Question on RMS Mean Free Path

In summary, Lambda(rms)=v(rms) * t(rms)--1. The mean free path and the RMS free path would actually be the same (even later on when used in the aforementioned Survival Equation).
  • #1
warhammer
151
31
Homework Statement
The distribution of free path is given as N = N(o) e^ (-x/lambda) where lambda is the mean free path, and N is the number of molecules that survive collision on travelling a distance x, and N(o)is the no. of molecules at distance x = 0. Show that the root mean square (rms) free path is given by √2*lambda
Relevant Equations
lambda (rms)= v(rms) * t(rms) where t(rms) is the relaxation time.
lambda (rms)= v(rms) * t(rms) -- 1

Now I assume here that t(rms)=1/(√2*n*π*d^2*v(rms))

But this cancels the v(rms) term when used in eq (1) so the mean free path and the RMS free path would actually be the same (even later on when used in the aforementioned Survival Equation)

I would like to state that I have not understood what is meant by v(rms) clearly.. I would be really obliged if someone would explain the concept behind the question as well as provide the guidance to complete the question.

(I even tried to use the original derivation that was used for Survival Eqn in context of Maxwellian gas but that also provided no real insight)
 
Physics news on Phys.org
  • #2
You are over-thinking things. x has an exponential distribution, for which it is a known property that the mean value of x is λ and the variance is λ2 so the mean value of x2 is 2λ2.
 
  • Like
Likes warhammer
  • #3
mjc123 said:
You are over-thinking things. x has an exponential distribution, for which it is a known property that the mean value of x is λ and the variance is λ2 so the mean value of x2 is 2λ2.
Thank you for your response. However I'm unable to gauge more clearly how it would entail for the succeeding step in this particular case here. Please provide a hint or elaborate on this.
 
  • #4
warhammer said:
I would like to state that I have not understood what is meant by v(rms) clearly.. I would be really obliged if someone would explain the concept behind the question as well as provide the guidance to complete the question.

A minor typo here. Please read it as Lambda(rms) or alternatively rms value of free path rather than v(rms).
 
  • #5
If you have a distribution ##N(x)##, the mean value of ##x## over an interval ##a\leq x \leq b## is $$\langle x \rangle = \frac{\int_a^b x N(x)~dx}{\int_a^b N(x)~dx}.$$ The mean-squared is $$\langle x^2 \rangle = \frac{\int_a^b x^2 N(x)~dx}{\int_a^b N(x)~dx}.$$The root-mean-squared is $$x_{\text{rms}}=\sqrt{\langle x^2 \rangle} =\sqrt{ \frac{\int_a^b x^2 N(x)~dx}{\int_a^b N(x)~dx}}.$$Just use the definition.
 
  • Informative
Likes warhammer
  • #6
kuruman said:
If you have a distribution ##N(x)##, the mean value of ##x## over an interval ##a\leq x \leq b## is $$\langle x \rangle = \frac{\int_a^b x N(x)~dx}{\int_a^b N(x)~dx}.$$ The mean-squared is $$\langle x^2 \rangle = \frac{\int_a^b x^2 N(x)~dx}{\int_a^b N(x)~dx}.$$The root-mean-squared is $$x_{\text{rms}}=\sqrt{\langle x^2 \rangle} =\sqrt{ \frac{\int_a^b x^2 N(x)~dx}{\int_a^b N(x)~dx}}.$$Just use the definition.

Sir what you're expressing quantitatively is not coming out and the remaining resultant text is indecipherable.

Request you to express it again..
 
  • #7
I have expressed it as clearly as I can. Just substitute and do the integral. If it doesn't "come out", please post what does come out and we will take it from there.
 
  • Like
Likes warhammer
  • #8
kuruman said:
I have expressed it as clearly as I can. Just substitute and do the integral. If it doesn't "come out", please post what does come out and we will take it from there.
First of all I would like to apologise for the silly question from my side. I was viewing the PF Site on my mobile and it was not rendering the relevant mathematics, just the written words. It was only after I opened the same via another device I was able to finally see the mathematics😅😅

Now I inputted everything via the survival equation & integrated the function using the "by parts technique". This resulted in the mean squared value to be 2*(lambda)^2. Using the same in the RMS definition specified by you, I obtained that RMS value=√2*lambda.

The answer has matched now but I would love if you could explain the mathematical definitions qualitatively if possible (the specific integrands in numerator and denominator, and I hope you would excuse this very trivial and silly probing).
 
  • #9
Apology accepted. Look up weighted average (or weighted mean). There are plenty of links and videos explaining it and it makes no sense to repeat that stuff when it already exists.
 
  • Like
Likes warhammer
  • #10
kuruman said:
Apology accepted. Look up weighted average (or weighted mean). There are plenty of links and videos explaining it and it makes no sense to repeat that stuff when it already exists.
Thank you tons for your help sir. You've been most kind!
 

1. What is the RMS mean free path?

The RMS mean free path is a measure of the average distance traveled by a particle in a gas before it collides with another particle. It is also known as the average free path or the mean free path.

2. How is the RMS mean free path calculated?

The RMS mean free path is calculated by taking the square root of the mean squared displacement of a particle in a gas. This can be determined using the kinetic theory of gases and considering the average speed and collision frequency of the particles.

3. What factors affect the RMS mean free path?

The RMS mean free path is affected by the temperature, pressure, and composition of the gas. It also depends on the size and shape of the particles and the interactions between them.

4. Why is the RMS mean free path important in gas dynamics?

The RMS mean free path is important in gas dynamics because it helps to understand the behavior of gases at the molecular level. It is also used to calculate the transport properties of gases, such as thermal conductivity and viscosity.

5. Can the RMS mean free path be used to predict the behavior of gases?

While the RMS mean free path provides important information about the behavior of gases, it cannot be used to predict their behavior with complete accuracy. Other factors, such as external forces and non-ideal conditions, also play a role in the behavior of gases.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
781
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
982
  • Electrical Engineering
Replies
18
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
718
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
12
Views
5K
Back
Top