# Does a gas in a container lose kinetic energy?

When a gas is in a container, it frequently collides with the container wall, exerting pressure. However, with a collision, kinetic energy ought to be transferred from the gas molecule to the container wall. Does that mean a gas isolated in a container will lose kinetic energy on standing, and its temperature will gradually decrease?

• Some of the answer below discuss real containers and at least one discusses the PHYS 101 (and CHEM 101 for that matter) notion of an ideal gas where the container has some basically magical properties to simplify the analysis. You seem to be interested in the former, but the level of situational detail you provide is suited to the latter. What properties do you want to container to have? Jul 23, 2015 at 14:38

• When you say "conserves the kinetic energy" do you mean (1) the KE of the molecule doesn't change, or (2) the $\Delta KE$ of the molecule is the negative of the $\Delta KE$ of the molecules in the wall? And what does an ideal gas have to do with this? Jul 23, 2015 at 14:31
• I could be wrong, but I seem to remember that the ideal gas assumptions are: 1) molecules undergo elastic collisions with themselves 2) molecules undergo elastic collisions with the wall. So to reply to your question, the $KE$ of the molecule does not change. Jul 23, 2015 at 15:36
• I agree with you. In this case let's consider an elastic collision of a molecule with the wall. Before the collision the molecule has $KE=\frac{1}{2}mv^2$. The wall has $KE=0$. After the collision the total $KE$ should be unchanged. Since the wall is still at rest and has $KE=0$ this implies that the $KE$ of the molecule is still the same. Of course in the real world molecules in the wall also get some energy. Indeed ideal gas it's just a (useful) model. Jul 24, 2015 at 12:22