# The Free Path - The "Friendly" In front of your party of ideal gas particles lies a crossroads - three roads, all adorned with bent, creaking, weathered signs. Which path will you take, adventurer? Now, you're not the first one to have adventured through these woods. In fact, you're a small fish in the pond - physics majors, stars in their eyes have all tread this path - some have never returned. As a result, your party has devised a system to determine the most optimal path forward - using the **Mean Free Path** of those who came before you. Time for some maths! # The Logic - You, Cylindrical You! In a thermodynamic system, particles tend to move. When [[Kinetic Theory & Ideal Gases (Thermodynamics)#Translational Kinetic Energy|kinetic energy]] is directly dependant on temperature to be defined, there's always going to be some level of movement, given that [[The Zeroth (0th) Law of Thermodynamics#Temperature and Absolute Zero|absolute zero]] can't be reached. With all these particles flying around each other at different angles, surely we can take a single, reference particle to monitor average collisions along any path? ![[Pasted image 20231214222001.png]] Let's take our blue friend, the particle "P" as our reference. It's moving with a velocity $v$ across the container. How could we model the number of atoms it comes into contact with? ![[Pasted image 20231214223212.png]] Any particle that has an area which overlaps the dotted area we'll consider as being 'in the way' of sorts. This means that the distance the particle sweeps through is just dependant on the velocity vector and the time it takes: $d = v\Delta t$ To find the total number of collisions, we need to see the number of particles in that area. We'll assume that the area the particle sweeps out is a cylinder in this regard: ![[Pasted image 20231214223914.png]] Any free path - without collisions thus has to be *inversely* proportional to this cross-sectional area. We then have to determine the surrounding molecular density of the gas given that collisions also depend on the amount of stuff around it. This we can do using basic gas laws: $\rho_{mol} = \frac{N}{V}$ Where $N$ is the number of particles in and $V$ is the total volume of the system. So, to get the maximum distance a particle can travel, we can assemble the following equation: $d_{max} = \frac{v\Delta t}{\pi \ v \Delta t \ d^2 N /V} = \frac{1}{\pi d^2 N / V}$ Mind you, this is an *inertial* solution to the mean free path problem, where none of the other particles are moving. With [[statstics]] and simulations, we can get an accurate value for a constant - $\frac{1}{\sqrt{2}}$. Also, I'm replacing the $d_{max}$ variable with a $\lambda_{max}$. It just looks better - like the number eight! $\lambda_{max} = \frac{1}{\sqrt{ 2 }} \frac{1}{\pi d^2 N /V}$ With that done, you and your party decide to go down the path heading south, namely, the **Liouville Lane.** However, why not take a short rest? Yes, 'short'... >[!Example]- Problem Taster - Module 19-5 Problem 28 >At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at 1.0 atm pressure and 0.00 $^\odot{C}$? The molecular diameter is $3 \times 10^{-8}$ cm. > >**Solution:** >Let's first set the stage with the problem - we're trying to find the frequency, and we've been kind enough to have been given the molecular diameter. Remember - the cross-sectional area scales with the *radius*, not the diameter, so let's find it! > >$A = \pi {\frac{d}{2}}^2 = 7.07 \times 10^{-20} m^2$ > >We'll correct this value to ensure SI units when going ahead with our calculations! > >1 atmosphere (atm) of pressure is equal to 101325 pascals. Let's substitute our mean free path expression with the [[Kinetic Theory & Ideal Gases (Thermodynamics) (rework needed very barebones!)#The Ideal Gas Law|ideal gas law]] to get: > >$\frac{1}{\sqrt{ 2 } \times 7.07 \times 10^{-20} N/V} = \frac{kT}{\sqrt{ 2 } \times 7.07 \times 10^{-20} \times [](Kinetic%20Theory%20&%20Ideal%20Gases%20(Thermodynamics).md#The%20Ideal%20Gas%20Law)he maximum wavelength - one step closer! > >$\frac{{{1.38 \times 10^{23}} \times {273}}}{7.07 \times 10^{-20} \times \sqrt{ 2 } \times 101325} = 3.72 \times 10^{-7} m$ > >A value that might be realistic? Impossible! Remember from [[Waves Foundations|waves]] that $v=f\lambda$. We can just it to get our value, given that the speed of sound in air is $330\ ms^{-1}$ >$\frac{330}{3.72 \times 10^{-7}} = f$ >$f = 8.874 \times 10^8 Hz$ >Should be anywhere around that range. Amazing! # Example Problems # What's Next? This is a much, much simpler prototype to the rest of the [[StatMech - Welcome to Probabilia|StatMech portal.]] Might even consider turning this into a fully-fledged campaign! Good luck! Navigate back to the statistics' mechanics page for now, there's nothing more to see here. - [[StatMech - Welcome to Probabilia]]