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Entropy Changes in Irreversible Processes

Entropy Changes in Irreversible Processes The efficiency of an irreversible Carnot cycle is always less than the efficiency of a reversible Carnot cycle operating between the same high and low temperature thermal reservoirs : reversible efficiency > irreversible efficiency

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Entropy Changes in Irreversible Processes

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  1. Entropy Changes in Irreversible Processes The efficiency of an irreversible Carnot cycle is always less than the efficiency of a reversible Carnot cycle operating between the same high and low temperature thermal reservoirs: reversible efficiency > irreversible efficiency Expressing these efficiencies in different ways for a single Carnot cycle we can write: 1 - TL / TH > 1 + qL / qH 0 > qL / TL + qH / TH Now using the same arguments that we used before in extending this result to a general cyclic process we develop the inequality of Clausius:  dS = 0 >  dqirr / T Can you justify the left hand equlity in this expression? which allows us to write that for the entropy change in an irreversible process that: dS > dqirr / T 25.1

  2. We can now express the 2nd Law of Thermodynamics for both reversible and irreversible processes as: irreversible dS > dq / T reversible For an isolated system, dq = 0 and, since the universe is the ultimate isolated system: dS isolated system = dS universe = dS total> 0 / T > 0 Finally, since all real processes are irreversible and only approach reversible processes as limiting cases, we have yet another and perhaps the most intriguing and potent statement of the 2nd Law of Thermodynamics: dS total, real processes > 0 which in essense states that: the entropy of the universse continually increases What does this statement of the 2nd Law of Thermodynamics say about the ultimate fate of the universe? 25.2

  3. DS irreversible path > dqirr / T (state 2) (state 1) DS irreversible path > dqirr / T (state 2) (state 1) DS reversible path = dqrev / T Since the differential entropy change in an irreversible process is related by an inequality to differential irreversible heat transfer in that process: the question arises as to how can we calculate the entropy change for an irreversible process? Fortunately, since entropy is a state function and DS is independent of path: the entropy change for the actual irreversible path will be equal to the entropy change calculated along an imagined or hypothetical reversible path between the same two states: DS actual irreversible = DS hypothetical reversible= dqrev / T 25.3

  4. dq = 0 (8.00 atm, 100.0 oC) (2.00 atm, 21.3 oC) irreversible reversible constant pressure cooling of an ideal gas (2.00 atm, 100.0 oC) 1.000 mole of an ideal gas initially at 8.00 atm and 100.0 oC expands adiabatically against a constant external pressure of 1.50 atm until its pressure is 2.00 atm and its temperature is 21.3 oC. What is DS for the gas? Is the process reversible or irreversible? Can we evaluate DS for the gas by calculating: DSgas = dqirr / T = 0 / T = 0 Is DS for the gas greater than, equal to, or less than zero? To calculate DS for the gaswe have to construct a hypothetical reversible path between the inital and final states: reversible isothermal expansion of an ideal gas DS irreversible = DS reversible expansion + DS reversible cooling = n R ln (P1 / P2) + n Cp ln (T2 / T1) = (1.000 mole) (8.314 J/mole K) [ln (8.00 atm / 2.00 atm) + (5/2) ln (251.9 K / 373.2 K)] = + 11.53 J / K - 3.268 J / K = + 8.26 J / K 25.4

  5. DSovap, 298.2 K = ? H2O (l) -----------------------------> H2O (g) 298.2 K, 1.00 atm DSo1 DSo3 DSovap, 373.2 K = ? H2O (l) -----------------------------> H2O (g) 373.2 K, 1.00 atm 1.000 mole of liquid water is vaporized at 25.0 oC and 1.00 atm: DSovap, 298 K H2O (l) -----------------------------> H2O (g) 298 K, 1.00 bar Is this phase change reversible and, if not, why not? An alternate reversible path via which we can calculate the DSo298 K is sketched below: Step 1: In this step the liquid water is reversibly heated from 298.2 K to 373.2 K at a constant pressure of 1.00 atm: DSo1 = 298.2 K 373.2 K n Cp, H2O (l) dT / T = (1.000 mole) (18.07 cal / mole K) ln (373.2 K / 298.2 K) = + 4.05 cal / K 25.5

  6. Step 2: In this step the liquid water is reversibly vaporized at 373.2 K and 1.00 atm: DSovap, 373.2 K = DHovap, 373.2 K / 373.2 K = + 9,720 cal/ 373.2 K = + 26.04 cal / K Why did we choose to vaporize the water at 373.2 K? Step 3: In this step the water vapor is reversibly cooled from 373.2 K to the initial temperature of 298.2 K at 1.00 atm: DSo3 = 373.2 K 298.2 K n Cp, H2O (g) dT = (1.000 mole) 373.2 K 298.2 K[7.17 - 2.56x10-3 T + 8x10+3 T-2] dT / T = (1.000 mole) [ + 7.17 ln (298.2 K / 373.2 K) + 2.56x10-3 (298.2 K - 373.2 K) - (8x10+3 / 2) [1 / (298.2 K)+2 - 1 / (373.2 K)+2 ] = - 1.82 cal / K The entropy change for the irreversible vaporization at 298.2 K and 1.00 atm is therefore: DSovap, 298.2 = DSo1 + DSovap, 373.2 + DSo3 = (+ 4.05 cal / K) + (+ 26.04 cal / K) + (- 1.82 cal / K) = + 28.27 cal /K 25.6

  7. With great care liquid water can be cooled below its normal freezing point of 0.00 oC at 1.00 atm. In fact liquid water has been cooled to - 40.0 oC! This supercooled liquid water is in a metastable phase and can spontaneously and irreversibly freeze to ice under these conditions: - 40.0 oC and 1.00 atm H2O (l) ---------------> H2O (s) What is DS for 1.000 mole of liquid water irreversibly freezing to ice at - 40.0 oC and 1.00 atm? Some data that you will need: DHofreezing, - 40.0 oC [H2O] = - 6,025 J / mole Cp [H2O (l)] = 76.29 J / mole K Cp [H2O (s)] = 35.30 J / mole K 25.7

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