# Calculating velocity using maxwell distribution

1. Oct 8, 2016

### kingkong23

1. The problem statement, all variables and given/known data
Consider helium as ideal gas with a Maxwell distribution of speeds.

(a) Investigate the maximal value Fmax at the peak of the Maxwell distribution F(v):

Calculate this value for He at T = 300 K and at 600 K, and for N2 at 300 K

b) For He at 300 K, obtain speeds v1 and v2 for which F(v) = Fmax/2, and then calculate the fraction of particles with speeds v between v1 and v2

2. Relevant equations
F=4*(M/(2*Pi*R*T))^(3/2)*Pi*v^2*exp(-M*v^2/(2*R*T))

3. The attempt at a solution

Okay so I got A) already, however for B I am stuck. For B I found the Fmax (using the equation above and plugging in most probable velocity as V, i am on the right track?) and then I divided by 2. So now I have the value (fmax/2) and I tried rearranging this equation to find V1. Am I on the right track because when I tried isolating for V^2 I ended up with v^2 ln v^2 and I don't can seem to get it into more simpler form.

I apologize if I posted this in wrong section, this is my first time posting here and I have no idea where this belongs. This question was from the subject:physical chemistry

Last edited: Oct 8, 2016
2. Oct 8, 2016

You find the maximum of $F(v)$ which I think occurs at $v_{max}=(2 kT/m)^{1/2}$ (set $dF(v)/dv=0$ ). Next, find $v$ where $F(v)=(1/2) F(v_{max})$. There will be two solutions for $v$ where this occurs. Also, note your $R$, which is the universal gas constant is $R=Nk$. $\$ I should write for you $v_{max}=(2RT/M)^{1/2}$ where $M$ is the mass of a mole of helium atoms. $(M=4 \, g=.004 \, kg. \ R=1.987 cal \cdot (4.184 \, joule/cal)/(mole \, deg \, K))$ . My $m$ above is the mass of one helium atom. ( $N$ is Avagadro's number)

Last edited: Oct 8, 2016
3. Oct 8, 2016

### BvU

Hello KK,

Well, what did you find for the most probable v (let's call it vmp) ?
Can you post your work for part b?

4. Oct 8, 2016

### kingkong23

oh I hope its like that then I can do it easily. But isn't asking to count the Y axis (or have I interpreted this thing wrong?). Does Fmax refer to X axis? If it did wouldn't it just say calculate the most probable velocity ?

5. Oct 8, 2016

### kingkong23

Well not much, I tried isolating for V from that equation I provided above but I couldn't get it by it self. And, the V(most probable) I got = 1116.74m/s

6. Oct 8, 2016

### BvU

No. You were doing the right thing: solving for v at the v where the y (F) is half the Fmax

7. Oct 8, 2016

The most probable velocity $v_{mp}$ is a better name for it, and it occurs at the maximum of the function $F(v)$. You can find the $v_{mp}$ and $F(v_{mp})$ by setting $dF(v)/dv=0$ and solving for $v$. This gives you $v_{mp}$. $F_{max}=F(v_{mp})$.

8. Oct 8, 2016

### BvU

Numerical value isn't interesting yet.
Can you post your work for part (b) ?

9. Oct 8, 2016

### BvU

I gather he/she has done that - worked it out numerically.

10. Oct 8, 2016

I'm responding to the OP's question in post #4. Note to the OP: $F(v)$ is the distribution (density ) function for the speeds of the atoms.

11. Oct 8, 2016

### BvU

Yes and you gave part a away in post #2. I'd like to read what KK found there -- and I also would like to have the full text of the exercise. Why do they ask for Fmax ? Or don't they ? What's KKs answer (including the dimension) ?

12. Oct 8, 2016

### kingkong23

Yea I did that for part A) but how am i suppose to count velocity at Fmax/2

13. Oct 8, 2016

It should be easy enough to compute. Set $F(v)=(1/2) F(v_{mp})$ and solve for v. I think I'm giving you about as much or more than the Physics Forum rules permit. You need to put a little effort into it and also show your calculations please.

14. Oct 8, 2016

### kingkong23

alright so this is what I have so far

Original equation and doing some rearranging, I get:

got this far?? Is this correct so far? I cant seem to isolate V

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15. Oct 8, 2016

### kingkong23

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16. Oct 8, 2016

The constant in front of the expression will divide out (cancel) on both sides. Try using the expression $F(v)=(1/2)F(v_{mp})$ and solving for $v$. You can at least get a numerical answer. You should get two roots $v_1$ and $v_2$ for the temperature they gave you. I don't know that you will be able to get a simple algebraic expression for $v$. A suggestion would be to sketch the graph of $F(v)$ and estimate what $v_1$ and $v_2$ are. If you tabulate the function $F(v)$ using a spreadsheet and about 1000 points, that may be your simplest way of solving this.

17. Oct 8, 2016

### kingkong23

yea I did use that equation and when I rearrange it would leave me with this:

so now I just gotta solve for V tho? I am on the right track? correct?

18. Oct 8, 2016

Please read my post # 18. I edited it slightly, so be sure to read the last couple of sentences.

19. Oct 8, 2016

### kingkong23

yes that makes sense but I am not sure we are allowed to do that. But still ,I will use your method if I don't get it resolved by tomorrow. Anyways, do you think there is a way to to isolate for V (from my work above)? you don't have to tell me the answer, yes/no will suffice as I don't really know what is allowed to be answered here and whats not.

20. Oct 8, 2016

My first impression is no, there is no simple algebraic form. I only looked at it quickly, but the form $v^2 exp^{-av^2}=C$ doesn't (as far as I can tell) have a simple algebraic solution for $v$. A sketch will allow you to estimate the $v_1$ and $v_2$ points very quickly, and a good tabulation could get you very accurate numbers for $v_1$ and $v_2$. Once you estimate $v_1$ and $v_2$, you could tabulate with higher resolution around those points, by first getting $v_{mp}$ and $F(v_{mp})$ to high precision.