Two-bulb experiment for measuring gas diffusivity

In summary, the conversation discussed a system used to determine the diffusivity of a binary mixture of gases. The goal was to obtain an expression for x_A^+ as a function of time. A method for plotting the experimental data was suggested and the question of measuring the mole fraction of a species in a gas sample using basic lab equipment was raised. The participant also asked if the steady state concentrations of oxygen and nitrogen would be the same as in air after carrying out the experiment and if the concentrations of O2 and N2 in air depend only on the diffusivity of the O2-N2 pair or if they depend on other factors. The conclusion was reached that diffusivity is an important parameter in transport processes/phenomena,
  • #1
MexChemE
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Hi, PF! I recently solved a problem from BSL which asked to analyze the following system used for determining the diffusivity of a binary mixture of gases.
diffusivity.png

The left portion of the system, from the left bulb up to the stopcock at the middle of the tube, is filled with gas A. The right portion of the system is filled with gas B. At t = 0, the stopcock is opened a the gases start to diffuse. This is a quasi-steady state process. First we derive an expression for the molar flux of A through the tube using a steady state molar balance, and then we make an unsteady state molar balance for species A on the left bulb. The goal is to obtain an expression for [itex]x_A^+[/itex] as a function of time. The function is
[tex]\ln \left(\frac{\tfrac{1}{2} - x_A^+}{\tfrac{1}{2}} \right) = - \frac{SD_{AB} t}{LV}[/tex]
Where S is the cross-section area of the tube. What got my attention is that the last part of the problem asked to suggest a method of plotting the experimental data in order to find the diffusivity. What I suggested was to define
[tex]y = \ln \left(\frac{\tfrac{1}{2} - x_A^+}{\tfrac{1}{2}} \right)[/tex]
[tex]m = - \frac{SD_{AB}}{LV}[/tex]
Then we can make a linear regression with the data from the experiment and find the slope, then obtain [itex]D_{AB}[/itex] from it.

My actual doubt lies within the implementation of the experiment. Specifically, calculating the mole fraction of gas A in the mixture. Is there a way to do that without using advanced equipment? I.e. can you calculate the mole fraction of A in the mixture using only basic lab equipment?

Thanks in advance for any input!
 
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  • #2
The question is, "how do you measure the mole fraction of a species in a gas sample?" It depends on the situation. If it's water vapor in air, for example, you just condense out the water vapor. In other situations, it might be much more difficult.

Chet
 
  • #3
Chestermiller said:
The question is, "how do you measure the mole fraction of a species in a gas sample?"
Yes, that sounds better. I thought it would be a nice experiment to do at home or school, but measuring mole fractions would be a big problem without the use of sophisticated techniques or equipment.

Here's another question, probably a big misconception though. If I fill one half of the system with oxygen, and the other half with nitrogen, will the steady state concentrations of each species be the same as in air after carrying out the experiment?

Edit: Another related question, if not the same. Do the concentrations of O2 and N2 in air depend only on the diffusivity of the O2-N2 pair or do they depend on other factors also?
 
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  • #4
MexChemE said:
Yes, that sounds better. I thought it would be a nice experiment to do at home or school, but measuring mole fractions would be a big problem without the use of sophisticated techniques or equipment.

Here's another question, probably a big misconception though. If I fill one half of the system with oxygen, and the other half with nitrogen, will the steady state concentrations of each species be the same as in air after carrying out the experiment?
What are your thoughts on this?
Edit: Another related question, if not the same. Do the concentrations of O2 and N2 in air depend only on the diffusivity of the O2-N2 pair or do they depend on other factors also?
What are your thoughts on this?
 
  • #5
Chestermiller said:
What are your thoughts on this?
Well, for the first question, in order for the system to be at constant temperature and pressure, there must be the same amount of moles of each gas in each half of the system. So I guess no, the mole fraction of both oxygen and nitrogen will be 0.5 when the system reaches steady state. This is also what the obtained model tells us. So, in this case, diffusivity affects how fast diffusion will happen, but not the extent of mixing. Therefore, if we want the final mixture to be like air, we should fill one half with 0.21 moles of oxygen and the other half with 0.79 moles of nitrogen, or a proportional multiple of these quantities.

For the second question, I guess diffusivity again plays an insignificant role in the nature of the concentration distribution of air, so it depends in other factors like temperature, pressure and the inherent properties of each gas.

Bottom line: Diffusivity is an important parameter in transport processes/phenomena, but not so useful or significant in equilibrium thermodynamics, right?
 
  • #6
MexChemE said:
Well, for the first question, in order for the system to be at constant temperature and pressure, there must be the same amount of moles of each gas in each half of the system. So I guess no, the mole fraction of both oxygen and nitrogen will be 0.5 when the system reaches steady state. This is also what the obtained model tells us. So, in this case, diffusivity affects how fast diffusion will happen, but not the extent of mixing. Therefore, if we want the final mixture to be like air, we should fill one half with 0.21 moles of oxygen and the other half with 0.79 moles of nitrogen, or a proportional multiple of these quantities.

For the second question, I guess diffusivity again plays an insignificant role in the nature of the concentration distribution of air, so it depends in other factors like temperature, pressure and the inherent properties of each gas.
Mainly, the naturally occurring amounts of the gases.
Bottom line: Diffusivity is an important parameter in transport processes/phenomena, but not so useful or significant in equilibrium thermodynamics, right?
Of course not. It's only a transport parameter, and doesn't figure in equilibrium thermodynamics in any way.

Chet
 
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1. What is a two-bulb experiment for measuring gas diffusivity?

A two-bulb experiment for measuring gas diffusivity is a scientific method used to determine the rate at which a gas diffuses through a medium. It involves measuring the change in pressure inside two connected bulbs as a gas diffuses from one bulb to the other.

2. How does a two-bulb experiment work?

In a two-bulb experiment, two bulbs are connected by a tube filled with a gas. One bulb is filled with the gas being studied, while the other is kept at a constant pressure. The pressure inside each bulb is measured over time, and the data is used to calculate the gas diffusivity.

3. What is the purpose of a two-bulb experiment for measuring gas diffusivity?

The purpose of a two-bulb experiment is to determine the rate at which a gas diffuses through a medium, which can provide valuable information about the properties of the gas and the medium. This information can be used in various scientific and industrial applications.

4. What factors can affect the results of a two-bulb experiment for measuring gas diffusivity?

The results of a two-bulb experiment can be affected by various factors such as the temperature, pressure, and type of gas being studied. The size and shape of the bulbs, as well as the length and diameter of the connecting tube, can also impact the results.

5. How is the gas diffusivity calculated in a two-bulb experiment?

The gas diffusivity can be calculated using the following formula: D = (V/t)(l/ΔP), where D is the gas diffusivity, V is the volume of the gas, t is the time it takes for the gas to diffuse, l is the length of the connecting tube, and ΔP is the change in pressure between the two bulbs. This formula is based on Fick's law of diffusion.

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