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Chemistry LibreTexts

20.2: Nonequilibrium Isolated Systems Evolve in a Direction That Increases Their Probability

  • Page ID
    13715
  • Suppose we have a small reversible change \(dU\) in the energy of an ideal gas. We know that \(U\) only depends on temperature:

    \[dU = C_vdT\]

    We also know that any reversible work would be volume work.

    \[δw_{rev} = -PdV\]

    This means that we can write:

    \[δq_{rev} = dU - δw_{rev}\]

    \[δq_{rev} = C_vdT + PdV\]

    Let us examine if this represents an exact differential. If that were so we could take cross derivatives and arrive at the same answer:

    • \(\dfrac{\partial C_v}{\partial V}\) should be equal to \(\dfrac{\partial P}{\partial T}\).
    • \(\dfrac{\partial C_v}{\partial V}=0 \) because \(C_v\) does not depend on volume (only \(T\), just like \(U\): it is its derivative).

    However,

    \[\dfrac{\partial P}{\partial T} = \dfrac{\partial nRT}{\partial T} = \dfrac{nR}{V}\]

    which is not zero!!

    Clearly, \(δq_{rev}\) is not a state function, but look what happens if we multiply everything with an 'integration factor' \(1/T\):

    \[\dfrac{δq_{rev}}{T} = \dfrac{C_v}{T}dT + \dfrac{P}{T}dV\]

    \[\dfrac{\partial C_v/T}{\partial V} = 0\]

    because \(\dfrac{C_v}{T}\) does not depend on volume.

    However,

    \[\dfrac{\partial (P/T)}{\partial T} = \dfrac{\partial (nR/V)}{\partial ∂T} = 0\]

    Thus, the quantity \(dS = \dfrac{δq_{rev}}{T}\) is an exact differential, so \(S\) is a state function and it is called entropy.