Pressure Drop In Bulb : Application of Graham's Law

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SUMMARY

The discussion focuses on calculating the molar ratio of oxygen and another gas in a bulb after a pressure drop, applying Graham's Law of Effusion. Initially, the bulb contained oxygen and a gas with a molecular weight of 72 at a total pressure of 6000 mm Hg. After 70 minutes, the calculated pressure of oxygen was 2300 mm Hg, while the pressure of the other gas was determined to be 7600/3 mm Hg. The correct molar ratio of the gases was found to be 39/46, highlighting a common misconception regarding the relationship between pressure and diffusion rates.

PREREQUISITES
  • Understanding of Graham's Law of Effusion
  • Knowledge of partial pressure calculations
  • Familiarity with the concept of molar ratios
  • Basic principles of kinetic theory
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  • Study the application of Graham's Law in various gas mixtures
  • Learn about the kinetic theory of gases and its implications on diffusion
  • Explore advanced pressure and effusion calculations in chemistry
  • Investigate the effects of temperature on gas behavior and diffusion rates
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Chemistry students, educators, and professionals involved in gas behavior analysis and effusion processes will benefit from this discussion.

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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