Effusion through hole of box

In summary, the conversation discusses the relationship between temperature and effusion of molecules between two vessels, with one vessel initially at room temperature and the other at a higher temperature. The average kinetic energy of particles is found and used to calculate the temperature of the vessels, with the final temperature and time taken also being determined. It is shown that the average translational kinetic energy of molecules traveling directly from one vessel to the other is 2kT0.
  • #1
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Homework Statement



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Part (a): Explain what A and f means. Describe an experiment that shows this relation is true.
Part (b):Find the temperature of vessel 1.
Part (c): Why is temperature of vessel initially the same as vessel 1? Calculate the final temperature of vessel 2. Find the time taken for vessel 2 to reach this temperature.

Homework Equations


The Attempt at a Solution



Part (b)
Letting ##T_0 = 300 K## be room temperature, and ##\alpha = \frac{m}{2kT}##,
the speed distribution of effusing molecules from 1 into 2 is:

[tex]g_{(v)} = 2 \alpha^2 v^3 \space exp\left(-\alpha v^2\right)[/tex]

The average energy of these particles is:

[tex]\langle E'\rangle = \frac{1}{2}m \int_0^{\infty} 2\alpha^2 v^5 \space exp\left(-\alpha v^2\right)[/tex]
[tex] = 2kT_0[/tex]

Now, in general the average KE of a system with temperature ##T## is ##\frac{3}{2}kT##.

[tex]\langle E' \rangle = \frac{3}{2}k\left(\frac{4}{3}T_0\right)[/tex]
[tex] = \frac{3}{2}kT_1[/tex]

This implies that temperature of box 1 is ##T_1 = \frac{4}{3}T_0 = 400 K##.

Part (c)
In the beginning, the average KE of molecules entering box 2 should be higher than the average KE of particles in box 1, since particles effusing have higher KE? I'm not sure why it would be the same.

Now the same as above,
[tex]T_2 = \frac{4}{3}T_1 = 533 K[/tex]

The flux of particles effusing from box 1 is given by:
[tex]\phi = \frac{1}{4}n \langle v_1\rangle = \frac{1}{4} \frac{P}{kT} \sqrt{\frac{8kT_1}{\pi m}}[/tex]

Thus number of particles per unit time is ##\phi A = \pi r^2## where ##r^2## is the area of the hole.

Not sure how to find the time taken.
 
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  • #2
Your result for part (b) looks correct for a monatomic gas where KEavg = (3/2)kT. (But note that at the end of the statement of the problem, it mentions nitrogen which is diatomic.)

For part (c), you are looking for the initial temperature in box 2 a short time after hole 1 is opened. Almost all of the molecules that make it through hole 2 initially are those molecules that travel directly from hole 1 through hole 2 without any collisions on the way. See if you can show that the average translational KE of such a molecule is 2kT0.
 
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  • #3
TSny said:
Your result for part (b) looks correct for a monatomic gas where KEavg = (3/2)kT. (But note that at the end of the statement of the problem, it mentions nitrogen which is diatomic.)

For part (c), you are looking for the initial temperature in box 2 a short time after hole 1 is opened. Almost all of the molecules that make it through hole 2 initially are those molecules that travel directly from hole 1 through hole 2 without any collisions on the way. See if you can show that the average translational KE of such a molecule is 2kT0.

Yes, I worked through the question and I got the right answers. Thanks alot!
 

Related to Effusion through hole of box

What is meant by "effusion through hole of box"?

"Effusion through hole of box" refers to the process of gas molecules escaping from a confined space, such as a box, through a small opening or hole.

What causes effusion through a hole?

Effusion through a hole is caused by a difference in pressure on either side of the hole. Gas molecules will naturally move from an area of higher pressure to an area of lower pressure, resulting in effusion through the hole.

What factors affect the rate of effusion through a hole?

The rate of effusion through a hole is affected by several factors, including the size of the hole, the pressure difference between the two sides of the hole, and the temperature of the gas.

How is effusion through a hole related to the kinetic theory of gases?

The kinetic theory of gases states that gas molecules are in constant, random motion. When gas molecules collide with the walls of a container, they exert pressure. In the case of effusion through a hole, the gas molecules are able to escape through the hole due to their high speed and random motion.

What applications does the concept of effusion through a hole have?

Effusion through a hole has practical applications in gas separation processes, such as in the production of industrial gases and in the separation of isotopes. It is also important in the design and functioning of certain devices, such as gas masks and pressure relief valves.

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