Lec41 - Lecture 41 Friday 12/11/09 Chapter 20: The Second...

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Unformatted text preview: Lecture 41 Friday 12/11/09 Chapter 20: The Second Law of Thermodynamics Entropy • Entropy S: A state function that measures the amount of disorder in a system. • Change in S: dS = dQ / T S= 2 1 dQ / T • Entropy statement of the 2nd Law: In an isolated system, the entropy always either stays the same (reversible process) or increases (irreversible process), but never decreases -- The engine and refrigerator statements of the 2nd law are derivable from this one -- Determines the arrow of time Consequences of ISU Closure on Weds, December 9 ISU was closed on Wednesday Dec 9 due to the bad weather 1. Labs that would have been held on Dec 9 are canceled. Students are NOT required to make up this lab. The score for the missed lab and for the missed prelab will be the average of the other 6 labs and 6 prelabs, respectively, for each student. 2. The lecture on Friday Dec 11 is Lecture 41 on entropy 3. The review lecture 42 for the final exam will be held Saturday, Dec 12, at 9 a.m. (normal 8 a.m. lecture), at 10 a.m. (normal 9 a.m. lecture), or 11 a.m. (normal 10 a.m. lecture) in Room 5 Physics Hall Reversible and Irreversible Thermodynamic Processes Reversible thermodynamic process: • Is always infinitesimally close to mechanical and thermal equilibrium • Can be reversed by a small change in the process parameters • Can be represented by a path on a p -V diagram Reversible and Irreversible Thermodynamic Processes Irreversible thermodynamic process: • All real processes are irreversible to some extent • Cannot be represented by a path on a p -V diagram • Increases the disorder (entropy) of an isolated system • A movie of an irreversible process run backwards would look impossible • The universe is evolving towards increasing disorder • Determines the direction of time s arrow Second Law of Thermodynamics Engine statement of the 2nd law: It is impossible for a heat engine operating in a cycle to completely convert heat into work (the thermal efficiency e of the engine must satisfy 0 e < 1) Refrigerator statement of the 2nd law: Heat cannot spontaneously flow from a cold object to a hot object without the input of work (the coefficient of performance of the refrigerator must satisfy 0 K < Entropy statement of the 2nd law: The entropy of an isolated system either stays the same (reversible process) or increases (irreversible process) for any thermodynamic process: S 0 • This definition is the most general one, and can be used to derive the engine and refrigerator statements of the 2nd law • All real processes are irreversible: the entropy of the universe increases with time. This fixes the direction of time’s arrow Question TK = TC + 273 H2O: Lf = 334 J/g QH2O = mH2OLf QH2 O Solution = m H 2 O Lf = (1.00 g)(334 J/g) = 334 J The “system” is 1 gram of solid H2O ice at 0 ˚C. Heat is added to melt it to liquid water at 0 ˚C. The entropy of the system changes by __ J/K. 1. 2. 3. 4. 0 1.2 2.4 3.6 334 J S= = = 1.22 J / K . 273 K T Since the entropy change is positive, this says that water is more disordered than ice at the same temperature. The ice is not an isolated system, so we don't know if the process was reversible or not. QH2 O Question Solution An isolated system consists of 1 gram of solid H2O at 0 ˚C plus the temperature reservoir at slightly above 0 ˚C that is used to melt the ice. The heat to melt the ice is QH2O = 334 J. The entropy of the system changes by __ J/K when the ice is melted by adding heat. 1. 2. 3. 4. 0 1.2 2.4 3.6 The isolated system consists of both the ice and the temperature reservoir. Both are at almost the same T . Ssystem = Sice + Sreservoir Isolated system: Qreservoir = Qice . Ssystem = Qice Qreservoir + T T Qice Qice = + =0 273 K 273 K Therefore the process is reversible. Question TK = TC + 273 H2O: QH2O = 334 J Solution The isolated system consists of both the ice and the temperature reservoir. The isolated system is 1 gram of Both are not at the same T. solid H2O at 0 ˚C plus the Ssystem = Sice + Sreservoir temperature reservoir at 20 ˚C Isolated system: that is used to melt the ice. Qreservoir = Qice . The entropy of the system Qice Qreservoir Ssystem = + changes by __ J/K when the ice Tice Treservoir is melted by adding heat. Qice Qice = + Tice Treservoir 1. 0 2. 0.1 334 J 334 J = = 0.08 J / K . 3. 0.2 273 K 293 K 4. 0.3 Therefore the process is not reversible. Summary of the 3 Answers to the Questions Treservoir (˚C) S (J/K) System Ice ------(not isolated system) Ice + T reservoir (isolated system) Ice + T reservoir (isolated system) 0 20 1.2 0 (reversible process) 0.08 (irreversible process) (positive sign corresponds to correct heat flow direction) • These results are consistent with the 2nd law: S 0 for any thermodynamic process in an isolated system • S < 0 in the last isolated system would correspond to heat spontaneously flowing from a cold object (ice at 0 ˚C) to a hot object (T reservoir at 20˚C) (never happens in real life) Example: Entropy of the Ideal Gas 1 Law: dQ = dU + dW = nCV dT + p dV nRT , so But p = V dV . dQ = nCV dT + nRT V Increment of entropy dS: dV dQ dT = nCV + nR . dS = V T T Now assuming a process is reversible, we can integrate both sides from state 1 to state 2: st 2 T2 S = dS = nCV 1 dV dT + nR V T T1 V1 V2 T2 V2 S = nCV ln , + nR ln T1 V1 which is only a function of the initial and final states of the gas. Since S is a state function, this expression is valid for either reversible or irreversible processes. Example: Isothermal Expansion of Ideal Gas Ideal gas only : Sgas = nCV ln T2 V + nR ln 2 T1 V1 Isolated system = gas + T reservoir (heat source) The heat absorbed by the gas is Qgas = Sgas T = nRT ln V2 . V1 Reversible Isothermal Expansion (V2 > V1 ) T2 = T1 = T , so ln(T2 / T1 ) = 0 and Sgas = nR ln V2 V1 where V2 > V1 The heat absorbed by the reservoir is Qreservoir = Qgas . The change in entropy of the system is Ssystem = Sgas + Sreservoir = = 0. Qgas T + Qgas T But Sgas > 0, so doesn't this mean that the process is not reversible? Answer: No. One must consider only isolated systems to answer that. In order to keep T constant during the expansion, an external temperature reservoir at temperature T in contact with the gas must provide heat. Thus the process is reversible. Example: Adiabatic Free Expansion of Ideal Gas Two moles of an ideal gas undergo an adiabatic free expansion from V1 = 1.00 L to V2 = e L = 2.72 L. (The gas itself is an isolated system). The change in the entropy of the gas is ___ J/K. ( U = nCV T, R = 8.31 J/mol·K) In an adiabatic free expansion of an ideal gas, T2 / T1 = 1. Therefore, Sgas V2 = nR ln V1 = (2.00)(8.31) ln 2.72 J/K 1.00 = 16.6 J / mol·K . Since S > 0 in this isolated system, this process is not reversible , as one certainly expects. Solution Ideal Gas : Sgas = nCV ln T2 V + nR ln 2 T1 V1 Example: Reversible Adiabatic Expansion of Ideal Gas Ideal Gas : Sgas = nCV ln T2 V + nR ln 2 T1 V1 1 Since (1) Reversible Adiabatic Expansion (V2 > V1 ) TV 1 = constant = T1V1 1 1 = T2V2 1 1 1 , so T2 V1 = T1 V2 = V1 V2 1 = 1 V2 V1 =( and V2 . V1 CV + R , we get CV CV C +R R 1= V 1= and CV CV CV ( 1) = R. (3) Substituting (3) into (2) gives = Cp = Sgas = 0 . Note: since the gas is not an isolated system, this result does not by itself show that the process is reversible. However, equation (2) was derived assuming a reversible process. T V ln 2 = ( 1)ln 2 T1 V1 1)ln Substituting this expression into Eq. (1) gives Sgas = n [ CV ( 1) + R ] ln V2 . V1 (2) ...
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This note was uploaded on 10/21/2010 for the course PHYS 221 taught by Professor Herrera-siklody during the Fall '08 term at Iowa State.

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