### Further Reading

# Kinetic theory: The ideal gas law

In trying to understand what we see at the macroscopic level in terms of the microscopic properties of a system made up of atoms and molecules, we'll start by looking at the ideal gas law. We choose ideal gases because they're comparatively simple. Since the molecules are spaced relatively far apart you don't really have to worry too much about the interactions between molecules — except to realize that they allow the molecules to share energy through collisions. Mostly the motion is just that of free particles (gravity plays a very small role since the molecules move so fast). And you're already familiar with the macroscopic form of the ideal gas law, *pV = nRT*, from chemistry.

## A simple model of a gas

What is a gas? A according to an idea developed by James Clerk Maxwell in the middle of the 19th century, It's lots of individual molecules moving around freely and bouncing off the walls of the container (or the room, etc.). Each individual molecule basically obeys Newton's laws. So if we wanted, we could model the behavior of a gas by looking at each molecule separately, and figure out the position and velocity of each one. But there's one problem: we're talking about a **lot** of molecules. One mole of gas (or of anything) has 6x10^{23} molecules. To make things easier, it's more useful to think about bulk quantities like temperature and pressure, which represent averages over the entire population of molecules.

To make the connection between the micro and macro pictures, we can look at one representative particle and treat it as if it is going in a random direction. The motion of this one particle is not actually random: it is very definite, and influenced by the forces acting on that one particle. But because we're randomly picking it out of a larger set of particles (moving in every possible direction), its motion looks random. These ideas of randomness and probability will become very significant in the coming weeks.

Where does pressure come from? In our model of many molecules moving very fast, it comes from the molecules of the gas hitting a wall. Imagine a single gas molecule colliding with a wall. It bounces off and reverses direction, which means that the wall is exerting a force on the molecule to change its velocity. By Newton's 3rd Law, we know that the molecule must be exerting an equal and opposite force on the wall. The sum of all the forces on the wall, from all the molecules (per unit area) is the pressure of the gas. All we have to do is figure out:

- How much force does a single molecule exert on a wall when it bounces off it?
- How many molecules hit the wall per second?

To do this, all we need is our basic equation of velocity and Newton's second and third laws — plus a model of the randomness of the motion of many molecules. Here are the only equations we need. (Plus, we need to know the ideal gas law, *pV = nRT*.)

$$F_{wall \rightarrow molecule} = m { {\Delta v_x} \over {\Delta t} } = -F_{molecule \rightarrow wall}$$

$$\langle v_x \rangle = { \Delta x \over \Delta t}$$

Now let's use them.

## How much force does a single molecule exert on a wall when it bounces off it?

Imagine a wall with the gas on its left. Now consider one molecule bouncing off the wall. In the figure at the right, the molecule's path is in red. We have chosen our coordinates so that the positive x direction is in the direction of the normal to the wall pointing outward from the region where the gas is.

When the molecule hits the wall, the x component of its velocity is reversed, but the y and z components stay the same. Because the collision is assumed to be elastic, the molecule keeps the same amount of kinetic energy, so the magnitude of the velocity stays the same; only the direction changes.

Now let's find the magnitude of the force that the molecule exerts on the wall. It's equal to the mass of the molecule times its acceleration during the collision, and the acceleration is its change in velocity divided by the time for one collision

$$F_{wall \rightarrow molecule} = m { {\Delta v_x} \over {\Delta t} } = -F_{molecule \rightarrow wall}$$

How big is $\Delta v_{x}$? If it had some value $v_{x}$ before the collision, then it was $-v_{x}$ after the collision (same magnitude, opposite direction), so the magnitude of the change is $-v_{x} - (v_{x}) = -2v_{x}$_{.} So the magnitude of the force on the wall from each collision is ${2mv_{x} \over \Delta t}$.

## How many molecules hit the wall per second?

Our result above is just from one molecule. How can we figure out how many molecules are hitting the wall? The result depends on the density of molecules. Let *n* be the number of molecules per unit volume, $A$ be the area of the wall we are considering, and consider a small time interval $Δt$*.* This time should be long enough that many molecules will hit the wall, but short enough so that not much changes in a macroscopic sense in that time.

To figure out how many molecules will hit our area in a time Δ*t, *look at the molecules moving near the wall. Only those that are moving towards the wall are going to possibly hit it. And only those that are close enough are going to hit. To see how this works, let's make even a simpler model. Instead of having the molecules move randomly in all directions, let's take all the molecules as moving randomly either left or right. Further, we'll give all the molecules the same speed. Although this seems overly simplistic, the up and down motions (y direction) don't contribute to the force so we can ignore them. We'll handle the randomness in speed later by replacing our assumed constant velocity by an average velocity.

So our model is the following: lots of molecules, spread out in the y direction, but only moving left or right with a uniform speed $v_x$ and with a density of $n$ molecules per unit volume. In this simplified model our molecules near the wall look like the figure at the right.* *

In this simplified model, only the molecules in the cylinder with area $A$ and height $v_xΔt$ will make it to hit the area $A$ in the time $Δt$*. *Since the volume of the cylinder is $Av_xΔt$, the number of molecules in the cylinder is $nAv_xΔt$*, *and the number that will hit in the time $Δt$ is $½nAv_xΔt$*.*

(Is this the same $Δt$ we were using before? Yes, if it's the **average** force, averaged over the entire time, understanding the "time for one collision" to also include the time **between **collisions.)* *

So if we put it all together, the total force on the left wall of the room is (the number of molecules hitting it) times (the force exerted by each molecule), which comes out to:

$$F = {1 \over 2} \left( n A v_x \Delta t \right) \left( {{2 m v_x} \over \Delta t} \right) = n m v^2_x A $$

## Handling the randomness

The molecules in the gas are not all traveling at the same speed and are not all traveling in the x direction. How do we handle that? Well, the y (and z) directions don't matter. Some of our molecules will have $v_y$ velocities. So some will move out of our box in the time $Δt$. But just as many will move in as will move out, so we can ignore that.

But what about the fact that all our molecules are not moving at the same speed? Well, since some will move faster and some will move slower, we can try working with the molecule's average speed $\langle v \rangle$. How does that connect to the average value of $\langle v_x \rangle$, which is what we need? By the Pythagorean theorem, the magnitude of the average speed is given by

$$\langle v^2 \rangle = \langle v^2_x \rangle + \langle v^2_y \rangle + \langle v^2_z \rangle $$

and there's no particular reason why the x, y, and z directions should be different from each other, so the average x component, the average y component, and the average z component of the velocities of the gas molecules should all be equal. This tells us that $\langle v^2 \rangle = 3\langle v_x^2 \rangle$, so the average value of $\langle v_x^2 \rangle$^{ }(which appears in equation (1)) is $v^2/3$ (where $v$ is the average speed).

## Making the connection to the Ideal Gas Law

Remember that pressure is force divided by area, so continuing from the equation for the force the molecules exert on the wall, we find that $p = nmv_x^2$. Since $n$* *is the total number of molecules, $N$, divided by the volume $V$, we can plug these in and get:

$$p = \left( {N \over V} \right) m \left( v^2 \over 3 \right) $$

$$p V = N \left( {1 \over 3} m v^2 \right) = N {2 \over 3} \left( {1 \over 2} m v^2 \right) $$

What do we have here? $pV$ = the number of molecules times something proportional to the average kinetic energy of each molecule! Where have we seen anything like this before? Oh right, everyone who has ever taken chemistry knows the Ideal Gas Law in what we call *chemist's form*:

$$pV = n_{moles} RT,$$

where $R$ is a constant that doesn't depend on what type of gas it is, and $T$ is the temperature. This relationship was determined experimentally from macroscopic experiments. Perhaps you've done this experiment (with balloons or something) in chemistry lab.

Since this representation of the Ideal Gas Law includes the number of moles, we'll need to convert this into the number of **molecules** if we want to match it up to what we have in our equation. You may recall that a mole of gas (or a mole of anything) contains 6.02x10^{23} molecules (Avogadro's number, $N_A$), so the number of moles is equal to $N$ (the number of molecules) divided by $N_A$. So the Ideal Gas Law now looks like

$$ pV = \left( { {N} \over {N_{A}} }\right) RT. $$

To simplify this, we'll combine the two constants and write it in what we call *physicists' form* of the ideal gas law:

$$ pV = N k_B T ,$$

where $k_B$ (known as the Boltzmann constant) is the gas constant $R$ divided by Avogadro's number.

The reason the chemists like moles (and $R$) is that moles are what combine naturally chemically. The reason that the physicists like number of molecules (and $k_B$) is that molecules are what count in creating forces and therefore pressure and temperature. Both are reasonable and they are just different choices of units (and therefore of constants).

## Interpreting temperature

One of the very nice things that this derivation does for us is to give us a mechanical definition of the meaning of temperature.

Combining our results, we see that

$$ { 2\over3 }\left( {1\over2} mv^2 \right) = k_B T , $$

so we can solve and find that the average kinetic energy ${1\over2} mv^2 $ of each molecule is ${3\over2} k_B T $, and the total kinetic energy of all $N$ molecules is ${3\over2} N k_B T $. Don't lose any sleep over the factor of 3/2; it's just an artifact of living in a three-dimensional world. The point here is that we can build a bridge between temperature (a macroscopic property) and kinetic energy (a property of individual molecules). The Boltzmann constant (which we'll see in a few other places) functions as this bridge. To find its value, just divide $R$ by $N_A$. You may have seen it in other units, but in SI units, $R$ = 8.31 J/mol-K. (Yes, pressure time volume has units of energy! Work this out for yourself if you'd like more practice with dimensional analysis.) $N_A = 6.02 x 10^{23}$, and its units are 1/mol (since it's the number of things — any type of things — in a mole). Divide these, and we get $k_B = 1.38 \times 10^{-23} \mathrm{J/K}$. It has units of joules per kelvin, so it is like a conversion factor between energy and temperature. This looks like a very small number, but don't forget that this represents the energy of a single molecule!

This kinetic energy of all the molecules together is what we call "thermal energy". It's not actually a different type of energy — deep down, it's just kinetic energy (in the same way that "chemical energy" is just potential energy), but we give it another name, because if you just look at an object with thermal energy, it doesn't look like it's moving! Remember that energy is a scalar, not a vector (energy has no direction), so when you add up the energies of all the molecules, even though they're going in all different directions, they don't add up to zero, the magnitudes just add up.

This detailed analysis was specific to gases, but it's the same general idea for liquids and solids. The math is more complicated, since the molecules are close together and interacting — there's no "Ideal Liquid Law". But in those cases too, the temperature is still proportional to the average kinetic energy of the molecules. So the total thermal energy is proportional to the temperature times the number of molecules. In solids, the molecules aren't free to move about, so the kinetic energy we're talking about is associated with vibration.

Ben Dreyfus and Joe Redish 11/26/11

#### Follow-ons

Last Modified: July 5, 2019