This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
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.
 Spring '07
 Burnell

Click to edit the document details