Class content > The micro to macro connection

Prerequisites:

• The Micro to Macro Connection
• Restating Newton's 2^nd law: momentum
• Newton's 3^rd 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.  You should be familiar, from your introductory chemistry course, with the macroscopic form of the ideal gas law, pV = nRT where p is pressure exerted by the gas, V is the Volume of the gas, n is the number of molecules in units of "moles", T is temperature, and R is a constant.  This simple relation is very powerful, it tells us, for example, that a gas a higher temperature but with the same volume must be at higher pressure.

On this page we will show you how you can use the basic physical principles of Newton's laws, applied to the colliding molecules in a gas, to find the ideal gas law.  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 momentum through collisions (the molecules also share energy, as we discuss later).  Mostly the motion is just that of free particles (gravity plays a very small role since the molecules move so fast and don't have time to fall very far between collisions). 

A simple model of a gas

What is a gas?  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 6x10²³ 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 and in straight lines between collisions (by Newton's 1st law), pressure 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 force the wall feels due to the pressure of the gas. (The fact that we don't say that it IS pressure is a subtlety with the concept of pressure. ) Then 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.)

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:

How big is Δv_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 .

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 Δt, 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 than that all the molecules are moving randomly in all directions. Let's take all the molecules as moving randomly in a left or right direction and all with the same speed. Although this seems overly simplistic, the up and down motions (y direction) don't contribute to the force so we can actually ignore them. And 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 vx and 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 n(Av_xΔt), and the number that will hit in the time Δt is ½nAv_xΔt. The ½ in the equation comes the fact that half the molecules in the volume are moving to the left and won't hit the wall. Only the right-going ones will.*

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:

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 <v>.  How does that connect to the average value of <v_x>, which is what we need? By the Pythagorean theorem, the magnitude of the average speed is given by

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 <v>² = 3<v_x²>, so the average value of v_x² (which appears in equation (1)) is v²/3 (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_x². Since n is the total number of molecules N divided by the volume V, we can plug this in and get:

What do we have here?  pV are both macroscopic quantities.  The right hand side of the equation though still contains something microscopic  - 1/2 m v² .   We will see later this semester that this is simply the average kinetic energy of a molecule!  As we will also learn soon, the macroscopic quantity that depends on the average kinetic energy of a molecule is the temperature! 

where k_B is the Boltzmann constant).

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_molesRT,

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 6.02x10²³ 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

                         . 

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

                         ,

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).

It seems a bit tedious to have to calculate molecules colliding with a container just to arrive at the simple ideal gas law!  Indeed for a dilute gas, the ideal gas law will work just fine!  But if you need to understand more complicated materials or living systems, the ideal gas law no longer applies.   However, you can still simulate how molecules move and collide with each other and use this physical picture to infer macroscopic rules.  In modern biology and chemistry this is done with molecular dynamics simulations that - as the name implies - simulate the dynamics of a large number of molecules.    

* 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.

Ben Dreyfus and Joe Redish 11/26/11

Revised by Joe Redish and Wolfgang Losert 10/24/12