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B Confusion regarding the First Law of Thermodynamics

  1. Apr 24, 2017 #1
    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.
     
  2. jcsd
  3. Apr 24, 2017 #2

    DrClaude

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    Staff: Mentor

    No, it corresponds to work done by the system. It is equivalent to reducing the external pressure and letting the system do the work.
     
  4. Apr 24, 2017 #3
    Thanks very much, that makes sense.
     
  5. Apr 24, 2017 #4

    hilbert2

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    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##.
     
  6. May 1, 2017 #5

    Andrew Mason

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    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.

    AM
     
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