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Proving Gibbs energy equals work

  1. Mar 28, 2017 #1
    Hi,

    I'm preparing for my exams in a few weeks, of which one covers Thermodynamics.

    I was trying to solve a question, where I noticed the Gibb's free energy had to equal the (negative) work. I kind of came to an answer, but was not sure if I did it the right way. All steps are reversible.

    First, from the ideal gas law : $$ pV = nRT$$
    Then, if I differentiate both sides: $$d(pV) = d(nRT)$$
    $$ dpV = -pdV$$

    So, let's hold that thought. Now for the Gibb's free energy (with T = constant): $$ dG = dH - TdS$$
    And because ##H = U + pV##: $$ dG = dU + d(pV) - TdS$$
    Now, ## U = q + w = q - pdV##: $$dG = q - pdV + dpV + pdV - TdS$$

    Because ##q = q_{rev} = TdS##: $$ dG = q_{rev} + dpV - q_{rev} = dpV$$

    Then, from the beginning: $$dG = dpV = -pdV = - w_{rev}$$

    So my question, am I right with this reasoning?

    Thanks in advance :)
     
  2. jcsd
  3. Mar 29, 2017 #2

    Mapes

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    Your third equation (which would be more clearly written ##V\,dp=-p\,dV##) already assumes constant temperature, right? Otherwise, I don't know how you're getting it.

    Also, your first equation for Gibbs free energy is incorrect; you've mixed up ##T## and ##S##.
     
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