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 : 1.8 )`

Explanation

Rate of effusion is expressed as `-(dp)/(dt) = (kp)/(sqrtM)`
K = constant, p = instantaneous pressure `-(dp)/(p) = (k dt)/(sqrtM)`
Integration of above equation gives in `((p_(o))/(p))= (kt)/(sqrtM)`
Using first information : in `((2000)/(1500)) = (k47)/(sqrt32) rArr k = (sqrt32)/(47) ln ((4)/(3))` ... (i)
Now in mixture, initially gases are taken in equal mole ratio, hence they have same initial partial pressure of 2000 mm of Hg each.
After 74 min: For `O_(2) ln ((2000)/(P_(O_(2)))) = (74k)/(sqrt32)`
Substituting k from Eq. (i) gives `ln ((2000)/(P_(O_(2)))) = (74)/(sqrt32) xx (sqrt32)/(47) ln ((4)/(3)), ln ((2000)/(P_(O_(2)))) = (74)/(47) ln((4)/(3))`
Solving k from Eq. (i) gives `ln ((2000)/(p_(g))) = (74)/(sqrt79) xx (sqrt32)/(47) ln ((4)/(3))`
Solving gives : `p_(g) = 1500 mm rArr` After 74 min `p(O_(2)) : p(g) = 1271.5 : 1500`
Also, in a mixture, partial pressure `prop` number of moles `rArr n (O_(2)) : n(g) = 1 : 1.18`