Confusion regarding the First Law of Thermodynamics

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Forcing a closed system to expand, such as pulling out a piston, involves work done by the system rather than work done to it. This expansion results in cooling, but if the external force does all the work, the gas does not cool, as it is not performing work on the surroundings. In an adiabatic process where Q = 0, the first law of thermodynamics indicates that ΔU = 0 if no work is done. The confusion arises from the differing sign conventions used by physicists and chemists regarding work. Ultimately, understanding the context of work and energy transfer is crucial for applying the first law correctly.
Smith
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Is forcing a closed system to expand (e.g. by pulling out a piston), causing it to cool, work done to the system or work done by the system? I assume it was work done to the system, but that means the first law of thermodynamics formula no longer balances if you assume an adiabatic change. With adaibatic, Q = 0, so:
-w = + delta u.
-w is equal to work done to the system, but delta u should be negative, not positive as the formula would make it.

Any ideas what I'm doing wrong.
 
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Smith said:
Is forcing a closed system to expand (e.g. by pulling out a piston), causing it to cool, work done to the system or work done by the system? I assume it was work done to the system
No, it corresponds to work done by the system. It is equivalent to reducing the external pressure and letting the system do the work.
 
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Thanks very much, that makes sense.
 
Physicists and physical chemists use opposite signs for the work done by the system. But that has to be accounted for when writing the first law as ##\Delta E = q + w## or ##\Delta E = q - w##.
 
Smith said:
Is forcing a closed system to expand (e.g. by pulling out a piston), causing it to cool, work done to the system or work done by the system? I assume it was work done to the system, but that means the first law of thermodynamics formula no longer balances if you assume an adiabatic change. With adaibatic, Q = 0, so:
-w = + delta u.
-w is equal to work done to the system, but delta u should be negative, not positive as the formula would make it.

Any ideas what I'm doing wrong.
It is not clear from your scenario whether the expanding gas does any work. If it is the external force rather than the gas that does all the work on the surroundings in expanding the volume for the gas, the gas will not cool. The gas freely expands doing no work. By the first law if Q = 0 and W = 0, then ΔU = 0 which means ΔT = 0. In that case, the work is not done to the system by the external force. Rather the work is done on the surroundings.

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