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Quick question - internal energy

  1. Aug 22, 2011 #1
    So, is the internal energy of a system always constant?
    I mean, if work done on the system is negative by convention.
    dU = dQ - pdV
    Then if work is done on the system, does that mean that it has to absorb heat to keep the internal energy constant? and if the system does work, it has to release heat?
     
  2. jcsd
  3. Aug 22, 2011 #2
    If the system is adiabatic it won't be able to release, absorb heat, and then the internal energy will not be constant. That is, if you compress(give work) to an adiabatic system it will increase U and it will get hotter.
    And I think you got it backwards, if the system does work it needs to absorb heat to maintain U and if it receives work it must release heat.
     
  4. Aug 22, 2011 #3
    Right, actually I think the convention confuses me a little. If the system does work, would that change the minus into a plus (since it's positive)?
    Or is it, dU = dQ - (-pdV) if work is done on the system, and dU = dQ - pdV if the system does work?
    Another small question, how can an adiabatic process be reversible (isentropic)?
    I mean, for a process to be reversible it has to happen in infinitely small steps, so that the system enters thermodynamic equillibrium in each step, right?
    But for an adiabatic process, it happens in one big jump, so how can it be reversible?
     
  5. Aug 22, 2011 #4
    I'm not a thermo master or anything but let's see
    You don't need to do it in one big jump, yes when you do it fast it's adiabatic because there will be no time to exchange heat with the surroundings but if you use an insulating container you will also not exchange heat and it will be adiabatic, in an ideal situation of course.
    Real processes are neither completely adiabatic nor reversible.



    About the convention, in physics it's normally dW= PdV, without the minus.
    And dU= dQ - dW
    So when the system does work on the surrounding,expands , that is final volume is bigger than initial volume, the integral is positive. W is positive.
    So -dW is negative, the system loses internal energy.

    In chemistry, physical chemistry, chemical engineering...it's backwards.
    Yes, it's confusing, hope I explained it right.
     
  6. Aug 22, 2011 #5
    Oh right, so isentropic processes ideally occur only in isolated systems then?
    I understand now, not used to the thinking in thermo yet.
    Thanks!
     
  7. Aug 22, 2011 #6
    Thermo is kinda complex, since it's not based on simple laws like mechanics that work on every situation
     
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