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handout12 - 4.10 Spontaneity and Free Energies AIM Explain...

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4.10 Spontaneity and Free Energies AIM: Explain why Δ G < 0 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 internal-external 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

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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 > 0 ; ( P - P ext ) d V > 0 If P < P ext , then the gas will contract d V < 0 ; ( P - P ext ) d V > 0 So then d Q rev - d Q > 0 (4.97) T d S - d Q > 0 (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 0 (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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