 2000 x X_(B))/( 74) =6.77 rArr X_(B) = 1498.96`
As `P prop n`
`rArr` Ratio of moles is given as `: X_(O) : X_(B) =1 :1235`