Kinetic theory and transformation exercises
We need to demonstrate the formula below, which links together micro and macro properties. How can we do this?
$$ <K>=\frac{3}{2}k_BT $$
Our model for this demonstration
Each side of the cube has length l and we have an N amount of particles in the container that move equiprobably in all three directions. All collisions are elastic and particles don’t interact at all with each other.
At first, we will take into account just one particle. Let’s say that this particle is moving with a certain velocity towards one of the walls. Being a vector, we can split the velocity in its two components: $v=(v_x; v_y)$. After hitting the wall, the particle will come back with the exact same vy and the same vx but in the opposite direction: $v=(-v_x;v_y)$. This gives us the following change in velocity: $dv=(-2v_x;0)$, which means that the momentum of the particle also changed in this way: $dq=(-2mv_x;0)$. Since all collisions in this model are elastic, such a change in the momentum must have gone somewhere! It was transferred to the wall, which acquired a momentum of $q=(2mv_x;0)$ or $q=2m<v>$.
<aside> ❗ Notice how we used <v> instead of v. This means that we’re talking about the mean velocity of all particles. Not all particles will move at the same velocity, so if we want to talk about the properties of the gas as a whole we will need to talk about its mean micro properties (we will see this again with kinetic energy, which will be called <K> instead of K).
</aside>
With the momentum, we can calculate the force exerted by the particle on the wall. The force is equal to the change of momentum over the change of time, like below ⬇️
$$ F=\frac{dq}{dt} $$
The change of time is equal to the time it takes for the particle to go from one side of the cube to the other and back, so it will be $dt=\frac{2l}{<v>}$. Our force now looks like this ⬇️
$$ F=\frac{2m<v>}{\frac{2l}{<v>}}=\frac{m}{l}<v>^2 $$
To find the force that all particles apply on the wall we need to know how many particles are hitting the wall. At the start of the demonstration, we said that particles move equiprobably in all three directions, meaning that on average N/3 particles are applying force on one of the walls. Pressure is one of the main macro properties we talk about when dealing with gases, so let’s find it! Hint: remember that $p=\frac{F}{A}$ where A is the surface area and that $<K>=\frac{1}{2}m<v>^2$.
$$ p=\frac{\frac{N}{3}\frac{m}{l}<v>^2}{l^2}=\frac{2}{3}\frac{N}{V}<K> $$
We’re almost at the end... it’s now time to rewrite our equation so that it matches what we wanted to demonstrate. We have this ⬇️
$$ pV=\frac{2}{3}N<K> $$
Remember: pV=nRT.
$$ <K>=\frac{3}{2}\frac{nR}{N}T $$