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).

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- 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!!

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.

Can the entropy of a (particular) system ever decrease? Yes!! but only at the expense of (at least an equal) increase in another system 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?

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.

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.

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)