How is this argument of Feynman in his lecture on Kinetic Theory correct?

In this chapter on Kinetic Theory of gases, Feynman is trying to find a formula for pressure of a gas in terms of the average kinetic energy of the gas particles. He does so by assuming a piston, which is free to move, and a gas enclosed within some volume. He is using Newtonian laws in his explanations. First, he considers a single particle which is colliding with the piston. Suppose $m$ is the mass of the particle, and $v_x$ is the component of its velocity along the $x-$axis. Also, he assumes that the piston is fixed, and there is no loss of energy to the piston in the form of heat. So, after colliding, the particles velocity becomes $-v_x$, i.e it goes out exactly the way it comes in. Now, Feynman, according to me, uses the law of conservation of momentum to deduce the following fact: If the particle moves out that way, then after the collision the momentum of the piston is $2mv_x$. However, this way, the law of conservation of energy is violated: If we sum the energies of the particle and the piston, then it will be greater than the energy before the piston.

What is wrong here? Is he doing something else?

• He probably assumes the process is quasistatic, which means that the piston moves slow enough such that the system remains in equilibrium. – eranreches Nov 18 '17 at 12:32

I will put it in a more simple way than NowIGetTolearnWhatA.

He is talking of elastic scattering . In the simple case of particles in the laboratory This further link shows the possible cases for different masses, your case is case III: m1 much smaller than m2. Note the ~ in the values.

It illustrates what NowIGetTolearnWhatA shows with a formula. Very little velocity is necessary to be taken up by the piston which has an enormous mass in comparison with the molecules, so as to balance momntum. The approximation of the 2mv is approximately correct.

Feynman at first assumes the Piston is moveable to establish that the Piston will get pushed out of the cylinder is no force is applied. Then he decides to determine what force is needed to keep the Piston fixed:

To keep it from moving, we must pour back into it the same amount of momentum per second from our force.

Essentially we apply by hand an impulse of $2mv_x$ every time an atom hits the Piston. Another way of thinking about it is that the Piston is infinitely massive.

You are right that is you don't make this assumption, then you get a different answer as can be found using the partition function:

$$\langle V \rangle =\frac{\int_0^\infty V \exp[N \log V- PV/T]}{\int_0^\infty \exp[N \log V- PV/T]}=\frac{T\Gamma(N+2)}{P\Gamma(N+1)}$$.

This answer for the average volume is slightly larger than what the simplified argument gives and therefore a slightly larger pressure is actually needed to maintain the volume when you consider recoils.

• @Thnx a lot for the help, but I accepted Anna's answer as it is a bit more intuitive. +1 – codetalker Nov 18 '17 at 14:15

The assumption is that the system is already at equilibrium. The force on the piston exactly cancels out the pressure of the gas, so that the piston does not move as it reflects atoms of the gas. So the only kinetic energy involved is that of the gas atom, which of course is conserved by this process, since its speed is not changed.

Put another way, the atom imparts $2mv_x$ of momentum to the piston, but the force on the piston in the same amount of time (on average, at least) imparts the same impulse in the opposite direction. And so the momentum of the piston is unchanged.

• I am still not completely satisfied. If the particle is colliding with the piston, then why would it rebound with the same velocity? The atom imparts some momentum to the piston, giving it some of its kinetic energy. Then the piston is acted by the force on it, taking away all that energy. So, the kinetic energy of the electron is less than before, so how can it completely reflect? And where does that extra energy go? – codetalker Nov 18 '17 at 12:40
• Because the piston doesn't move and thus conservation of momentum dictates total reflection. – eranreches Nov 18 '17 at 13:21
• @codetalker It's an assumption that those two things basically happen at the same time. You can think of it the other way: sometimes the force will occur first, and thus impart a little tiny amount of momentum to the atom. Since many, many, many atoms hit every second and we're only really concerned about averages, it's not a bad assumption. – Chris Nov 18 '17 at 21:13