Find Number of Nitrogen Molecules with Specific Kinetic Energies

In summary: I also suggest that you post this question in the Homework & Coursework Questions forum, and say explicitly that you are seeking help on your approach to the problem. That will bring in more responders, and you will get more help.
  • #1
Gauss
17
1

Homework Statement



A container holds N molecules of nitrogen gas at T = 280K. Find the number of molecules with kinetic energies between .0300 eV and .0312 eV.

dE = .0012 eV
E = .0306 eV
kT = 3.8668*10^-21 J or .02413 eV

Homework Equations



N(E) = (2N)(E^1/2)(e^(-E/kT)(π^-1/2)((kT)^-3/2) (defines the energy distribution of the gas)
dN = N(E)dE
dN = (2N)(E^1/2)(e^(-E/kT)(π^-1/2)((kT)^-3/2)dE

The Attempt at a Solution



I assumed that N(E) can be approximated to be linear therefore the above equation can be used. E was found to be the average energy between the two given values and dE was found by taking the difference between the two given energy values. kT was found by multiplying Boltzmann's constant (1.381*10^-23) and multiplying by the given temperature.

The answer is supposed to be in terms of N (the overall number of molecules in the volume).

I calculated the number to be
(2)(N)(.175 ev^1/2)(.281)(.5642)(266.79 ev^3/2)(.0012 eV) = .0178 N

However, the actual answer turns out to be 6.1 * 10^-6 N.

I can't figure out what I've done incorrectly and any help would be appreciated!
 
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  • #2
To solve the problem you just have to integrate the distribution between the stated energy limits.
 
  • #3
Since the interval was small, a Riemann sum approximation appeared to provide a result close enough to an integration.

I get that there would be some error from this approximation, but I don't think it would account for the four degrees of magnitude by which I'm off. For ease, we've been allowed to use this approximation by representing E as the midpoint and dE as the ΔE.

EDIT: After doing the integral I got a value of .222 which still seems incorrect.
 
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  • #4
Check your units, especially for Boltzmann's constant.
 
  • #5
I used 8.62*10^-5 evK for Boltzmann, but it still seemed to not work. Everything else is either unitless or in eV.
 
  • #6
How many particles between energy E and E+dE
How did you decide which distribution to use?
 
  • #7
The number of molecules between E and E+dE is some fraction of N and what we're trying to solve. The distribution equation was given earlier in the chapter and describes the likelihood of certain energies occurring in the system.
 
  • #8
You need to be able to make a concrete statement about what the energy distribution is telling you.
What do you mean by "likelihood of energies occurring"? What does that mean in relation to the equation you wrote down?
 
  • #9
I'm sorry, but I don't see what is wrong. The method you are using seems right, and gives reasonable answers in other cases. The answer you got (and I got) seems a little big, but that's just an eyeball estimate. The "correct" answer you give seems a lot too small.

Are you sure you have stated the problem correctly?
 
  • #10
I just double checked the problem and all the information I've written appears to be correct. Maybe the book answer is wrong, though this seems unlikely.
 
  • #11
N(E) refers to the distribution of molecules at respective energies, so the expected number of molecules to be found at a certain interval (between E and E+dE).
 
  • #12
N(E) refers to the distribution of molecules at respective energies, so the expected number of molecules to be found at a certain interval (between E and E+dE).
You are saying: the number of moecules between E and E+dE is given by
N(E) = (2N)(E^1/2)(e^(-E/kT)(π^-1/2)((kT)^-3/2)
... that reads to me like:$$n(E) = 2N_{0}E^{1/2}e^{-E/kT}π^{-1/2}(kT)^{-3/2} = \frac{2N_0 E^{1/2}}{\sqrt{π (kT)^{3}}}e^{-E/kT}$$You say you chose that distribution because it was the only one given earlier rather than because you thought it through and decided that was the appropriate one for the job?

What are the units of n(E) (or your N(E)) as written above?

...
Anyway, if n(E) is supposed to be the density of states,
then the number of molecules between ##E## and ##E+\text dE## would be ##n(E)\text dE##
and the total number of molecules between ##E_a## and ##E_b## would be:$$N=\int_{E_a}^{E_b}n(E)\;\text{d}E$$

refs.
http://hep.ph.liv.ac.uk/~hock/Teaching/StatisticalPhysics-Part3-Handout.pdf
http://www.physics.udel.edu/~glyde/PHYS813/Lectures/chapter_4.pdf
 
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  • #13
The distribution equation describes the distribution of molecular energies for a gas in thermal equilibrium. The derivation of the distribution equation is not done and N(E) is instead attributed to being given by the Maxwell-Boltzmann distribution.

What you wrote is the same as the given distribution for n(E).

dN = n(E)dE

So the number of molecules between two energy levels E and E+dE, would be the integral you wrote above. I did a midpoint Riemann Approximation as that was the advised method of computing the answer in the example. Additionally, after not getting the correct answer I did the integral and got another incorrect answer. All my energies and constants were in terms of eV and my temperature was in terms of K. I still can't figure out where I went wrong.
 
  • #14
Gauss said:
The distribution equation describes the distribution of molecular energies for a gas in thermal equilibrium. The derivation of the distribution equation is not done and N(E) is instead attributed to being given by the Maxwell-Boltzmann distribution.

What you wrote is the same as the given distribution for n(E).

dN = n(E)dE

I copied down what you wrote for N(E), but called it n(E) instead.
It that the same n(E) in dN=n(E)dE that you wrote just now?

So the number of molecules between two energy levels E and E+dE, would be the integral you wrote above.
I don't think so. Let's check:

The number of states between E and E+dE would be n(E)dE

##n(E)dE=g(E)f_EdE## where ##g(E)## is the density of states and ##f_E## is the probability density there is a particle with energy E. For Maxwell statistics: $$f_EdE=2\sqrt{\frac{E}{\pi}}\left(\frac{1}{kT}\right)^{3/2}e^{-E/kT}dE$$... looks a bit like yours right?

If the states are equally spaced, then ##n(E)dE=N_{tot}f_EdE## is the number of particles with energy between E and E+dE.

Does it make a difference that Nitrogen is a diatomic gas?

Lastly:
If you are certain that you have followed the correct procedure etc then you have two options. Either you have misunderstood the question or the model answer is wrong.
Check with fellow classmates, and with the person who set the question, to settle this matter.

Good luck.
 
  • #15
Simon Bridge said:
Does it make a difference that Nitrogen is a diatomic gas?
Wouldn't that just add more factors of ##1/kT##? Which would make the result larger.
 
  • #16
Certainly affects how the energy is partitioned off, but the question may only be concerned with translational states anyway.

I have a statistical mechanics crash course someplace... oh yes:
http://home.comcast.net/~szemengtan/StatisticalMechanics/IdealGas.pdf
Whole thing - but see 4.7, as relevant to above.
Well it may help.
 
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  • #17
Thank you both! I'll watch the video, re-read the section in the book dealing with this and talk with some friends. The help has been much appreciated.
 
  • #18
Hi! It's 2019 (haha) and I am now having the same issue. The book answer says 6.1*10^-6N. How do they arrive at this? I've tried using both eV and J throughout (made sure I was consistent about each one), but I still arrive at the same answer 0.178N (0.17778N). Has anyone come across any additional explanation of this?
 

1. What is the formula for calculating the number of nitrogen molecules with specific kinetic energies?

The formula for calculating the number of nitrogen molecules with specific kinetic energies is N = (n/V) * NA, where n is the number of molecules with the specified kinetic energy, V is the volume of the container, and NA is Avogadro's number (6.022 x 10^23).

2. How do you determine the kinetic energy of a nitrogen molecule?

The kinetic energy of a nitrogen molecule can be determined by using the formula KE = (1/2) * m * v2, where m is the mass of a nitrogen molecule (28 g/mol) and v is the velocity of the molecule.

3. What is the unit of measurement for kinetic energy?

The unit of measurement for kinetic energy is joules (J).

4. Can the number of nitrogen molecules with specific kinetic energies be calculated for a gas mixture?

Yes, the number of nitrogen molecules with specific kinetic energies can be calculated for a gas mixture. The formula for calculating the number of molecules remains the same, but the value for n would be the sum of the number of molecules with the specified kinetic energy for each gas present in the mixture.

5. How does temperature affect the number of nitrogen molecules with specific kinetic energies?

Temperature directly affects the kinetic energy of molecules, so an increase in temperature would result in an increase in the number of nitrogen molecules with specific kinetic energies. This is because the average speed of molecules increases with temperature, leading to more molecules having the specified kinetic energy.

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