Lecture3 - Lecture 03: State Functions Reading: Zumdahl...

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Lecture 03: State Functions Reading: Zumdahl 9.3, 9.4 Outline 1. Example of Thermo. Pathways 2. State Functions 3. (9.4…for laboratory) 1
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Thermodynamic Pathways: an Example • Example 9.2. We take 2.00 mol of an ideal monatomic gas undergo the following: – Initial (State A): P A = 2.00 atm, V A = 10.0 L – Final (State B): P B = 1.00 atm, V B = 30.0 L We’ll do this two ways: Path 1 : Expansion then Cooling Path 2 : Cooling then Expansion 2
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Thermodynamic Jargon • When doing pathways, we usually keep one variable constant. The language used to indicate what is held constant is: – Isobaric: Constant Pressure – Isothermal: Constant Temperature – Isochoric: Constant Volume 3
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Thermodynamic Path: A series of manipulations of a system that takes the system from an initial state to a final state isochoric isobaric 4
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Pathway 1 • Step 1. Constant pressure expansion (P = 2 atm) from 10.0 l to 30.0 l. P∆V = (2.00 atm)(30.0 l - 10.0 l) = 40.0 l.atm = (40.0 l.atm)(101.3 J/l.atm) = 4.0 x 10 3 J = -w And ∆T = P∆V/nR = 4.05 x 10 3 J/(2 mol)(8.314 J/mol.K) = 243.6 K 5
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• Step 1 is isobaric (constant P); therefore, q 1 = q P = nC p ∆T = (2mol)(5/2R)(243.6 K) = 1.0 x 10 4 J
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This note was uploaded on 01/18/2012 for the course CHEM 152B taught by Professor Zhang during the Winter '12 term at University of Washington.

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Lecture3 - Lecture 03: State Functions Reading: Zumdahl...

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