Entropy

Physics 1061 Entropy and the Second Law of Thermodynamics 1 Entropy We consider an ideal gas undergoing a cyclical process, as shown in Fig. 1. When evolving along the upper branch a → b the gas expands, thus doing positive work on the environment. When returning along the branch b → a to the initial state, the gas is compressed, so the environment is doing work on the gas. The area underneath the curve a → b is greater, than the area underneath the curve b → a , so the gas did more work than it was done on it. Thus the net work done by the gas, given by the area enclosed by the cycle, is a positive amount. According to the First Law of Thermodynamics the infinitesimal change of internal pressure volume a b area=net work done by gas Figure 1: A thermodynamic cycle proceeds clockwise. The enclosed area gives the net work done by the gas on the environment. energy of the gas is dE int = δW + δQ. Since the internal energy is a state function, the system has the same value of internal energy when it reaches state a again, after completing one full cycle. Thus integrating dE int over the complete cycle we get zero: I E int = 0 . We can say that the internal energy of the system is conserved over the full cycle. On the other hand, we have I δQ = - I δW = W net,output , namely the differential form δQ is not a full differential since the cyclical integral does not vanish. Integrating δQ over a cycle gives the total work done by the system on the environment. The only way the integral could vanish is if the process b → a exactly retraces the process a → b , but in reverse direction. In this case, the area enclosed by the cycle in pV-space is zero and the cyclic integrals vanish trivially. Nevertheless, we shall explore the idea of reversibility further. A quantity called entropy was introduced by Rudolf Clausius (1822-1888) in 1865, to describe quantita- tively the irreversibility in thermodynamic processes. By definition, the infinitesimal amount of entropy is defined as the full differential dS = δQ T , 1

where δQ is the amount absorbed by the system reversibly from its environment. Mathematically, the factor 1 /T serves as an integrating factor , which upon multiplication transforms the differential form δQ into a full differential dS . For any reversible cyclic process we have I dS = I δQ T = 0 . Thus, we say that entropy is conserved over a full cycle - the starting and the initial values of the entropy of the system are the same. The finite change of entropy when the system evolves reversibly between two states, is given by the integral Δ S = S f - S i = Z f i δQ T . For an isothermal reversible process the entropy change is Δ S = Q T . We see that entropy is measured in SI units of J/K , the same as the units for the Boltzmann constant k and the heat capacity C . Example: A block of 1 . 0 kg ice melts completely at 0 ◦ C . The latent heat of fusion of water is 333 kJ/kg . Thus the entropy change of the ice is Δ S = Q T = mL T = (1 . 0 kg )(333 kJ/kg ) (273 . 15 K ) ≈ 1 . 2 kJ/K. The reversible thermodynamic processes are an useful abstraction. In reality, all processes are irreversible. The entropy change for irreversible processes is calculated by using a reversible process that connects the same initial and final states. Example: Consider an ideal gas that is confined to the left half of a thermally insulated container, by a closed stopcock. When the stopcock is opened, the gas rushes to fill the entire container, doing no work in the process (no piston to push). The process is irreversible. The gas molecules will never return to the left half on their own. Since no heat was absorbed, and no work was done, according to the First Law of Thermodynamics, the internal energy, and respectively the temperature of the gas, will remain constant. To calculate the entropy for the adiabatic free expansion, we cannot simply use Δ S = S f - S i = Z f i δQ T . since this expression applies only to reversible processes connecting the initial and the final states. If we blindly plug in Q = 0, we will get that the entropy of the gas did not change, which will be incorrect. Instead, we use the fact that the gas temperature did not change, and use a reversible isothermal expansion that doubles the volume of the gas. For the reversible isothermal expansion we can write Δ S = Q T = - W T = nRT ln( V f /V i ) T = nR ln 2 = Nk ln 2 . We see that the change of entropy is positive. It is also proportional to the number of gas molecules, which means that the entropy is an extensive thermodynamic property, much like energy and volume. If a system is made of two subsystems A and B , then its entropy is the sum of the entropy of the subsystems: S = S A + S B . 2

Figure 2: An ideal gas undergoing an adiabatic free expansion. The process is clearly irreversible. For an ideal gas, undergoing a reversible process, we can write the infinitesimal amount of heat as δQ = TdS. Then the First Law of Thermodynamics takes the form dE int = TdS - pdV. For a general reversible process, involving the ideal gas, we can then use dS = 1 T ( dE int + pdV ) to calculate the entropy change. Using the Ideal Gas Law pV = nRT to express the pressure, we obtain dS = nC V dT T + nR dV V . Integrating from some initial state to some final state, we obtain for the finite change of entropy Δ S = S f - S i = nC V ln V f V i + nR ln T f T i . We see that, in general, if the volume or the temperature of the gas change, the entropy will also change. When heating liquid and solid uniform substances, the change of entropy can be calculated by integrating dS = δQ T = mcdT T from the initial to the final state. The specific heat capacity may be a function of the temperature. Example: We consider two paper cups holding an equal amount of water but at different temperatures T H and T C , where T H > T C . The cups are placed in a thermally insulated container, separated by a movable 3

Figure 3: (a) The cups hold equal amounts of water, but at different temperatures. (b) When the shutter is removed, the cups exchange heat and come to a final state, both with the same temperature. shutter. Once the shutter is removed, the cups begin to exchange thermal energy via thermal radiation, till they reach the same final temperature T f . This process is clearly irreversible. The two cups will never return on their own to their initial states, once they reach their final temperature. We can imagine that instead of undergoing the irreversible heat exchange process, just described, the cups instead were respectively heated and cooled, reversibly. This may be accomplished by placing the cup in contact with a heat reservoir. A heat reservoir is a body that has enormous heat capacity. Then adding heat to the reservoir would not change its temperature significantly. Earth’s oceans may be considered approximate heat reservoirs. We place one cup in contact with the reservoir and very, very slowly raise its temperature from T C to T f . Similarly, we place the hot cup in contact with the heat reservoir and this time very, very slowly cool it down to the final temperature. Figure 4: The cups can evolve from their initial to their final state in a reversible way, if we use a heat reservoir with controllable temperature. Proceeding reversibly, we can calculate the entropy change of each cup of water. For the hot cup we obtain Δ S H = Z T f T H mcdT T = mc ln T f T H . For the cold cup, we get Δ S C = Z T f T C mcdT T = mc ln T f T C . Assuming a perfect thermal balance, we have Q H + Q C = 0 , or mc ( T f - T C ) + mc ( T f - T H ) = 0 . Since the water amount in each cup is the same, we see that T H - T f = T f - T C . 4

Thus the equilibrium final temperature will be right in the middle between the hot and cold temperatures. T f = 1 2 ( T H + T C ) Then we can express the entropy change of the hot cup as Δ S H = mc ln T f T H = mc ln T H + T C 2 T H . For the cold cup we get Δ S C = mc ln T f T C = mc ln T H + T C 2 T C . The total entropy change is Δ S = Δ S H + Δ S C = mc ln T H + T C 2 T H + mc ln T H + T C 2 T C = mc ln ( T H + T C ) 2 4 T H T C . The argument of the logarithm is greater than 1. To show that, we test the inequality ( T H + T C ) 2 > 4 T H T C . Expanding the left-hand side gives T 2 H + 2 T H T C + T 2 C > 4 T H T C , which leads to ( T H - T C ) 2 > 0 , which is true. Thus we obtain that the entropy change of the system of two cups is a positive quantity. Notice that the thermal energy got distributed equally between the two cups and both reached the same final temperature. The hot cup lost heat, so its entropy decreased. The cold cup gained heat, so its entropy increased. But overall, the total entropy increased, regardless of the exact values for the temperatures. This and similar findings, eventually resulted in the formulation of the so called Entropy Principle : If an irreversible process occurs in a closed system, the entropy of that system always increases; it never decreases. 2 The ideal heat engine A heat engine is a device, operating on a cycle, that extracts heat from its environment and performs useful work. Let during the cycle the device absorbs heat Q input from sources in its environment, and delivers net amount of work W net,output to its environment. The thermal efficiency of the heat engine is defined as the ratio = Q input W net,output . The first scientific study on the efficiency of heat engines was conducted in 1824 by Nicolas L´ eonard Sadi Carnot (1796-1832), a French military engineer. In his work “Reflections on the Motive Power of Fire”, he set for himself the goal to answer two questions: 1. Is the work available from a heat source potentially unbounded? 2. Can heat engines, in principle, be improved by replacing the steam with some other working fluid? 5