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Chapter 20 Entropy and the Second Law of Thermodynamics 20.1 Some one-way processes Which is closer to ‘common’ sense? Ink diffusing in a beaker of water.

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Presentation on theme: "Chapter 20 Entropy and the Second Law of Thermodynamics 20.1 Some one-way processes Which is closer to ‘common’ sense? Ink diffusing in a beaker of water."— Presentation transcript:

1 Chapter 20 Entropy and the Second Law of Thermodynamics 20.1 Some one-way processes Which is closer to ‘common’ sense? Ink diffusing in a beaker of water or diffused ink in a beaker concentrating out of solution?? Although we would not be violating energy conservation, we would be violating the postulate for the change in entropy, which states: For an irreversible process in a closed system, the entropy always increases. What is entropy? Entropy is a state function which is a measure of the disorder in a system. Highly disordered systems (e.g. gases) have more entropy than ordered systems (e.g. solid crystals).

2 The world behaves as if we can not treat work and heat on an “equal” footing!! 20.3 Change in entropy Actually, strictly speaking, all real [macroscopic] processes are irreversible!! Many real processes are very close to being reversible. Reversibility of processes are only an approximation!! A process is almost reversible when it occurs very slowly so that the system is virtually always in equilibrium (e.g. adding grains to a piston in isothermal contact). Entropy is a state variable. To calculate the change in entropy between any two states (i & f): 1- Find a reversible process between initial and final states. 2- Calculate: dS = dQ r /T for infinitesimal steps in the process.

3 3- Take the integral between initial and final states:  S = i  f dQ r /T It is crucial to distinguish between Q and Q r. What if the process is irreversible?! It does not matter!! Entropy is a state function. It depends on the state not the process!! CP #1; Problem 20-1 Special cases: 1- Reversible process for an ideal gas:  S = n R ln(V f /V i ) + n c v ln(T f /T i ) 2- Melting:  S = m L F /T m 3-  S for a reversible adiabatic process: zero! 4-  S for (an arbitrary) cyclic process: ZERO!!

4 5- Heat conduction:  S = Q/T L - Q/T H 6- Adiabatic (isolated) free expansion:  S = n R ln(V f /V i ) 7- Irreversible heat transfer (w/o mixing):  S = m 1 c 1 ln(T f /T 1 ) + m 2 c 2 ln(T f /T 2 ) 20-4 Second law of thermodynamic If a process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible processes. That is, the entropy of a closed system never decreases.  S closed =  S sys +  S res > 0[irreversible]  S closed =  S sys +  S res = 0[reversible] Notice that isolated systems tend toward disorder; i.e. the entropy of the universe increases in all natural processes.

5 Can the entropy of a (particular) system ever decrease? Yes!! but only at the expense of (at least an equal) increase in another system. 20-5 Entropy in the real world: Engines Heat engine/ engine/ working substance/ cycle/ strokes/ diagram with Q,T,W. Ideal engine: is an engine in which all processes are reversible and no wasteful energy transfers occur due to friction or turbulence or otherwise. Note: Real engines are not ideal, but “very” good engines are approximately ideal. A Carnot engine is an ideal engine, the cycle of which consists of four strokes: two idiabatics and two isothermals. How does this look on a P-V diagram? How does this look on a T-S diagram?

6 Note that for a Carnot engine: (can you prove this?) T H /T L = |Q H |/ |Q L | How do we calculate the work of a Carnot cycle? W c = |Q H | - |Q L | = area inside the T-S cycle. Efficiency (e) of an engine is defined to be: e = W/ Q H For a Carnot engine, the efficiency (e c ) is: e c = W/ Q H = 1- |Q L |/ Q H = 1- T L / T H How can one increase the efficiency? Note that since T L > 0 and T H < ∞, e c is always less the unity. Therefore: Even the ideal engine is not “perfect”!! Real engines have even lower efficiencies (e ~< 40 %) than that of Carnot’s.

7 One way to express the second law of thermodynamic is that: It is “impossible” for a machine to transfer thermal energy completely into other forms of energy in any cyclic process. Or, we can say: Second law of thermodynamic: It is impossible to construct a heat engine that, operating in a cycle, produces no other effect than the absorption of thermal energy from a reservoir and the performance of an equal amount of work. (Kelvin-Plank statement) 21-5 Entropy in the real world Refrigerators Refrigerators and heat pumps are heat engines running in reverse; they move thermal energy from a region at lower temperature to a region at higher temperature (used for cooling or heating). diagram: Q,T,W Can this be done with no work?! Work must be done on the working substance.

8 Coefficient of performance (COP or K): K =Heat transferred/ Work done For a refrigerator: K = Q L /W [refrigerator] K c = T L /(T H - T L ) [refrigerator] For a heater: K = |Q H |/W[heat pump] K c = T H /(T H - T L ) [heat pump] Second law of thermodynamic: It is impossible to construct a machine operating in a cycle that produces no other effect than to transfer thermal energy continuously from one object to another object at a higher temperature. (Clausius statement)


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