The pressure in a bulb of constant volume dropped from 2000 to 1500 mm of Hg in 47 minutes when the contained He leaked through a small hole. Calculate the pressure of CO2in the same vessel after 20 minutes if the initial pressure of CO2 in the vessel is 2000 mm of Hg. Assume the gases are leaking through the same hole to vacuum. Also assume the rate of leakage of gas is directly proportional to the pressure of gas in the vessel under given conditions.
Answers
Answered by
2
Total pressure of the mixture of oxygen and another gas in the molar ratio of 1 : 1 is 4000 mm of Hg, thus pressure of each gas is 2000 mm of Hg. In 47 min change in pressure of oxygen = (2000 - 1500) mm Hg = 500 mm Hg .'. After 74 min the decrease in pressure of oxygen = 500 x 74/47 = 787.2 mm Hg :. After 74 min, remaining pressure of oxygen = 2000 - 787.2 = 1212.8 mm Hg From Graham's law of effusion Thus, after 74 min, the remaining pressure of another gas = 2000 - 500.8 = 1499.2 mn Hg
I hope it's helpful marked me as brainlist plz
Similar questions