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Chem 300 - Ch 20/#2 Today’s To Do List

Chem 300 - Ch 20/#2 Today’s To Do List. 2nd Law Entropy Changes. D S and Isolated Systems. Isolated System D U = 0 For Spontaneous Process dS > 0 For Reversible Process dS = 0. Isolated 2-Compartment System. Isolated 2-Compartment System. U A + U B = constant

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Chem 300 - Ch 20/#2 Today’s To Do List

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  1. Chem 300 - Ch 20/#2 Today’s To Do List • 2nd Law • Entropy Changes

  2. DS and Isolated Systems • Isolated System • DU = 0 • For Spontaneous Process • dS > 0 • For Reversible Process • dS = 0

  3. Isolated 2-Compartment System

  4. Isolated 2-Compartment System • UA + UB = constant • VA & VB are constant • S = SA + SB • Show that dStot > 0 if TA>or< TB

  5. Attainment of Equilibrium

  6. NonIsolated Systems • Consider dS in 2 parts: • dSprod = produced by the process • dSexch = due to heat exchange • dS = dSprod + dSexch = dSprod + dq/T • For rev process: dq = dqrev • dSprod = 0 • dS = dqrev/T

  7. continued • For irrev process: • dSprod > 0 dSexch =dqirr/T • dS > dqirr/T • Inequality of Clausius • True in general

  8. Cyclic Process: S = 0

  9. Clausius • Die Energie der Welt ist Konstant • Die Entropie der Welt Strebt einem Maximum zu

  10. Dr. Frank Lambert on Entropy http://www.entropysite.com

  11. Lambert Summarized •    Entropy measures the energy dispersion in a system divided by temperature (Dq/t) • Entropy measures the energy distribution by molecules on quantized levels in a system: the number of ways in which energetic molecules in a system are distributed among the energy levels available to them. • Entropy is NOT a driving force; it measures the driving force.

  12. To calculate DS • Only need initial & final states • Devise a reversible path • Expansion into vacuum • Isothermal expansion • Mixing of 2 gases • Heat flow between 2 metals

  13. Heat Engine Model

  14. Sadi Carnot • Born: 1 June 1796 in Paris, FranceDied: 24 Aug 1832 in Paris, France (of Cholera) • Carnot Cycle (Heat Engine)

  15. Carnot Cycle • Duengine = w + qrev,h + qrev,c = 0 • -w = qrev,h + qrev,c • Define Max efficiency (emax): • emax = -w/qrev,h = (qrev,h + qrev,c)/qrev,h Dsengine = qrev,h/Th + qrev,c/Tc = 0 Rearrange: qrev,c = -qrev,h(Tc/Th)

  16. Carnot Continued • emax = 1 – Tc/Th = (Th – Tc)/Th • Indep of: • Engine design • Working material • Dependent upon: • Operating T • Example: Gas Turbine • Th = 600 K Tc = 300 K • emax = 1 – 300/600 = 0.5 = 50%

  17. Next Time • Start Chapter 21 • S(T) • 3rd Law • Absolute Zero

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