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The pressure of bulb dropped from 2000mmHg to 1500mmHg in 47 min, when a contained O2 gas is leaved through a small hole, the bulb was then evaculated and the mixture of O2 and another gas (m=79)in the mole ratio 1 :1 at total pressue of 4000mmHg was introduced. Find the molar ratio of two gases of the bulb after 74 min ?

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Solution

NOTE THIS STEP: Since the evacuated bulb contains a mixture of oxygen and another gas in the molar ratio of 1 : 1 at a total pressure of 4000 mm, the partial pressure of each gas is 2000 mm. The drop in the pressure of oxygen after 74 minutes = 500/47 = 787.2 mm of Hg ׵ After ͹Ͷ minutes, the pressure of oxygen = 2000 – 787.2 = 1212.8 mm of Hg Let the rate of diffusion of other gas be rn, then 2 32 79 n O r r  ׵ Drop in pressure for the other gas = 787.2 * 32/79 = 501.01 mm of Hg ׵ pressure of the other gas after ͹Ͷ minutes = 2000 – 501.01 mm = 1498.99 mm of Hg Molar ratio = Moles of unknown gas/Moles of O2 = 1498.99/1212.8 = 1.24/1 = 1.24 : 1. [Partial pressure ∝ mole fraction]

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