# Help me understand this problem please

1. Nov 29, 2004

### Spectre5

I need a little help on this problem. Attached is the problem statement...I had just finished typing it out in this box when I accidentally closed the window....didn't wanna type it again cuz its a long statment...

Anyways, for part (a) and (b) I used these two equations:

$$v_{rms}=\sqrt{\frac{3K_bT}{m}}$$

and

$$T=\frac{mV_{rms}^2}{3k_b}$$

I used the escape velocity of eath to be 11,200 m/s
and get:
T(nitrogen) = 14081.6576275 K
T(hydrogen) = 10057.90411625 K

Then for part (c), I am really unsure on what I am supposed to do...can someone explain to me what is being asked and how to start it. What I have tried is this:
KE = (1/2)m*v^2
I used that to calculate the escape kinetic energy for each nitrogen and hydrogen

Then I used
average translational KE = (3/2) k_b*T
to find the average translational KE of each of the gasses...using their escape temperatures for T

But the KE and avg translational KE for nitrogen is the exact same...and they are the same for hydrogen as well.

I know I am misunderstanding something, can anyone help out? Thanks...

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2. Nov 29, 2004

### marlon

No, you are correct,... Just use the rule that the kinetic energy is less then one percent of the earth's escape velocity...

rms means root mean square and can be seen as a avg kinetic energy. The formula 3/2k_b*T is the right one...

You are correct...

marlon

3. Nov 29, 2004

### Spectre5

the translational KE for nitrogen is:
(3/2)*(1.38 x 10^(-23))*(140832) = 2.916 x 10^-18 J

and the translational KE for hydrogen is:
(3/2)*(1.38 x 10^(-23))*(10059.4) = 2.083 x 10^-18 J

Now what do I compare that to?? When I find the KE using (1/2)m*vrms^2 I get the same values

4. Nov 29, 2004

### marlon

well you need to check whether the velocity corresponding to these numbers are less that one percent of the escape velocity of the earth...

It is logic that 1/2mv²=3/2kT...you know the right hand side and solve this for v in order to find the velocities i was talking about...

marlon

5. Nov 29, 2004

### Spectre5

I thought the escape velocity of earth was known though from the first part, to be 11200 m/s.....I understand what you are saying and why I do it....I understand that
1/2mv^2 = 3/2 kb T

Solving this though, using the right hand side to be equal to the parts in my previous post and then solving for v yields, expectedly, 11200 m/s.

But the problem says that the translational KE should be less than 1% of the escape KE....but it seems to me that they are the same?

6. Nov 29, 2004

### Spectre5

I thought that the translational KE was 3/2 kb T
(using the temp for escape conditions)

and that the escape KE was the 1/2m v^2
(using escape velocity)

7. Nov 29, 2004

### marlon

nono...escape velocity is the velocity that particles have in order to overcome the earth's gravitational pull...

regards
marlon

8. Nov 29, 2004

### marlon

nono...escape velocity is the velocity that particles have in order to overcome the earth's gravitational pull...expressed by the potential energy

regards
marlon

9. Nov 29, 2004

### Spectre5

here is what I have done...

I calculated the average translational KE by using (3/2)*kb*T
I used T = 293 K, so that it is approximately what the temperature would be in the air.
thus I get average translational KE = 6.0651 x 10^-21 J

Then I calculated the KE needed for escape of each particle...using 1/2 m*v^2
thus....

For Nitrogen:
1/2m*v^2 = (1/2)*(28*u)*(11200)^2 = 2.915 x 10^-18 J
And 1% of this is 29.15 x 10^-21 J

For Hydrogen:
1/2m*v^2 = (1/2)*(2*u)*(11200)^2 = 2.08 x 10^-19 J
And 1% of this is 2.08 x 10^-21 J

THUS:
Using the rule of thumb stated in the problem, Nitrogen is contained in the atmosphere becuase 1% of the KE needed to escape is greater than the average translational KE and Hydrogen is not contained becuase 1% of the KE needed to escape is less than the average translational KE.

I believe this is correct.

Last edited: Nov 29, 2004