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

Suppose `p_(O_(2))` and p be the pressure-drops per minute for `O_(2)` and an unknown gas x (say) respectively.
`therefore p_(O_(2)) = (2000 - 1500)/(47) = 10.64` mm/min.
`therefore (r_(O_(2)))/(r_(x)) = (p_(O_(2)))/(p_(x)) = sqrt((M_(x))/(M_(O_(2))))`
`(10.64)/(P_(x)) = sqrt((79)/(32)), p_(x) = 6.77` mm/min.
Sicne the bulb is now refilled with equal number of moles of `O_(2)` and x, the partial pressures of each gas will be 2000 mm as the total pressure is 4000 mm.
`therefore` pressure of `O_(2)` after 75 min
= partial press. of `O_(2)` - press. drop in 74 min
`= 2000 - (10.64 xx 74)`
= 1212.64 mm
and pressure of x after 74 min
= partial press. of x - press. drop after 74 min.
`("moles of " O_(2) " left after 74 min")/("moles of x left after 74 min") = ("pressure of " O_(2) " after 74 min")/("pressure of x after 74 min")`
`(1212.64)/(1499.02) = 0.8089`.
Hence, molar ratio of `O_(2)` and x after 74 minutes is 0.8089 : 1.