Aufgabe: In einem Behälter mit der Grundfläche A = 100 1/cm², dessen obere Begrenzung ein lose aufgesetzter Kolben bildet, läuft eine chemische Reaktion ab, in deren Verlauf der Kolben um 10 cm gegen den äußeren Druck p = 1.0 atm angehoben wird. Welche Arbeit wird dabei vom System verrichtet?
Ansatz:
Skizze:
Nun berechnet sich die Arbeit durch:
\[ W = - \int_{V_A}^{V_E} p_{ex} dV \qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad \qquad\qquad\qquad \]
Die Volumenänderung dV lässt sich berechnen durch:
\[ dV = A \cdot ds \qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad \qquad\qquad\qquad \]
\[ dV = 100 cm^2 \cdot 10 cm \qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad \qquad\qquad\qquad \]
\[ dV = 1 \cdot 10^{-3} \ m^3 \qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad \qquad\qquad\qquad \]
Eingesetzt in:
\[ W = - p_{ex} \int_{V_A}^{V_E} \ dV \qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad \qquad\qquad\qquad \]
\[ W = - 1,013 \ bar \cdot 1 \cdot 10^{-3} \ m^3 \qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad \qquad\qquad\qquad \]
\[ W \approx - 1 \cdot 10^{-3} \ J \qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad \qquad\qquad\qquad \]
Zwei Fehler im Lehrbuch (Atkins); die Einheit der Grundfläche muss cm² und nicht 1/cm² sein und die Musterlösung, W = 100 J, ist offensichtlich genauso falsch.