# Doubt in the derivation of pressure of an Ideal Gas Using Kinetic Theory Of Gasses

Link of the book https://www.feynmanlectures.caltech.edu/I_39.html Chapter:- Kinetic theory of gasses

We would like to find out what force on the piston results from the fact that there are atoms in this box.

Then the book explains simply the force will be due the the bombardments of particles due to which the momentum will be transferred to piston. So force will be equal to the momentum per second delivered to the piston by the colliding molecules.

To calculate the momentum per second is easy—we can do it in two parts: first, we find the momentum delivered to the piston by one particular atom in a collision with the piston, then we have to multiply by the number of collisions per second that the atoms have with the wall. The force will be the product of these two factors.

First part (understood completely)

If $$\mathbf{v}$$ is the velocity of an atom, and $$v_x$$ is the $$x$$-component of $$\mathbf{v}$$, then $$mv_x$$ is the $$x$$-component of momentum “in”; but we also have an equal component of momentum “out,” and so the total momentum delivered to the piston by the particle, in one collision, is $$2mv_x$$ , because it is “reflected.”

Second part (Doubt)

Now, we need the number of collisions made by the atoms in a second, or in a certain amount of time $$dt$$; then we divide by $$dt$$. How many atoms are hitting? Let us suppose that there are $$N$$ atoms in the volume $$V$$, or $$n=N/V$$ in each unit volume. To find how many atoms hit the piston, we note that, given a certain amount of time $$t$$, if a particle has a certain velocity toward the piston it will hit during the time $$t$$, provided it is close enough. If it is too far away, it goes only part way toward the piston in the time $$t$$, but does not reach the piston. Therefore it is clear that only those molecules which are within a distance vxt from the piston are going to hit the piston in the time $$t$$. Thus the number of collisions in a time $$t$$ is equal to the number of atoms which are in the region within a distance $$v_xt$$, and since the area of the piston is $$A$$, the volume occupied by the atoms which are going to hit the piston is $$v_xtA$$. But the number of atoms that are going to hit the piston is that volume times the number of atoms per unit volume, $$nv_xtA$$. Of course we do not want the number that hit in a time $$t$$, we want the number that hit per second, so we divide by the time $$t$$, to get $$nv_xA$$. (This time $$t$$ could be made very short; if we feel we want to be more elegant, we call it $$dt$$, then differentiate, but it is the same thing.)

So we find that the force is $$F=nv_xA\cdot 2mv_x.$$

My Doubt - Not all particles in the volume $$v_xtA$$ have velocity along $$x$$-axis but those particles are considered in the equation $$nv_xtA$$, since $$n$$ is number of particles per unit volume so when we multiplying $$n$$ in the equation we are also considering those particles who have no velocity along $$x$$-axis but are present in the region $$v_xtA$$. I think $$N$$ should be number of particles having $$x$$-component of their velocity be $$v_x$$.

Edit - I understand till the part that each molecule will transfer $$2mv_xt$$ momentum after one collision can you please explain after that and how do we get equation P=nm⟨$$v^2_x$$

• You are just misunderstanding what he is trying to say. $v_x$ is a variable that is different for each particle; He is talking about a volume that is different for each particle. By definition, then, as long as the particle is in the volume $v_x t A$ of the wall, then it necessarily must have a collision with the wall within time $t$, so your doubt is just wrong. $v_x$ is the specific "velocity along x axis" and thus, by definition, there will not be particles "who have no velocity along x axis but are present in the region $v_x t A$" because its volume would be zero in that case. Commented Sep 7, 2023 at 18:26
• If that is is case then all the molecules are going to hit the wall but it is written in book only those particles which are within the distance vxt are going to hit. Commented Sep 9, 2023 at 6:01
• You are reading it wrong; the text is correct. $v_xt$ is actually going to be tiny, and particle dependent. $t$ is going to be taken to be infinitesimal. Most of the particles will be too far away from the wall to collide with the wall. Commented Sep 9, 2023 at 9:09

You are right to say that $$v_x$$ would be different for different particles. What one should instead look at is the average velocity in the $$x$$-direction, i.e., $$\langle v_x \rangle$$.

This would give the pressure as (note I have discarded the 2 in front as only half the particles would be moving toward the piston)

$$P = nm \langle v_x^2 \rangle$$

The link you provided also suggests this in the following paragraph to the one you quoted.

If you know the temperature $$T$$, then you can relate the average velocity to the temperate as (for a mono-atomic gas)

$$\frac{3}{2} kT = \frac{1}{2}m \langle v^2 \rangle ,$$

where $$\langle v^2 \rangle = \frac{1}{3}(\langle v_x^2 \rangle + \langle v_x^2 \rangle + \langle v_x^2 \rangle)$$

this gives $$\langle v_x^2 \rangle = \frac{1}{3}\langle v^2 \rangle$$