Pressure Drop In Bulb : Application of Graham's Law

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The discussion focuses on calculating the molar ratio of oxygen to another gas in a bulb after a pressure drop due to leakage. Initially, the bulb contained oxygen and another gas at equal partial pressures of 3000 mm Hg. Over 70 minutes, the pressure of oxygen decreased by 700 mm Hg, while the other gas's pressure decreased at a different rate according to Graham's Law. The calculations led to an incorrect molar ratio of 76/69, while the correct answer is 39/46. The importance of considering the proportionality of diffusion rates to pressure in effusion scenarios is emphasized.
Sanchayan Dutta
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Homework Statement


The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 50 minutes when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight 72 in the molar ratio of 1 : 1 at a total pressure of 6000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 70 minutes.

Homework Equations


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The Attempt at a Solution


Since the evacuated bulb contains a mixture of oxygen and another gas in the molar ratio of 1 : 1 at a total pressure of 6000 mm, the partial pressure of each gas is 3000 mm.The drop in the pressure of oxygen after 70 minutes= 500/50*70 = 700 mm of Hg∴ After 70 minutes, the pressure of oxygen= 3000-700=2300 mm of HgLet the rate of diffusion of other gas be rn, thenRate of O2/Rate of gas =1.5 (graham's formula)∴ Drop in pressure for the other gas = 700/1.5=1400/3 mm Hg∴ pressure of the other gas after 70 minutes= 3000 – 1400/3 mm = 7600/3 mm of HgMolar ratio = Moles of unknown gas/Moles of O2= (7600/3)/(2300) = 76/69[Partial pressure ∝ mole fraction]

But the answer is 39/46.

Where am I going wrong?
 
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Sanchayan Dutta said:

The Attempt at a Solution


Since the evacuated bulb contains a mixture of oxygen and another gas in the molar ratio of 1 : 1 at a total pressure of 6000 mm, the partial pressure of each gas is 3000 mm.

The drop in the pressure of oxygen after 70 minutes

= 500/50*70 = 700 mm of Hg
When the initial pressure of the oxygen is 2000 mm Hg, the pressure drops by 500 mm Hg after 50 minutes.

If the initial pressure of the oxygen is changed to 3000 mm Hg, would the pressure still drop by 500 mm after 50 minutes?
 
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So should I use that rate of diffusion is directly proportional to pressure?
 
Sanchayan Dutta said:
So should I use that rate of diffusion is directly proportional to pressure?
Yes. Here you are dealing with "effusion". Kinetic theory shows the rate is proportional to the pressure.
https://en.wikipedia.org/wiki/Effusion
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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