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Unformatted text preview: 4.10 Spontaneity and Free Energies AIM: Explain why Δ G < is a criterion of spontaneity. • Consider a gas cylinder with an internal pressure P and a constant external pressure P ext . ext P P • Depending on the internalexternal pressure difference, the piston will move in one or the other direction. • We apply the First Law in differential form d U = d Q P ext d V . (4.91) NOTE: This equation is valid for both reversible and irreversible processes. • If, in addition, the process is reversible, then we can use the relationship in Eq. 4.24 in differential form: d Q rev = T d S (4.92) to write the First Law as: d U = d Q rev P d V (4.93) = T d S P d V . (4.94) • Because U is a function of state, d U is path independent, and the equation holds for both reversible and irreversible processes. 4–35 • Now we combine the rhs of Eqs. 4.91 and 4.94: d Q rev P d V = d Q P ext d V (4.95) d Q rev d Q = ( P P ext ) d V . (4.96) • Now lets imagine what might happen, depending on the pressures: ‹ If P > P ext , then the gas will expand ⇒ d V > ; ⇒ ( P P ext ) d V > › If P < P ext , then the gas will contract ⇒ d V < ; ⇒ ( P P ext ) d V > • So then d Q rev d Q > (4.97) T d S d Q > (4.98) T d S > d Q (4.99) • This relationship is known as the Clausius Inequality for irreversible processes and is commonly written d S > d Q T . (4.100) NOTE: For reversible processes, we have equality as in Eq. 4.92. 4–36 • Now lets use this inequality to illustrate the relationship between free energy changes and spontaneity. We combine the Clausius Inequality with the First Law: T d S ≥ d Q = d U d W d U + d W + T d S ≥ (4.101) d U P d V + d W nonexp + T d S ≥ 0 , (4.102) where we have divided the total work into that due to a volume change and nonexpansion work d W nonexp ....
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 Winter '09
 scotty
 Thermodynamics, Entropy, dW nonexp, Pext dV

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