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
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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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