What is the Temperature of Vessel 2 After Effusion?

  • Thread starter Thread starter unscientific
  • Start date Start date
  • Tags Tags
    Box Effusion Hole
unscientific
Messages
1,728
Reaction score
13

Homework Statement



28r047t.png


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.
 
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.
 
Last edited:
  • Like
Likes   Reactions: 1 person
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 a lot!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
5
Views
3K
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 59 ·
2
Replies
59
Views
14K
  • · Replies 5 ·
Replies
5
Views
3K