Caloric properties of ideal-gas mixtures

5.2 Caloric properties of ideal-gas mixtures

To calculate the chemical potential \(\mu_i\) of one component \(i\) we combine Eq. (3.78) and \(\mu = \Delta G_m\)

\begin{equation*} \mu_i = \mu_i^0 + R\,T \ln \frac{p_i}{p^0} \quad \mbox{with}\quad p_i = x_i\,p_{tot} \quad \Rightarrow \quad \mu_i = \mu_i^0 + R\,T \ln x_i + R\,T \ln \frac{p_{tot}}{p^0} \label{eq:mu_ideal_2} \end{equation*}

(5.1)

Figure 5.1: Scheme for mixing of two ideal gases.

This is Dalton’s law. Normally \(p_{tot}\) and \(p^0\) (the standard pressure) are selected 1 bar, thus, the second ln-term is vanishing.
For an extensive property \(\Phi\) the difference by mixing compared to the pure components is

\begin{equation*} \Delta_{mix} \Phi^{id}=\Phi^{id}(mixture) - \sum_i n_i \Phi^{id}_m(pure) \label{eq:DmixPhi_ideal} \end{equation*}

(5.2)

Since no interaction between the ideal gas molecules exist we find

\begin{equation*} \Delta_{mix} H^{id} = \Delta_{mix} V^{id} = 0 \label{eq:DmixH_ideal} \end{equation*}

(5.3)

To calculate the effect of mixing on the entropy we apply the scheme of Fig. 5.1. Since \(S\) is a state function we can replace the calculation of mixing by expansion performed separately for each component. Since this ideal mixing is isothermal (\(0 = dU = T \,dS - p\, dV\))

\begin{equation*} dS = \frac{p}{T} dV = \frac{n\,R\,T}{VT} dV = n\,R\, d\ln V \label{eq:dS_isothermal_2} \end{equation*}

(5.4)

\begin{equation*} \begin{split} &\;\; \Delta_{mix} S_1^{id}=n_1\,R\ln \frac{V_1+V_2}{V_1} , \qquad \Delta_{mix} S_2^{id}=n_2\,R\ln \frac{V_1+V_2}{V_2} \\ \Rightarrow &\;\; \Delta_{mix} S_1^{id}=- n_1\,R \ln x_1 , \qquad \Delta_{mix} S_2^{id}=- n_2\,R \ln x_2 \\ \Rightarrow &\;\; \Delta_{mix} S^{id} = \Delta_{mix} S_1^{id} + \Delta_{mix} S_2^{id} = - \sum_i n_i \, R \ln x_i\\ \end{split} \label{eq:DmixS_ideal} \end{equation*}

(5.5)

Thus

\begin{equation*} \Delta_{mix} G^{id} = \sum_i n_i \, R \, T \ln x_i \quad \mbox{i.e.} \quad \Delta_{mix} G_m^{id} = - T\; \Delta_{mix} S_m^{id} = \sum_i x_i \, R \, T \ln x_i \label{eq:DmixG_ideal} \end{equation*}

(5.6)

Since \(\Delta_{mix} S \gt 0\) and \(\Delta_{mix} G \lt 0\) mixing of gases is SPONTANEOUS, i.e. an irreversible process, even for ideal gases for which no energy contributions exist.