Question

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`.

Answer 1

`2000` mm pressure of `O_(2) overset(t=47 "min")rarr 1500 mm`
`4000 mm` pressure of mixture `overset(t=74 "min") rarr1:1`
`(O_(2) +gas)`
For Pure `O_(2) (P_(1))/(P_(2)) = (n_(1))/(n_(2))`
When `n_(1)` and `n_(2)` are original on of mole of `O_(2)` and mole of `O_(2)` after `47` minute
`:. (n_(1))/(n_(2)) = (2000)/(1500)`
`:. n_(2) = (3//4)n_(1)`
or mole of `O_(2)` diffused in `47 "min" = n_(1) - (3n_(1))/(4) = (n_(1))/(4)`
or mole of `O_(2)` diffused in `74 "min" = (n_(1) xx 74)/(47 xx 4)`
`=0.3936 ("if" N_(1) =1)`
Since diffusion of `O_(2)` in mixture ialso occurs at partial pressure of `2000 mm`
The ratio of gas and `O_(2)` being 1:1
Now gas and `O_(2)` both diffusing in form of mixture through same orifice at the partial pressure pf 2000 mm each
`:.(n_(O_(2)))/(74) xx (74)/(n_(g)) =sqrt(((79)/32))`
`:. n_(g) = n_(O_(2)) xx sqrt(((32)/(79))) = (74)/(188) xx sqrt(((32)/(79)))=0.249`
`:.` Mole of gas left after 74 minate `= 1 -0.249 =0.7510`
`:. O_(2) : gas :: 0.6064 : 0.7510`
`: : 1:1.236` .