# Pressure exerted by a particle of ideal gas

There is a step I fail to grasp in the proof that for a monatomic ideal gas, $$U=\frac{3}{2}PV$$, where $$U$$ denotes the internal energy of the gas. The proof involves considering a particle with mass $$m$$ travelling with velocity $$v$$ inside a cube of side $$x$$. The particle is assumed to travel perpendicularly towards one of the sides before bouncing back off it. The force exerted by the particle at any instant is opposite to the force exerted by the side of the cube on the particle. The latter can be worked out using Newton's second law, prior to which we need to calculate the acceleration of the particle during the collision. Assuming the collision to be perfectly elastic, the change in velocity before and after the collision with the wall is $$\Delta v=2v$$. To work out average acceleration, according to me, one would need the duration $$\Delta t$$ of the collision, since assuming it to be instantaneous would imply infinite acceleration. However, the various proofs I've read assign to $$\Delta t$$ the time that is needed for the particle to bounce off the wall and return after colliding with the opposite wall, in other words $$\Delta t=\frac{2x}{v}$$. I really can't see how that $$\Delta t$$ is relevant to the force exerted by the particle on the wall during the collision.

[Edit :] If there are no considerations about duration of the collision, then theoretically, no two particles would ever collide simultaneously with the wall and therefore the wall experiences very frequent collisions with single particles.

• The formula U=3/2PV cannot be true, as for the same P and V, U depends on on type of molecule and non translational active degrees of freedom that do not contribute to pressure. – Poutnik Jun 8 '20 at 3:14
• @Poutnik I edited it to say monatomic ideal gas, to address your comment. – theorist Jun 8 '20 at 3:19

You need to consider two different things:

1. The momentum transfer per particle per collision. There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that complication is completely unnecessary for introductory kinetic theory, given that there is a simple and direct way to calculate the effect of each collision.] And the simple and direct way is to use the change in momentum, which is $$2mv_x$$.

Note: Your statement that "The particle is assumed to travel perpendicularly towards one of the sides before bouncing back off it" is incorrect. Rather, in determining the momentum transfer per collision, we use the component of the velocity perpendicular to the wall, namely $$v_x$$.

1. The number of collisions per unit time. Here is where the average time it takes for the particle to travel from collision to collision is relevant. This is where $${\Delta t}$$ is being used -- not to calculate the momentum transfer/collision, but to determine the frequency of collisions per particle. And you need the average momentum transfer/collision, and the frequency of collisions per particle, to determine the average force per unit area contributed per particle.

So, in summary: For momentum transfer per collision, we don't consider time, since we can get this directly from $$2mv_x$$. For number of collisions per unit time, we need to account for the time between collisions.

• @JamesWell The average force on the wall over some finite timespan is equal to $\Delta p/\Delta t$, where $\Delta p$ is the total momentum transfer to the wall over the time $\Delta t$ from all collisions that may have occurred during that time. Increasing the frequency of collisions increases the total number of collisions during that timespan $\Delta t$, meaning that it increases $\Delta p$, so it increases the average force. – probably_someone Jun 8 '20 at 21:16
• @JamesWell The question then becomes, "How much does the force fluctuate about the average value?" We can get an idea of this by assuming that the number of collisions on the wall in a particular time interval $\Delta t$ is roughly a Poisson distribution with mean $f\Delta t$, for average collision frequency $f$. Then the standard deviation is $\sqrt{f\Delta t}$, and the magnitude of the fluctuations compared to the mean is roughly $\sqrt{f\Delta t}/f\Delta t=1/\sqrt{f\Delta t}$. – probably_someone Jun 8 '20 at 21:25
• @JamesWell So, if you have a very, very dilute ideal gas, or you measure over very, very short time intervals, you might notice some fluctuation in the force that you measure over time. But for most typical experimental situations, $f$ is very high (around $10^{20}$ collisions per second wouldn't be unreasonable) and $\Delta t$ is not that small; so it wouldn't be unreasonable to expect that the measured forces at different times will agree to a part per billion or better, certainly more precise than the precision of the usual devices you use to measure this quantity. – probably_someone Jun 8 '20 at 21:31
• @JamesWell You raise an interesting point. It sounds like you're noting that, if each collision is instantaneous, then only one particle collides at a time, and the force on the wall must thus be constantly varying between zero and infinity. And yes, that would be the case for truly instantantaneous collisions. The resolution is that, in real gases, the collisions aren't instantaneous, and there are so many gas particles colliding per unit time that we typically will have many interacting with the wall at the same time..... – theorist Jun 9 '20 at 0:24
• But if we had to actually account for the duration of collisions (i.e., make a more realistic model) that would make our calculations both much more complex, and less universal (gas-dependent). The nice thing about kinetic theory is that we can make these idealizing assumptions that allow the theory to be both mathematically and conceptually simple, while at the same time capturing the essence of why all gases exert pressure, and enabling us to calculate (for ideal gases) what that pressure is directly from a microscopic picture. – theorist Jun 9 '20 at 0:24

In ideal gases no intermolecular forces, therefore no potential energy. Thus, internal energy is equal to total kinetic energy (KE) of the system. Consider $$N$$ monoatomic particles in a cubical box of side $$\ell$$ (assumption: ideal gasses consist of monoatomic point particles). The amount of ideal gas in the box is $$\frac{N}{N_A} = n \ \pu{mol}$$ where $$N_A$$ is Avogadro number. Suppose one particle travel in $$y$$-direction with velocity $$v_y$$. If mass of the particle is $$m$$, its momentum is $$mv_y$$:

If it collides on right $$xz$$-plane, it bounce back with same velocity but in but travelling opposite direction (assuming a 100% elastic collision). Therefore momentum change ($$\Delta p_y$$) is:

$$\Delta p_y = mv_y - (-mv_y) = 2mv_y \tag1$$

However, from Newton's second law of motion ($$F = ma$$ where $$a$$ is acceleration), if the particle's applied force on the plane is $$F$$:

$$F = ma = m \frac{\Delta v_y}{\Delta t} = \frac{m\Delta v_y}{\Delta t} = \frac{\Delta p_y}{\Delta t} = \frac{2mv_y}{\Delta t}\tag2$$

Here, $$\Delta t$$ is time difference between two collisions. That means the time required for particle to travel $$2\ell$$ distance with $$|v_y|$$ velocity that is $$\Delta t = \frac{2\ell}{v_y}$$. Applying that in equation $$(2)$$ gives:

$$F = \frac{2mv_y}{\Delta t} = \frac{2mv_y}{\frac{2\ell}{v_y}} = \frac{mv_y^2}{\ell} \tag3$$

The pressure ($$P_i$$) applied by this single particle on one plane is $$\frac{F}{A} = \frac{F}{\ell^2}$$:

$$P_i = F = \frac{F}{\ell^2} = \frac{\frac{mv_y^2}{\ell}}{\ell^2} = \frac{mv_y^2}{\ell^3} \tag4$$

The total pressure on one wall: $$P = \sum^N_1 P_i = \sum^N_1 \left(\frac{mv_{yi}^2}{\ell^3}\right) = \frac{m}{\ell^3}\sum^N_1 {v_{yi}^2} = \frac{Nmv_{y(m.s.v.)}^2}{\ell^3}= \frac{Nmv_{y(m.s.v.)}^2}{V} \tag5$$

where $$v^2_{y(m.s.v.)} = \frac{1}{N}\sum^N_1(v_{yi}^2)$$ and $$V = \ell^3$$ ($$v^2_{y(m.s.v.)} =$$ mean squire velocity in $$y$$-direction). Now, the equation $$(5)$$ cab be rewritten as:

$$PV = Nmv_{y(m.s.v.)}^2 \tag6$$

Yet, we still are considering only $$y$$-fraction of velocity. As a system, we have to consider effective velocity of all $$x,y,z$$ fractions. According to law of vectors, if the effective velocity is $$v$$, $$v^2 = v_x^2 + v_y^2 + v_z^2$$. In our box, the possibility of particle to travel all directions are same, thus:

$$v_{(m.s.v.)}^2 = v_{x(m.s.v.)}^2 + v_{y(m.s.v.)}^2 + v_{z(m.s.v.)}^2 = 3v_{y(m.s.v.)}^2 \ \Rightarrow \ v_{y(m.s.v.)}^2 = \frac{1}{3}v_{(m.s.v.)}^2$$

Therefore, from the equation $$(6)$$:

$$PV = Nmv_{y(m.s.v.)}^2 = \frac{1}{3}Nmv^2_{(m.s.v.)}\tag7$$

Note that $$mv^2_{(m.s.v.)}$$ is two times KE of the particle. Thus, $$Nmv^2_{(m.s.v.)}$$ is twice KE of the system. Since, KE of the system for ideal gas is its internal energy $$Nmv^2_{(m.s.v.)} = 2U$$:

$$\therefore \ PV = \frac{1}{3}Nmv^2_{(m.s.v.)} = \frac{2}{3}U \ \bbox[yellow]{\Rightarrow \ U = \frac{3}{2}PV} \tag8$$