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Mehrkomponentige ideale Gase
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====Mischungsentropie==== Zwei ideale Gase befinden sich in einem Wärmebad T Trennwand entfernt → Durchmischung!! <u>Vor </u>der Durchmischung: Entropie <math>S={{n}_{1}}{{s}_{1}}+{{n}_{2}}{{s}_{2}}</math> mit <math>\begin{align} & {{s}_{i}}=\int_{{{T}_{0}}}^{T}{{}}dT\acute{\ }\frac{{{c}_{vi}}(T\acute{\ })}{T\acute{\ }}+R\ln \frac{{{V}_{i}}}{{{n}_{i}}}+const. \\ & i=1,2 \\ \end{align}</math> Nach der Durchmischung Entropie <math>S={{n}_{1}}{{s}_{1}}\acute{\ }+{{n}_{2}}{{s}_{2}}\acute{\ }</math> mit <math>\begin{align} & {{s}_{i}}\acute{\ }=\int_{{{T}_{0}}}^{T}{{}}dT\acute{\ }\frac{{{c}_{vi}}(T\acute{\ })}{T\acute{\ }}+R\ln \frac{V}{{{n}_{i}}}+const. \\ & i=1,2 \\ \end{align}</math> Also ergibt sich als Entropie- Differenz: :<math>\begin{align} & \Delta S={{n}_{1}}\left( {{s}_{1}}\acute{\ }-{{s}_{1}} \right)+{{n}_{2}}\left( {{s}_{2}}\acute{\ }-{{s}_{2}} \right)={{n}_{1}}R\ln \frac{V}{{{V}_{1}}}+{{n}_{2}}R\ln \frac{V}{{{V}_{2}}} \\ & \frac{V}{{{V}_{i}}}>1 \\ & \Rightarrow \Delta S>0 \\ \end{align}</math> * der Mischungsvorgang ist irreversibel!
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