3.4 Calculation of \(\Delta S\) for irreversible change of state

As already stated above the change of entropy for an irreversible change of state can be calculated for a reversible substitute since entropy is a state function and does not depend on the path from one state to the other. Here we will discuss two examples:

• The adjustment of the thermodynamic equilibrium between two identical metal blocks, but with different temperatures. So we have to calculate \(\Delta S\) when two blocks of the same metal are contacted (\(m_1 = m_2\), \(C_{p,1} = C_ {p,2}\), but \(T_1 \neq T_2\)). As a reversible substitute we split up the system into two subsystems

  \begin{equation*} \begin{split} \Delta S = &\int_{T_1}^{T_{f}} \frac{C_p}{T} dT + \int_{T_2}^{T_{f}} \frac{C_p}{T} dT \quad \mbox{with} \quad T_{f}= \frac{T_1+T_2}{2}\\ \Delta S = & C_p \ln \left( \frac{\left(T_1 + T_2\right)^2}{4\,T_1\,T_2} \right) \geq 0 \end{split} \label{eq:DS_T1_T2} \end{equation*}

  (3.6)

  So reaching thermodynamic equilibrium by reaching the same temperature everywhere in the system is always a spontaneous, i.e. irreversible process since only for \(T_1 = T_2\) we get \(\Delta S = 0\).

• Calculate \(\Delta S\) when one mol of a perfect gas is irreversibly expanded in an adiabatic process: 300 K, 1 bar \(\rightarrow \) 240 K, 0.5 bar. As a reversible change in state we can choose

  • Rev. isothermal expansion: 300 K, 1 bar \(\rightarrow\) 300 K, 0.5 bar, \(\Delta S_1 \gt 0\)

  • Rev. isobaric cooling: 300 K, 0.5 bar \(\rightarrow\) 240 K, 0.5 bar, \(\Delta S_2 \lt 0\)

  \(\Delta S_{tot} = \Delta S_1 + \Delta S_2 \gt 0\) (!), despite the fact that the complete process is adiabatic. \(\Delta S_{tot}\) represents the excess entropy based on the irreversibility of the process.

© J. Carstensen (TD Kin I)