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Derivation of dA=-PdV?

  1. Jul 5, 2011 #1
    I know ΔA=ΔU-TΔS
    how does this lead to ΔA=-integral(PdV)? (which seems like work)
     
  2. jcsd
  3. Jul 5, 2011 #2

    rock.freak667

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    What exactly does 'ΔA' mean? That looks like the first law of thermodynamics just that you replaced W12 with ΔA
     
  4. Jul 5, 2011 #3

    danago

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    Are you talking about an isothermal process?
     
  5. Jul 5, 2011 #4
    I mean Helmholtz free energy.
    Thanks!
     
  6. Jul 6, 2011 #5

    Mapes

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    ΔA=ΔU-TΔS is only correct for an isothermal process; the general equation is dA=dU-TdS, which is always true. (Try integrating it; you need to assume that T is constant to obtain the first equation.) Because dU=TdS-PdV for a closed system, ΔA=-integral(PdV). Does this make sense?
     
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