Calculating velocity using maxwell distribution

In summary, the conversation discusses investigating the maximal value of the Maxwell distribution of speeds for helium and N2 at different temperatures, as well as calculating the fraction of particles with speeds between two specific values. The conversation also mentions the universal gas constant and the mass of a mole of helium atoms. The conversation highlights the importance of finding the maximum of the distribution function and solving for the most probable velocity.
  • #1
kingkong23
17
2

Homework Statement


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[/B]

Homework Equations


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

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
 
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  • #2
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)
 
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  • #3
Hello KK, :welcome:

Well, what did you find for the most probable v (let's call it vmp) ?
Can you post your work for part b?
 
  • #4
Charles Link said:
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)

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
BvU said:
Hello KK, :welcome:

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

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
kingkong23 said:
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 ?
No. You were doing the right thing: solving for v at the v where the y (F) is half the Fmax
 
  • #7
kingkong23 said:
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 ?
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
kingkong23 said:
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
Numerical value isn't interesting yet.
Can you post your work for part (b) ?
 
  • #9
Charles Link said:
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}) ##.
I gather he/she has done that - worked it out numerically.
 
  • #10
BvU said:
I gather he/she has done that - worked it out numerically.
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
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
Charles Link said:
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}) ##.

Yea I did that for part A) but how am i suppose to count velocity at Fmax/2
 
  • #13
kingkong23 said:
Yea I did that for part A) but how am i suppose to count velocity at Fmax/2
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
Charles Link said:
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.

alright so this is what I have so far

Original equation and doing some rearranging, I get:
upload_2016-10-8_22-25-33.png

upload_2016-10-8_22-28-29.png

upload_2016-10-8_22-29-58.png

upload_2016-10-8_22-30-51.png

upload_2016-10-8_22-37-14.png

upload_2016-10-8_22-38-43.png


got this far?? Is this correct so far? I can't seem to isolate V
 

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  • #16
kingkong23 said:
alright so this is what I have so far

Original equation and doing some rearranging, I get:
View attachment 107167
View attachment 107169
View attachment 107170
View attachment 107171
View attachment 107172
View attachment 107173

got this far?? Is this correct so far? I can't seem to isolate V
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
Charles Link said:
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.

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

upload_2016-10-8_23-6-4.png


so now I just got to solve for V tho? I am on the right track? correct?
 
  • #18
kingkong23 said:
yea I did use that equation and when I rearrange it would leave me with this:

View attachment 107178

so now I just got to solve for V tho? I am on the right track? correct?
Please read my post # 18. I edited it slightly, so be sure to read the last couple of sentences.
 
  • #19
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 what's not.
 
  • #20
kingkong23 said:
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 what's not.
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
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 :rolleyes: ).​

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
BvU said:
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 :rolleyes: ).​

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 !

alright thank you for your help. And, I will post def. post the solution here
 
  • #23
Charles Link said:
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.
Thanks so much for your help.
 
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  • #24
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.
 
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  • #25
kingkong23 said:
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.
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.
 
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  • #27
kingkong23 said:
hey guys one more thing.
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.
More things to come ? Any chance to post the full text of the exercise ? (same question as in post #11).

In post #21 you worry about what you are allowed to do to come up with a solution for this exercise. Where does that come from ? Two helpers have no better idea than to find ##v_1## and ##v_2## numerically.
 
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What is the Maxwell distribution?

The Maxwell distribution is a probability distribution that describes the distribution of velocities for particles in a gas at a certain temperature. It is a type of Gaussian distribution that is used to calculate the average velocity of particles in a gas.

How is velocity calculated using the Maxwell distribution?

The velocity of particles in a gas can be calculated using the Maxwell distribution formula, which takes into account the temperature and mass of the gas particles. The formula is v = √(2kT/m), where v is the velocity, k is the Boltzmann constant, T is the temperature, and m is the mass of the particle.

What is the significance of the Maxwell distribution in physics?

The Maxwell distribution is important in understanding the behavior of particles in a gas and is used in many applications, such as predicting the speed of particles in a gas, determining the diffusion coefficient, and studying the kinetic theory of gases.

How does temperature affect the Maxwell distribution?

The temperature of a gas directly affects the average velocity of its particles, which is described by the Maxwell distribution. As temperature increases, the average velocity of particles also increases, and the distribution curve shifts to higher velocities.

What are some limitations of using the Maxwell distribution to calculate velocity?

The Maxwell distribution assumes that all particles in a gas have the same mass and that the gas is in thermal equilibrium. This may not be the case in real-world situations, which can lead to inaccuracies in the calculated velocity.

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