Lect+17+08 - Lecture 17 Gibbs Energies Standard State...

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Lecture 17 Gibbs Energies Standard State Entropies Clausius inequality Gibbs Free Energy is a state function which determines if a system at constant T and P is at equilibrium. G < 0 spontaneous process G = 0 reversible process, at equilibrium G > 0 non-spontaneous process 2200 G ° = H ° - T S ° Examples
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Entropy Changes with Changing Temperature Adiabatic process (Isentropic) Isochoric processes (dV=0) If C v constant over T i to T f then 2200 S = nC v lnT f /T i (dV=0) Isobaric process if C p constant T i to T f 2200 S = nC p lnT f /T i (dP=0) S = dq rev /T i f = 0.0 S = dq rev /T Ti Tf = (nC v )dT/T Ti Tf
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S for isobaric heating of a Liquid or Solid 2.00 moles of water are warmed from 0°C to 100°C. Calculate S warm for this process. Use: For liquid water, c P = 4.184 J g -1 K -1 = 75.35 J mol -1 K -1 S = nc P dT T T 1 T 2 = nc P ln e ( T 2 / T 1 ) S = (2.00 mol) (75.35 J mol -1 K -1 ) ln e (373/273) = 47.03 J mol -1 K -1 Note that c P for anything other than an ideal gas must be measured and is usually found in a thermodynamic table (eg Appendix D)
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Conceptual Problems 2.0 moles of ideal gas have Cv =3/2R and Cp = 5/2R. Assume both are constant over the temperature range of interest. 298 K isothermal expansion from 24.2 L to 48.4 L. What are q, w, E, S for the system? Isothermal expansion so E = 0, S greater than zero, and q=-w How do you calculate S? 2.0 moles of ideal gas with Cv =3/2R and Cp = 5/2 R. Assume both are constant over the temperature range of interest. 48.4 L constant volume heating from 298K to 596K. What are q, w, E, S for the system? Constant volume heating so E = q, w = 0, S greater than zero How do you calculate S?
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Clausius Inequality Irreversible Expansion of an Ideal Gas Constant T in a large heat bath (surroundings) E = 0, So q = -w Consider the rapid expansion of an ideal gas w = - P ext dV but during an irreversible expansion P ext must be less than P int ; - w irrev = P ext dV < P int dV = - w rev But for a reversible expansion the external and internal pressure must remain equal through the expansion SO….The work done during an irreversible expansion is always less than the work done during a reversible expansion
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Reversible and irreversible expansion between identical initial and final states Area under the curve is the work Reversible process always provides more work than irreversible process
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