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lecture 5

# lecture 5 - The second law — an important case constant T...

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The second law — an important case: constant T and P ( w 0 = 0 ) q = q P = Δ H Δ S surr = q surr , rev T = - q T = - Δ H T Δ S total = Δ S surr + Δ S = - Δ H T + Δ S 0 which can be rearranged (by multiplying by - T where T is always positive) to: Δ H - T Δ S 0 As T is constant, Δ H - T Δ S = Δ( H - TS ) 0 Because H - TS involves only changes in state func- tions of the system, we define a new state function, G , the Gibbs function or the Gibbs free energy G = H - TS G ) T,P 0 (1) < spontaneous; = reversible; > impossible. 1

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The second law (continued) — an example Given that: C P (ice)= 37 . 7 JK - 1 mol - 1 C P (water)= 75 . 3 JK - 1 mol - 1 Δ H fusion (ice,273 K)= 6025 J mol - 1 calculate Δ H , Δ S , Δ S surroundings , Δ S total and Δ G for the melting of 1 mole of ice at P = 1 atm and T = - 10 C. Is this process spontaneous, reversible or impossible? Explain your answer! 2
The meaning of Δ G Consider a reversible process for a simple system with w 0 6 = 0. Writing the first law for a reversible process dU = dq rev + dw rev = TdS - PdV + dw 0 rev and using the definitions H = U + PV and G = H - TS we can show that: dG = V dP - SdT + dw 0 rev (2)

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lecture 5 - The second law — an important case constant T...

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