When the volume of a container of gases is decreased, does the Root mean square speed increase or decrease? I'm a High School student learning the kinetic theory of gases, and this is something I'm a bit confused about. So say there's a container covered with a piston. When you lower the piston and therefore decrease the volume, what happens to the root mean square speed? If I think about it, when the piston is lowered, there will be more collisions so the speed will increase. In school, one of the equations we're learning is:
$$p = \frac{Nmv_{av}^2}{3V}$$
i.e
$$pV = \frac{Nmv_{av}^2}{3}$$
Where we can see that volume is proportional to the mean square speed of the atoms. Judging from the equation alone, we can say that as volume decrease, the average speed of the gases should also decrease. But my personal intuition states it's incorrect. Which one is right? Thanks in advance for the help!
 A: The equation
$$pV=\frac13 Nmc_{rms}^2$$
isn't enough to tell you whether or not decreasing the volume of a gas will increase the rms speed of the molecules. All it tells you is that the ratio $\frac{c_{rms}^2}p$ will decrease.
The extra information that you need is thermodynamic in nature. This is not as frightening as it sounds; all it means is that you consider flows of heat and work into or out of the gas.
Case 1. You push the piston in quickly, so that negligible heat flows into or out of the gas through the container walls. [We say that the process is adiabatic.] But we have done work on the gas, as we have moved the force we apply to the piston in the same direction as the force itself. As energy is conserved the internal energy of the gas has increased. But the internal energy of an ideal gas is simply the sum of its molecules' kinetic energies. So $c_{rms}$ has increased. We say that the gas's temperature has risen.
Case 2 We push the piston in more slowly. As the gas's temperature rises (see case 1) heat has time to escape by conduction through the container walls to the surrounding environment (assumed to be at the gas's starting temperature). Heat is a mode of energy transfer, so the internal energy of the gas doesn't rise as much as in case 1.
Case 3 We push the piston in very slowly indeed, so slowly that enough heat escapes for the temperature of the gas not to rise at all as we push in the piston. We say that the gas stays in thermal equilibrium with its surroundings, and that the compression of the gas is isothermal (constant temperature). You should be able to work out what happens – or doesn't happen – to (a) $c_{rms}^2$, (b) pressure, $p$.
A: Yes, the average speed of the molecules should decrease as the volume is decreased.
I think the reason you cannot intuitively see this result is that you are probably forgetting that you must keep pressure constant (for your proportionality relation to hold). Therefore if you are reducing the volume, there will be more collisions with the walls per second, so you would expect a higher pressure if the average speed of the molecules stayed the same, but if you say that the pressure must be the same, then the average speed of the molecules must decrease.
It may also help to think in terms of temperature. The average kinetic energy of the molecules is related directly to their temperature via the Boltzmann constant.
$$E_k = \frac{1}{2}m\overline{v^2} = \frac{3}{2}kT$$
This tells you that the only way to decrease the volume of a piston whilst keeping the pressure constant is to reduce the temperature of the gas. This intuitively makes sense to me.
