110c-lecture5 - Chapter 20 Problems 8, 9, 13, 14, 18, 25,...

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2 nd Law of Thermodynamics. Entropy (S) of an isolated system increases ( dS > 0) during a spontaneous process. Entropy is a Thermodynamic State Function. Carnot Heat Cycle H C H H C H T T T q q q = C C H H T q T q = 0 = T q rev δ T q dS rev = (reversible path only) dV V nR dT T C dV T P T dU T q V rev + = + = (ideal gas) 1 2 1 2 1 2 ln ln V V nR T T C S S S V + = = Δ (path independent) 1 2 2 1 ln ) ( T T C dT T T C S P T T P = = Δ (constant pressure and C P ) Chapter 20 – Problems 8, 9, 13, 14, 18, 25, 27, 31, 32, 35, 38
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H C H C T T Q Q = 1 1 H H C C T Q T Q = H H C C T q T q = C C H H T q T q + = 0 . 0 = T q rev δ T q dS rev = Entropy Derived from Carnot Cycle
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Entropy for Reversible Expansion of Ideal Gas Volume Pressure Final Initial V 2 V 1 Reversible Expansion (n-steps) q rev = -w rev = dV V nRT V V 2 1 = nRT*ln(V 2 /V 1 ) rev rev w q dU δ + = sys sys rev V dV P q dT C = (ideal gas) sys sys rev V dV V nRT q dT C = sys sys rev V dV V nR T q T dT C = sys sys V rev dV V nR T dT C T q + = + = 2 1 2 1 2 1 sys sys V dV V nR T dT C dS where T q dS rev = + = 2 1 2 1 2 1 sys sys V V dV nR T dT C dS 1 2 1 2 1 2 ln ln V V nR T T C S S S V + = = Δ T q T q rev irrev < T q dS irrev > T q dS rev = Regardless of the path: 1 2 1 2 ln ln V V nR T T C S V + = Δ isothermal
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Entropy Practice Problem Calculate Δ S for one mole of an ideal gas expanded from 20 dm 3 at 273K to 300 dm 3 at 400 K. 1 2 1 2 1 2 ln ln V V nR T T C S S S V + = = Δ T 1 = 273 K, V 1 = 20 dm 3 T 2 = 400 K, V 2 = 300 dm 3 For ideal gas, C V /n = 3R/2 = 3(8.314 J K -1 mol -1 )/2 = 12.471 J K -1 mol -1 Δ S = (3R/2)*ln(400/273) + (8.314 J K -1 mol -1 )*ln(300/20) = 12.471*ln(400/273) + 8.314*ln(300/20) = +27.27 J K -1 mol -1 Δ S is positive because the gas becomes more disordered during the expansion.
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q rev is Path Dependent and NOT a State Function Path A: isothermal ( Δ T = 0) Path B: adiabatic (q B = 0) Path C: const. V ( δ w= ±0) Path D: const. P Path E: const. V Path Condition q rev A T=const = = + = = 1 2 1 1 ln 0 V V nRT q dV V nRT q w q dU A δδ δ B q=0 0 = B q C V=const ( ) 2 1 T T C q dT C dU w q V C V = = = δ D P=const ( ) 1 3 T T C q dT C dH q P D P = = δ E V=const ( ) 3 1 0 T T C q dT C dU q V E V = = = δ Path A connects the same two points as Path (B+C) or Path (D+E) But C B A q q q + and also E D A q q q + , so as we said before q
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Path A: isothermal ( Δ T = 0) Path B: adiabatic (q B = 0) Path C: const. V ( δ w= ±0) Path D: const. P Path E: const. V
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110c-lecture5 - Chapter 20 Problems 8, 9, 13, 14, 18, 25,...

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