###### By Joshua N. Winn, Princeton University

## Let’s say we have a box with a volume *V*, and we fill it with a certain number of particles (like little billiard balls), capital *N*. We define the number density, little *n*, as the number per unit volume. And into the box we inject a total energy of *E*. What will happen?

### Temperature and Energy

The total energy of *E* will be the sum of all the kinetic energies of all the particles, which is constant in time, because the box is sealed up tight. If we let the particles knock around for a long time, their positions and velocities become randomized. A particle could turn up anywhere in the box, with equal probability. And the particles will come to share the total energy more-or-less equally.

That’s the conceptual basis of temperature. The temperature, *T*, is defined to be proportional to the average energy per particle. For an ideal gas, the proportionality constant turns out to be 3/2*k*, where *k* is Boltzmann’s constant. Whenever we see little *k* in an equation, we know we’re doing thermodynamics. The numerical value is 1.4 times 10 to the minus-23 Joules per Kelvin.

The general rule is that the average energy is 1/2 *kT* times the number of independent ways a particle can store or exhibit energy. The technical term is the number of degrees of freedom. Our billiard balls can move in 3 dimensions, so the kinetic energy has 3 terms, 1/2 *m* times (*vx* squared plus *vy *squared plus *vz* squared), and each one counts as a degree of freedom, so the average energy per particle is 3/2*kT*, which means the energy density, *u*, the total energy per unit volume, is equal to 3/2 times the number density times *kT*. So, the temperature of a gas is a scale for the energy associated with the random motions of the particles.

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### Particles’ Momentum

In addition to energy, the particles have momentum. And the scale for that is pressure. The particles are constantly knocking into the walls of the box, or any surface that we might insert in the gas. Those collisions exert a force on the surface: that’s pressure.

To simplify the math, we’re going to imagine a universe in which the particles can only move in one direction. They can move back and forth in, say, the *X* direction, but not any other. When a particle with speed *v* hits the wall, it reflects back with speed *v* in the opposite direction. Its momentum changes from plus *mv* to minus *mv*, a change of minus 2*mv*. Since momentum is conserved, the wall must have absorbed a momentum of plus 2*mv*. It feels a push. And that keeps happening, as more particles hit the wall. In time *Delta t*, how much momentum does the wall absorb?

### Momentum and Pressure

Let’s say all the particles have the same speed, *v*, and half are moving to the right, and half are moving to the left. The particles that hit the wall are the ones moving to the right, that start within a distance of *v-Delta-t* from the wall. If we focus on an area *Delta-A* of the wall, that singles out a box of volume *v-Delta-t *times *Delta-A*. So, the total momentum absorbed by the wall will be the momentum from each collision, 2*mv*, times the number of collisions, which is equal to *n* halves, the number density of particles moving to the right, times *v-Delta-t Delta-A*, the volume of the box.

Force is momentum per unit time, and pressure is the force per unit area. So, to get the pressure, we divide our equation by *Delta-t* and *Delta-A*. That gives *nmv* squared. And since *mv* squared is twice the kinetic energy per particle, *epsilon*, we can also write it as 2*n-epsilon.*

### Ideal Gas Law and Energy Loss

We’ve been assuming all the particles have the same speed, *v*. But, that’s not true. There’s a whole range of speeds. So, we should replace *epsilon* by its average value, which would be 1/2 *kT*, in a one-dimensional universe. So, in the end, the pressure 2*n* times 1/2 *kT*, or simply, *nkT*.

We just derived the ideal gas law. Pressure is proportional to number density and temperature. We did it for a one-dimensional gas, but in 3-D, we end up getting the same equation.

Now, suppose we pop a tiny hole in the wall, with an area *A*. Gas particles will start leaking out, and the gas will lose energy. So, what’s the rate of energy loss? In our 1-D universe, the number of particles that leak out in time *Delta-t* is equal to the number density of right-moving particles—that’s *n* over 2—times the volume of that same box, *v-Delta-t Delta-A*. Each one has an energy of epsilon. That gives *Delta-E* equals *epsilon* times *n* over 2 times *v-Delta-t Delta-A*.

Let’s divide by *Delta-A* and *Delta-t* to give the power per unit area—that’s the flux—of escaping energy. And, again, since there’s a range of speed and energies, we should take the average. The important thing is that it’s proportional to *n* *v epsilon*, which also turns out to be proportional to the temperature to the 3/2 power.

### Common Questions about the Characteristics of Particles

**Q: What is the proportionality constant for an ideal gas?**

The temperature, *T*, is defined to be proportional to the average energy per particle. For an ideal gas, the proportionality constant turns out to be 3/2*k*, where *k* is Boltzmann’s constant. The numerical value is 1.4 times 10 to the minus-23 Joules per Kelvin.

**Q: What happens to energy when a particle can move in 3 dimensions?**

If a particle can move in 3 dimensions, the kinetic energy has 3 terms, 1/2 *m* times (*vx* squared plus *vy *squared plus *vz* squared), and each one counts as a degree of freedom, so the average energy per particle is 3/2*kT*, which means the energy density, *u*, the total energy per unit volume, is equal to 3/2 times the number density times *kT*.

**Q: What are force and pressure?**

Force is momentum per unit time, and pressure is the force per unit area.