Pressure Drop In Bulb : Application of Graham's Law

In summary, the pressure in a bulb containing oxygen and another gas dropped by 500 mm of Hg after 50 minutes due to a small hole. After being evacuated, the bulb was refilled with a mixture of oxygen and another gas in a 1:1 molar ratio at a total pressure of 6000 mm of Hg. After 70 minutes, the pressure of oxygen in the bulb was 2300 mm of Hg and the pressure of the other gas was 7600/3 mm of Hg. The molar ratio of the two gases remaining in the bulb was found to be 76/69. However, using the rate of diffusion being directly proportional to pressure, the correct molar ratio after
  • #1
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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  • #2
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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  • #3
So should I use that rate of diffusion is directly proportional to pressure?
 
  • #4
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
 

1. What is Graham's Law?

Graham's Law, also known as Graham's Law of Effusion, states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass.

2. How is Graham's Law applied to pressure drop in a bulb?

In the context of pressure drop in a bulb, Graham's Law can be used to predict the rate of diffusion of a gas from one compartment to another, based on the difference in molar masses between the two gases.

3. What causes pressure drop in a bulb?

Pressure drop in a bulb is caused by the diffusion of gas from one compartment to another. This can occur due to a difference in concentration, temperature, or pressure between the two compartments.

4. How does pressure affect the rate of diffusion in a bulb?

According to Graham's Law, the rate of diffusion is inversely proportional to the pressure of the gas. This means that as pressure increases, the rate of diffusion decreases, and vice versa.

5. What factors can affect the pressure drop in a bulb?

The pressure drop in a bulb can be affected by various factors such as the molar masses and concentrations of the gases involved, the temperature and pressure of the system, and the size and shape of the bulb or container.

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