# Homework Help: 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.

21. Oct 9, 2016

### BvU

Hi folks,

I kept quiet for I also don't see an analytical way out of the equation, nor a decent way to integrate F from $v_1$ to $v_2$.

For what it's worth, you can check you outcome against what I found (using, ahem, excel): $v_1/v_{mp} = 0.48163$ and $v_2/v_{mp} = 1.63656$ (my share of giveaway ).​

Perhaps you can make use of some of the error function relationships like $\ x^2e^{-x^2} = \left [ - {d\over da}\left ( e^{-ax^2} \right ) \right ]_{a=1}\$ , but you would still need a table for the integral (perhaps this helps).

I would appreciate it if, afterwards, you post the path through this part of the exercise that teacher (or book) intended. Good luck !

22. Oct 9, 2016

### kingkong23

alright thank you for your help. And, I will post def. post the solution here

23. Oct 9, 2016

### kingkong23

Thanks so much for your help.

24. Oct 9, 2016

### kingkong23

hey guys one more thing.
For part 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.
Analyze and conclude if this fraction depends on T and m0, and if yes, how exactly by calculating it for He at 600 K and for N2 at 300 K, and by comparing the values.

I keep getting 0.779~ fraction value between V2 and V1. Is it normal? Doubling the temperature yields the almost the same value and increasing molar mass (going from helium to nitrogen) still keeps the value around ~0.779.

25. Oct 9, 2016

This sounds correct. Let me give you a couple hints for this next part. Is it possible to write $F(v)=A (v/v_{mp})^2 exp^{-(v/v_{mp})^2}$ ? ( Note : $A$ may depend on $v_{mp}$ but it is simply a normalization constant.) $F(v_{mp})=A e^{-1}$ and the $v_1$ and $v_2$ points will have $F(v_1)=F(v_2)=(1/2) F(v_{mp})$. If you want to find the fraction between $v_1$ and $v_2$, can you write that as an integral expression that is independent of $v_{mp}$? (and thereby independent of $M$ and $T$). Suggestion is to make the substitution $u=v/v_{mp}$ in your integral. I have sketched the proof, but you will need to fill in the details. It takes a somewhat good background in first year calculus to complete this part. The limits for this integral $u_1= v_1/v_{mp}$ and $u_2=v_2/v_{mp}$ can readily be shown to be independent of $v_{mp}$ but you need to show the how and why of this part. Incidentally, @BvU actually gives you what $u_1$ and $u_2$ are in post #23. I computed them as well myself and got nearly identical numbers.