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

The pressure in a bulb dropped from 2000 mm to 1500 mm Hg in 47 minite

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