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The thermodynamic identity

  1. Mar 4, 2012 #1
    dU = TdS - P dV

    Is in my book derived by viewing a proces of changing volume and energy in two separate steps. First you add energy with volume fixed, then change the volume.
    I'm just not sure that I understand why, you are allowed to do this. I know the changes are infinitesimal but why is it, that you are allowed to assume that the energy first changes, and then after that has happened the volume changes..
  2. jcsd
  3. Mar 4, 2012 #2


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    You can view it that way, but you don't have to. You have an initial state and a final state thats very close to the initial state. You can jump from the initial to the final in one little step, two little steps (as above), three little steps, even a big loop that takes you far away from the initial state and then back to the final state. That's what the law is saying - if you start at the initial state and wind up at the final state (that is very close to the initial state), it doesn't matter what steps you took to get there, that equation will hold. (as long as you do it reversibly).
    Last edited: Mar 4, 2012
  4. Mar 4, 2012 #3
    so for a gas that assumption would only hold, if it expands quasistatically?
    I'm not quite sure what you mean by reversible to be honest.
  5. Mar 4, 2012 #4
    Reversibility is a tricky concept. A quasi-static process may not be reversible, but a reversible process must be quasi-static because you must be able to track every previous state you had to access in order to get to the desired one.

    A process is reversible if you can control the heat your system exchanges with the exterior, that is, all forces in your system must be conservative. Therefore if you push a block which has friction with the ground, this process is not reversible because you cannot take the heat generated by friction and insert back in your block for it to "push itself".
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