Is it possible in principle to determine the temperature and pressure of the gas when a piston is half-way through the compression?

Question

A closed container with a piston at one end is filled with an ideal gas. The piston compresses the gas rapidly to half its initial volume. Is it possible in principle to determine the temperature and pressure of the gas when the piston is half-way through the compression? Explain.

Answer 1

The problem here is that what we define as "temperature" and "pressure" are quantities from equilibrium thermodynamics. This implies that by definition these two quantities are only defined for a system in equilibrium. If you quickly move your piston, at any intermediate point, the piston will not be in equilibrium, hence the two concepts don't apply.

Now that's only part of the answer, because only because the quantities as they are defined don't apply doesn't really rule out the possibility that you can make them work or that in your particular case the usual math still gives a correct answer by accident, so let's go a bit further:

Let's talk about pressure first, because it's easier to visualise. Pressure is the result of particles bumping against a wall and it increases if you have more or quicker particles. Now let's have a look at the piston. It moves and thus decreases the volume of your box. Clearly, this increases the density of particles - but only near the piston. Initially, the density is not evenly distributed and therefore the "pressure" would be uneven throughout the system - the system is not in equilibrium. At this point, defining a pressure for the system just doesn't make sense. The same is also true for temperature.

However, "equilibrium" is more of an approximation anyway, so does it really matter? Is the difference between the pressure at the different points really important? If you move your piston slow enough, it isn't. You can then model the process as a finite number of steps between equilibrium states and the usual thermodynamic rules apply to any state in between - for example the state where you pressed the piston half of the way. Now this is an approximation, but it's very good when you move the system slowly and gets worse the faster you move the piston.

At some point, it really doesn't make any sense to say that the current system has "a" temperature (imagine a box the size of our solar system and a piston moving close to the speed of light for instance). What is this point?

An ideal gas is a gas of randomly moving particles with only elastic collisions. Therefore, I'd say that a good estimate for the time it takes the gas to relax into a state that is approximately in equilibrium is the time it takes one particle of the gas to traverse the chamber. Hence the speed of your piston must be quite a lot smaller than this for you to be able to speak about thermal equilibrium. Let's make an order of magnitude evaluation: If you assume that your particles have the same weight as, say, carbon, then the mean kinetic energy is given by $1/2 mv^2 = 3/2 kT$. At room temperature, $kT\approx 1/40~eV$ and the mass of carbon is about $12~GeV$ which implies that the speed is approximately $10^{-6} c$ or about a thousand metres per second. So unless you are moving the piston at a speed of a hundred of metres per second or more or you have an exceptionally large box, you should be fine with assuming a quasistatic or adiabatic process and you can apply normal thermodynamics at any intermediate point.

So it all comes down to this: What do you mean by "fast"?

Answer 2

For rapid (irreversible) movement of the piston, one could calculate the temperature and pressure distributions in the cylinder using gas dynamics (which considers the fluid mechanics of the system). The results of the calculations would be that the temperature and pressure would be predicted to vary with spatial position in the cylinder during the compression. And the gas dynamics equations would include local viscous dissipation of mechanical energy to internal energy, such that the state of stress within the gas locally would not be describable with an isotropic pressure. Measurements of local temperatures and pressures in rapid compressions confirm the results of gas dynamics calculations.