4.3
Equilibrium conditions for pure substances and second law
The general condition for equilibrium is \((dS)_{U,V,n} = 0\), i.e. \(S\) at maximum. We have
|
| | \begin{equation*} dU(S,V,n) = -p\,dV
+ T\, dS + \sum_i \mu_i\, dn_i \Rightarrow dS(U,V,n) = \frac{dU}{T}+ \frac{p}{T}\,dV - \frac{1}{T} \sum_i \mu_i\, dn_i \label{eq:dU_dS_general}
\end{equation*} | (4.2) |
Assuming two phases of a pure substance (\(\alpha\), \(\beta\))
inside an ISOLATED system (const. \(U, V, n\)) we get
|
| | \begin{equation*} \begin{split}
\Rightarrow dS(U,V,n) = &\;\; dS^{\alpha} + dS^{\beta} = 0\\ = &\;\; \frac{dU^{\alpha}}{T^{\alpha}}+ \frac{p^{\alpha}}{T^{\alpha}}\,dV^{\alpha}
- \frac{\mu^{\alpha}}{T^{\alpha}} \, dn^{\alpha} + \frac{dU^{\beta}}{T^{\beta}}+ \frac{p^{\beta}}{T^{\beta}}\,dV^{\beta}
- \frac{\mu^{\beta}}{T^{\beta}} \, dn^{\beta}\\ &\;\; U = U^{\alpha} + U^{\beta} = const. \Rightarrow dU^{\alpha} = - dU^{\beta};
\quad dV^{\alpha} = - dV^{\beta}; \quad dn^{\alpha} = - dn^{\beta}\\ \Rightarrow &\;\; \left(\frac{1}{T^{\alpha}} - \frac{1}{T^{\beta}}
\right) dU^{\alpha} + \left(\frac{p^{\alpha}}{T^{\alpha}} - \frac{p^{\beta}}{T^{\beta}} \right) dV^{\alpha} - \left(\frac{\mu^{\alpha}}{T^{\alpha}}
- \frac{\mu^{\beta}}{T^{\beta}} \right) dn^{\alpha} = 0\\ \Rightarrow &\;\; T^{\alpha} = T^{\beta} \quad \mbox{and} \quad
p^{\alpha} = p^{\beta} \quad \mbox{and} \quad \mu^{\alpha} = \mu^{\beta} \\ \end{split} \label{eq:two_phases_S} \end{equation*} | (4.3) |
which means that 1) thermal equilibrium, 2) mechanical equilibrium, and 3) chemical/phase equilibrium holds.
© J. Carstensen (TD Kin I)
