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?
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asked Jul 23, 2015 at 13:27
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1$\begingroup$ 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? $\endgroup$– dmckee --- ex-moderator kittenJul 23, 2015 at 14:38
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3 Answers
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The temperature of the gas will eventually reach equilibrium with the walls of the container, and since a perfect insulator is not possible, the gas, walls and outside environment will, given enough time, be at the same temperature.
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Yes, the temperature of the gas would decrease quite fast, given that the molecules in the container are still, which implies zero temperature for container. However, if the container's temperature is non-zero, it sometimes happens that gas molecules will instead gain energy because the molecule it collides with is moving fast enough in the opposite direction. It turns out that if the temperatures are equal, the average heat given to gas per time is equal to the heat it loses in the process mentioned here.
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If you assume that the gas is ideal then each collision of a molecule of gas with the wall conserves the kinetic energy. Hence the temperature will stay the same.
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$\begingroup$ 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? $\endgroup$– Bill NJul 23, 2015 at 14:31
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$\begingroup$ 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. $\endgroup$ Jul 23, 2015 at 15:36
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$\begingroup$ Elastic collision simply means that the total KE of the colliding system remains unchanged. The KE of the atom/molecule in the wall incvolved in the collision can change along with the gas molecule KE. Consider two objects of equal mass, one moving freely, the other at rest. They have a glancing but elastic collision. The KE of each changes, but the sum of KE doesn't. $\endgroup$– Bill NJul 23, 2015 at 16:16
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$\begingroup$ 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. $\endgroup$ Jul 24, 2015 at 12:22