PhysicsScience

The Physics Philes, lesson 126: Kinetic-Molecular Model of Ideal Gases, Part 1

As promised, this week we’ll start a discussion on the kinetic-molecular model of ideal gases. Be warned: we’ll have to take this in chunks.

Now that that’s out of the way…what is the kinetic-molecular model of idea gases? Basically, it’s a way to understand the properties of a bunch of molecules in terms of a material’s molecular structure. In this case, it models a bunch of gas molecules bouncing around a closed container and it’s a way to understand how the ideal gas equation of state relates to Newton’s laws of motion.

Before we can really dig into the kinetic-molecular model of ideal gases, we need to make some assumptions:

  • A container with a certain volume contains a very large number of identical gas molecules with a mass m;
  • The molecules behave as point particles. That is, their size is very small compared to the average distance between the particles and the size of the container;
  • The molecules are in constant motion, obey Newton’s laws of motion, and they occasionally hit the wall of the container. When the molecules hit the container wall, the collisions are perfectly elastic (the kinetic energy is the same before and after the collision).
  • The container walls are perfectly rigid, are infinitely massive, and do not move.

Now that those are out of the way, we can check out the relationship between collisions of gas molecules and the pressure of the gas.

You’ve played dodgeball, right? You know that, when you get hit, you feel it. You feel a force exerted by the ball onto you. Molecules behave the same way. The collisions exert forces on the walls of the container, this in turn creates pressure. A collision, typically, looks something like this:

Screen shot 2015-01-04 at 10.58.01 AM

The y-component (or the component parallel to the wall) of the velocity remains the same. The x-component (or the perpendicular component), on the other hand keeps the same magnitude but reverses direction.

We need to figure out the number of collisions per unit area of the wall and momentum change. This seems like an impossible task. How on Earth are we supposed to count gas molecules? Don’t worry. We don’t have to. We can use some math and a not-quite-right assumption.

Let’s assume for now that all of the molecules have the same velocity magnitude in the x-direction. For each collision, the x-component keeps its magnitude but reverses direction. Let’s use my terrible drawing above as a guide. Before the collision, the particle is moving in the negative x-direction. So when it reverses direction, it’s moving in the positive x-direction. That means that the momentum changes from negative to positive. The change in momentum, then, is twice the product of the mass and the magnitude of the x-velocity:

Screen shot 2015-01-04 at 11.15.29 AM

OK, now if the molecule is going to collide with a certain area of the wall during a very small time interval, the molecules must be with a distance of the absolute value of the x-velocity times the very small time interval. (And, of course, the molecule must be headed toward the wall.) The number of molecules that collide with a certain area of the wall during a very small time interval is the same as the number of molecules within a cylinder with a base area equal to the certain area of the wall and a length equal to the magnitude of the x-velocity times the very small time interval. The volume of that cylinder is the product of the area, magnitude of the x-velocity, and the very small time interval. If we assume that the number of molecules per unit volume is uniform, then the number of molecule is:

Screen shot 2015-01-04 at 11.25.10 AM

On average, only have of these molecules are actually moving toward the wall, so the number of collisions with a certain area of the container wall during a very small time interval is:

Screen shot 2015-01-04 at 11.28.06 AM

Now, for the system of all molecules in the box, the total change in momentum is the number of molecules times the change in momentum:

Screen shot 2015-01-04 at 11.32.25 AM

We can re-write this to represent the rate of change of the x-component of momentum:

Screen shot 2015-01-04 at 11.34.54 AM

Here is where Newton’s laws of motion come in. We know that, according to Newton’s second law of motion, the rate of change of momentum is equal to the force exerted by the wall area on the gas molecules. We also know that, according to Newton’s third law of motion, this force is equal and opposite to the force exerted on the wall by the molecules. The pressure is the magnitude of the force divided by the area, which means:

Screen shot 2015-01-04 at 11.38.35 AM

All of this is to say that the pressure exerted by the gas depends on the number of molecules per volume, the mass of the molecule, and speed of the molecules.

There is a lot to assimilate here, but it’s only one piece of the puzzle. We haven’t even touched on kinetic energy, and it’s right in the name of the model! We’ll fix that next week.

Featured image credit: Michael Gottwald via Flickr

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Mindy

Mindy

Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

1 Comment

  1. January 13, 2015 at 10:46 pm —

    If we let n be the number of molecules, and we remember that the speed of molecules is related somehow to the temperature T (through some stuff that Boltzman and others did), and if we say that the mass of the molecules and any needed conversion constants could be represented by an R (or a k, because why not), then we could say that P = nRT/V and then we could tell chemists that we have proven the derivation of the ideal gas law from the principles of mechanistic physics. That would be cool.

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