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

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Unformatted text preview: The second law an important case: constant T and P ( w = 0 ) q = q P = H S surr = q surr , rev T =- q T =- H T S total = S surr + S =- H T + S which can be rearranged (by multiplying by- T where T is always positive) to: H- T S As T is constant, H- T S = ( H- TS ) 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 (1) < spontaneous; = reversible; > impossible. 1 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 H fusion (ice,273 K)= 6025 J mol- 1 calculate H , S , S surroundings , S total and 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 G Consider a reversible process for a simple system with w 6 = 0. Writing the first law for a reversible process dU = dq rev + dw rev = TdS- PdV + dw...
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This note was uploaded on 02/05/2010 for the course CHEM 205 taught by Professor Burnell during the Spring '07 term at The University of British Columbia.

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

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