# Why we take time between consecutive collisions on the same wall in derivation of pressure exerted by ideal gas on container by one molecule?

My doubt is why we take time between two consecutive collision $$2L/v_{x}$$ on the same wall if distance between two parallel face of wall is $$L$$ instead of the time molecule takes to hit the wall and rebound which is very very small say $$t^{'}$$ which is almost instantaneous while calculating the pressure exerted by an ideal gas on container. I think when molecule collides with the wall in case of elastic collision it starts to move in opposite direction so change in momentum is $$-2mv_{x}$$ but according to the conservation of momentum change in momentum of wall(or momentum imparted as written in my book) is $$2mv_{x}$$ then force on the wall is $$2mv_{x}/t^{'}$$ but in my book it is given as $$mv^2_{x}/l$$ and I know this is right but I want to know why and what it actually mean by momentum imparted ?

now my explanation is as molecule hit the $$wall_{1}$$ of container it passes on momentum to the wall which is $$2mv_{x}$$ the moment molecule starts to move in opposite direction after collision say $$t_{1}$$ momentum of wall is zero(this is my second doubt that it is possible or not)? then to hit the $$wall_{1}$$ again it needs to travel 2L distance to hit the wall again at time $$t_{2}$$ which again passes momentum to the $$wall_{1}$$ equal to $$2mv_{x}$$ then change is momentum of $$wall_{1}$$ is ($$2mv_{x}$$-0) in time $$t_{2}$$-$$t_{1}$$=$$2l/v_{x}$$ then we calculate force on wall as $$2mv_{x}/t_{2}$$-$$t_{1}$$ by one molecule so if my explanation is right why we neglect the $$t^{'}$$

## 2 Answers

When we calculate things for a system that is in its most disorganised state, the best approach is to aim for average values. Have some patience and you'll know why.

# Backstory

Let's consider a container. Don't worry about the shape, it won't matter. There is an ideal gas filled in it. The container is sealed to its best to avoid any leakages.

# Averaging the stuffs

Now consider a molecule of the gas inside it. It has a randomly directed velocity. Now consider all the molecules of the gas. Their speeds might be same or different depending on multiple factors, though variance in speed will be limited and almost all the molecules will be moving in a fixed range os speeds. But their velocities remain randomly directed. And so will their momenta. Any randomly directed vector of magnitude in small variation from the mean value averaged over a very large number of observations is zero. So is their velocities and hence net momentum of the system. So if one molecule hits the wall of container in a direction, in the opposite direction(s), some other molecule(s) must be hitting the wall at almost the same time so that the net change in momentum of the gas is zero. This means "ideally" enough, the container would have a constant momentum, and if it were at rest in the beginning, you already know what would happen to it. It will stay in rest. Note the word "almost"! And since there is a very large population of molecules in the container, there will be many such "pairs" of molecules hitting the wall in almost every direction.

In the preceding statement, sometimes we might fall short of a few molecules hitting the wall and at the other times, there might be more than one molecule hitting the wall in a unit area ( if one molecule was supposed to hit one unit area of the wall ). So it's better to average the values over an infinitely large set of observations to obtain a near accurate result.

# Time taken while calculating the pressure

Consider a single molecule. I'll consider $$L$$ to be the mean free path i.e, average distance between the walls or the average distance covered by a molecule in between two collisions with walls. Say, it just met an elastic collision from a wall and is moving towards another wall $$L$$ distance away with a velocity $$v_x$$ towards the wall. Change in momentum of the molecule (of mass m) when it hits the wall is $$\Delta p = 2mv_x$$ (elastic collision). Now, for a time $$\Delta \tau=\frac{l}{v_x}$$ no collision will take place and hence no momentum change. This $$\Delta\tau$$ is called relaxation time between two successive collisions of the molecule with the wall. Then after a time period $$\Delta\tau$$ it will collide with another wall and have a momentum change of $$\Delta p= 2mv_x$$

This cycle will keep on repeating and repeating and repeating. So after a sufficiently long time has passed, can we say that net momentum change after a time period of $$\Delta \tau$$ is $$\Delta p$$? So, an average of $$F= \frac{\Delta p}{\Delta\tau}$$ force is acting on the molecule due to wall. Wouldn't this be the same force that is acting on the wall due to molecule (Newton's third law of motion). Thus $$F= \frac{mv_x^2}{l}$$ is acting on the wall due to each molecule.

PS: yes, you're right about the instantaneous force. But when calculating at the bigger picture, what will you do about all the major time a molecule spends being apart from the wall? That's why we deal with average values.

• Where is the 2 in the final expression of F. Shouldn't it be $F=\frac{2mv^2_x}{l}$ Commented Mar 30 at 14:46

(a) It's clearer, I believe, to consider the force exerted by the molecule on the wall than to consider "Momentum imparted to the wall". [The wall never seems to go anywhere!] The force exerted on the wall at any time during the impact is equal and opposite to the force that the wall exerts on the molecule.

(b) Now to the main point... Your $$2mv/t'$$ gives the magnitude of the mean force on the molecule, and therefore of the force that the molecule exerts on the wall during the time of impact. But that's not what we're concerned with when calculating the molecule's contribution to the pressure on the container wall. We need to know the mean force that the molecule exerts over all time. Since the time between collisions with the given wall is $$2l/v_x$$ there are $$v_x/2l$$ such collisions per unit time, so the mean force on the molecule from that wall is the rate of change of momentum, namely $$(v_x/2l) \times 2mv_x$$.

It may help to consider a force-time graph for the molecule. It will consist of a regular succession of 'spikes' due to its contact with the wall. The sum of the areas under these spikes over a period of time is the molecule's total change of momentum over that time. A continuous horizontal line on the graph, with an area underneath it equal to the sum of the areas under the spikes will represent the mean force by the wall on the molecule – and (in terms of magnitude) vice versa.