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Difference between internal E and enthalpy

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to understand two things:

    1. How internal energy is distinct from kinetic and potential energy

    2. The difference between internal energy and enthalpy.

    2. Relevant equations
    enthalpy ΔH = ΔU + PΔV
    internal energy ΔU = Q + W = Q - PΔV

    3. The attempt at a solution
    I have searched old posts. I found this. And this.

    My book asked a question: if an exothermic reaction occurs in a closed system, does internal energy change? The answer is no. So it seems that internal energy includes potential energy. Both pages that I linked say that E = KE + PE + U. This suggests that internal energy and potential energy are completely distinct, yet the book question suggests that they are the same! Also, how is internal energy different from enthalpy?

    thank you very much!
     
    Last edited: Mar 9, 2012
  2. jcsd
  3. Mar 10, 2012 #2

    Andrew Mason

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    For ideal gases, internal energy is the total kinetic energy of the molecules. For a real gas, it includes potential and kinetic energies.

    In an ideal gas, potential energy (which is energy due to the separation of molecules relative to others) does not exist. So it is only kinetic energy that you have to take into account. Translational kinetic energy is a function of temperature. But diatomic and polyatomic molecules can have kinetic energy other than translational kinetic energy (vibrational, rotational). These other kinetic energies are a function of temperature as well.

    For real gases, molecules exert forces on each other. This forces can be relatively strong (eg. water vapour) or weak (vanderwaals forces). It takes energy to overcome those forces to increase the separation of the molecules - ie. potential energy. So when energy is added and volume increases, some of the energy added goes into increasing potential energy, which means it does not go into increasing the kinetic energy of the molecules.
    In any process the change in internal energy is equal to the energy transfer that occurs. This energy transfer can result from Work (W) being done on or by the system or heat flow (Q) into or out of the system.

    Since H = U + PV,

    dH = dU + PdV + VdP

    For an ideal gas PdV = - VdP so dH = dU. But for real gases, PdV ≠ VdP so enthalpy becomes useful.

    AM
     
  4. Mar 10, 2012 #3
    thanks very much for the excellent response.
     
  5. Mar 10, 2012 #4

    Andrew Mason

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    The only problem is that the explanation for dH is not quite right!
    This is only true if T is constant, in which case dH = dU = 0.

    I'll have a look at this again later.

    AM
     
  6. Mar 10, 2012 #5
    appreciate it. I'm not very comfortable in a calculus based approach as my physics was only algebra based. So could you please accommodate me in your response? (ie, more conceptual, please)

    I didn't ask immediately because I didn't think this question gathered enough interest for follow up questions!
     
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