# Doubt about the idea of pressure exerted by a gas’ visualisation

In an educational video on derivation of the formula (3/2)PV=Kinetic energy of the gas, it was assumed that 1/3 of the particles are linearly translating along the three axes of the coordinate plane in the cubic container and there are N particles where P and V are the pressure exerted and volume of the gas, the pressure due to each particle of the gas on a wall was calculated as (mv^2)/x m is the mass of the particle, v is it’s velocity and x is the length of the cubic container, This when divided by area of one wall (x^2) gave the pressure exerted by one particle,p as (mv^2)/V this was then multiplied by N/3 to give the total pressure exerted on one of the wall(which is the total pressure of the wall), thereon the above formula is derived.

However my doubt is, how do we apply the assumption(emboldened assumption) to the general case for the gas’ pressure, since it’s applicable only if the N/3 particles are simultaneously hitting the wall at every instant, but even in this case it might be that only some are hitting the wall while others aren’t and still reaching the wall, and the distribution is uneven maybe 3 particles hit at one instant and 300 at another so how do we get the pressure all the time?

Further this reasoning can be extended to say that there will actually be no constancy in pressure ever, so how is it even a state function macro state? It should be impossible for a constant pressure to exist, be it an ideal or real gas. And assuming that it can be-taking an average pressure-how is one to find it out, and how can we use such a highly fluctuating quantity(even in equilibrium) in practice? Also it should be not possible to get a constant reading on a barometer, which however is not the case!

How do we solve this query?

The assumption that the video took was ill-founded. The derivation of the formula in question, $$PV=\frac 13 M_{total}{v_{rms}}^2$$ is based upon the classical Kinetic Theory of Gases. According to this theory, motion of a gas molecule is symmetrical in all directions, which means that, if we draw a coordinate system in a region of space, we would see that all three components of velocity are equal. Hence, if $$v$$ is the velocity of one of the gas molecules, the three components of velocity would be $$v_x=v_y=v_z=\frac {v}{\sqrt 3}$$. Using this, and impulse-momentum theory, the required expression can be derived by considering the collision of the gas molecules with a particular wall (since motion is symmetrical, pressure would be same at all other walls too). I leave this as an exercise to you.

• That is correct however, how do we get the proof? In most of my trials, I've got 3 instead of 3/2 in the formula. The reason seems to be the problem whilst taking the area for taking the pressure. Should it be 2*side^2 or 1*side^2? Since the change in momentum-that we take, that is, 2mv/root3-takes place only for one collision, thus only for one wall. And that gives 3 instead of the 3/2 in the formula Commented May 17, 2021 at 7:40
• Yes, but the time interval between $2$ collisions is $\frac {2l}{v_x}$, assuming collision along $x$ axis where container has dimension $l$. Commented May 17, 2021 at 7:49
• Would we be considering one collision to be the stating point, and the ending point to be the next collision then we would get the distance travelled for changing the velocity as 'l' only right? Commented May 17, 2021 at 10:50
• No, we are only considering collisions for a particular wall, of course. Commented May 17, 2021 at 10:51