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Gay-Lussac's law states that if you decrease the temperature of a gas at constant volume, the pressure decreases by a corresponding amount. This is easy to carry out experimentally. But in practice is it possible to perform this the other way round, i.e. decrease the pressure and thus cause the temperature to drop, while keeping volume constant?

It seems to me that in order to reduce the pressure, you would go about lowering the chamber's volume by removing air from it. However, this would also remove a proportional number of moles from the chamber, so the temperature would remain unchanged.

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When you decrease the temperature of an ideal gas held at constant volume, what you are doing is transferring energy as heat from the gas into the surroundings. You do this by keeping the surroundings at a targeted final temperature and placing the gas in thermal contact with the surroundings, allowing heat to dissipate. When you reach the final state, you measure the temperature and pressure of the gas and find, in agreement with Gay-Lussac's law, that both T and P have decreased. Then if you ask yourself, which did I lower first, you cannot say. They decreased simultaneously.

The pressure and temperature are coupled (dependent) properties of the ideal gas when the volume is held constant. This is a consequence of the thermodynamic state postulate that for a simple compressible system, the properties of the system are fixed once two independent ones have been specified. The fallacy in the question here is the assumption that one might be able to change the temperature and that a pressure change somehow follows after. The cooling experiment changes both simultaneously. You never actually choose to alter one variable rather than the other.

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You cannot cool a gas by reducing the pressure at constant volume. How would you reduce this pressure ? It is the contrary : you cannot reduce the pressure at constant volume without cooling it. And removing a part of the gas cannot be done at constant volume.

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