Pressure Drop In Bulb : Application of Graham's Law

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Homework Help Overview

The problem involves the pressure drop in a bulb containing a mixture of gases, specifically oxygen and another gas with a molecular weight of 72. The original poster seeks to determine the molar ratio of the two gases remaining in the bulb after a specified time, given initial conditions and pressure changes.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply Graham's Law to relate the rates of diffusion of the gases based on their partial pressures and molecular weights. Some participants question whether the rate of diffusion is directly proportional to pressure, while others explore the implications of changing initial pressures on the diffusion rates.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning the assumptions made regarding the relationship between pressure and diffusion rates. Some guidance has been offered regarding the concept of effusion and its relation to pressure.

Contextual Notes

There are indications of confusion regarding the application of Graham's Law and the effects of pressure changes on the rates of diffusion, as well as the initial conditions of the gases involved.

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
 

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