handout_L02_StateFunc_Hess_StdEnthalpy

# handout_L02_StateFunc_Hess_StdEnthalpy - To Date Ideal...

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To Date…. Ideal Monatomic Gas •C v = 3/2R •C p = C v + R = 5/2 R Polyatomic Gas •C v > 3/2R •C p > 5/2 R All Ideal Gases E = q + w w = -P ext V (for now) E = nC v T=q V H = nC p T= ±q P If T = 0, then E = 0 and q = -w Lecture 2A: State Functions • Reading: Zumdahl 9.3, 9.4 • Outline – Example of Thermo. Pathways – State Functions – (9.4…for laboratory) Thermodynamic Pathways: an Example • Example 9.2. We take 2.00 mol of an ideal monatomic gas undergo the followng: – 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 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 Thermodynamic Path: A series of manipulations of a system that takes the system from an initial state to a final state isochoric isobaric 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)

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Pathway 1 (cont.) • 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 = H 1 –And E 1 =nC v T = (2mol)(3/2R)(243.6 K) = 6.0 x 10 3 J (check: E 1 = q 1 + w 1 = (1.0 x 10 4 J) -(4.0 x 10 3 J) = 6.0 x 10 3 J ) Pathway 1 (cont.) • Step 2: Isochoric (const. V) cooling until pressure is reduced from 2.00 atm to 1.00 atm. • First, calculate T: –Now , ∆ T = PV/nR (note: P changes, not V) = (-1.00 atm)(30.0 l)/ (2 mol)(.0821 l.atm/mol K) = -182.7 K Pathway 1 (cont.) •q 2 =q v =nC v T = (2 mol)(3/2R)(-182.7 K) = - 4.6 x 10 3 J • and E 2 =nC v T = -4.6 x 10 3 J • and H 2 =nC p T = -7.6 x 10 3 J • Finally w 2 = 0 (isochoric…no V change) Pathway 1 (end) • Thermodynamic totals for this pathway are the sum of values for step 1 and step 2 q = q 1 + q 2 = 5.5 x 10 3 J w = w 1 + w 2 = -4.0 x 10 3 J E = E 1 + E 2 = 1.5 x 10 3 J H = H 1 + H 2 = 2.5 x 10 3 J Next Pathway • Now we will do the same calculations for the green path. Pathway 2 • Step 1: Isochoric cooling from P = 2.00 atm to P = 1.00 atm. • First, calculate T: T = PV/nR = (-1.00 atm)(10.0 l)/ (2 mol)R = -60.9 K
Pathway 2 (cont.) • Then, calculate the rest for Step 1: q 1 =q v =nC v T = (2 mol)(3/2 R)(-60.9 K) = -1.5 x 10 3 J = E 1 H 1 =nC P T = (2 mol)(5/2 R)(-60.9 K) = -2.5 x 10 3 J w 1 = 0 (constant volume) Pathway 2 (cont.) • Step 2: Isobaric (constant P) expansion at 1.0 atm from 10.0 l to 30.0 l. T = P

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handout_L02_StateFunc_Hess_StdEnthalpy - To Date Ideal...

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