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Unformatted text preview: Lecture 16 Second LAW Chapter 13 Sections 13.1, 13.2, 13.5 13.6, 13.7 • To calculate the entropy change ( ∆ S) between state 1 and state 2 we must construct a reversible path (at equilibrium) between state 1 and state 2 • Entropy changes – Isothermal – Changing T ∆ S = dq rev /T i f Calculating Entropy Changes Use the thermodynamic definition: dS = dq rev /T 1. Define clearly the initial and final states. 1. Find a reversible path linking those states . 1. If T = constant, (isothermal process): calculate q rev for the reversible process. 2. If T is not constant, you must integrate along the reversible path: ∆ S = dq rev T state 1 state 2 Plotting thermodynamic States and Processes Thermodynamic states can be plotted as functions of the thermodynamic variables P, V, T, and the number of moles of each component. The thermodynamic state of the system is defined by a specified set of thermodynamic variables. Any curve connecting two thermodynamic states is called a process . P V n = const T . State 1 . State 2 three different processes connecting State 1 and State 2 One path between two states is the reversible path . The other paths are irreversible . On the reversible path , the system remains arbitrarily close to thermodynamic equilibrium . Isothermal Entropy changes • Isothermal Reversible process • Isothermal Expansion of Ideal Gas • ∆ E = 0 because dT=0 • So along reversible path q rev = w rev and w rev =  PdV • =  nRT dV/V • 2200 ∆ S = nR ln(V f /V i ) ∆ S = dq rev /T i f =1/T dq rev =1/T dw rev =1/T nRT dV/V i f i f i f ∆ S = dq rev /T i f =1/T dq rev = q rev /T i f Isothermal Expansion of an Ideal Gas n 1 = 0.25 mole T 1 = 25°C V 1 = 2.00 L P 1 = 3.06 atm n 1 = 0.25 mole T 1 = 25°C V 1 = 10.00 L P 1 = 0.611 atm= 0....
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This note was uploaded on 04/01/2008 for the course CHEM 230 taught by Professor Sharp during the Winter '08 term at University of Michigan.
 Winter '08
 SHARP
 Equilibrium, pH

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