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?
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.
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.
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.