# 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.