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

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.

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