The KelvinPlanck statement of the second law


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1 The KelvinPlanck statement of the second law It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work Q W E =ΔE net net net, mass sys Q net W = net 0 ( Q H + W ) ( Q L + W ) = 0 in out Q H = W out Due to friction and dissipative effect, no heat engine can have a thermal efficiency of 100 percent A heat engine that violates the KelvinPlanck statement of the second law 1
2 The Causius statement of the second law It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a low temperature body to a higher  temperature body. Q W E =ΔE net net net, mass sys Q net = 0 Q = Q H Heat can not move from low temperature source to high temperature source without external work! L A refrigerator that violates the Clausius statement of the second law 2
3 Reversible or Irreversible process? Reversible process: a process that can be reversed without leaving any trace on the surroundings.  Both the system and the surroundings are returned to their initial states at the end of the reverse process W gas gas gas gas A irreversible process Q  For combined (original and reverse) process, if the net heat and net work exchange between the system and surrounding is zero, then it is a reversible process. Q net = 0 and W net = 0 3
4 Reversible or Irreversible process? Consider a heat reservoir giving up heat to a reversible heat engine, which in turn gives up heat to a pistoncylinder device. From 1 st law of Thermodynamics E E = ΔE in out c δq ( δw + δw ) = de R rev sys c δq δw = de δq T R R c c To prove reversible process, we assume a Q net = 0 R δ = δq = T δq QR TR T 4
5 Reversible or Irreversible process? Now the total work done is For a cycle and the total net work becomes If W c > 0 then a cyclic device exchanging energy with a single heat reservoir and producing an equivalent amount of work Thus, the KelvinPlanck statement of the second law is violated! Since T R > 0 (absolute temperature), we conclude Here Q is the net heat added to the system, Q net. or This equation is called the Clausius inequality. The equality holds for the reversible process and the inequality holds for the irreversible process. 5
6 Definition of Entropy In the reversed cycle case, all the quantities will have the same magnitude but the opposite sign. Therefore, the work W C, which could not be a positive quantity in the regular case, cannot be a negative quantity in the reversed case. Then W C,int rev = 0 since it cannot be a positive or negative quantity, and therefore for internally reversible cycles. Thus we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones. Clausius (1865) defined entropy, ds δq = T int rev 6
7 Entropy Consider the cycle shown below composed of two reversible processes A and B. Apply the Clausius inequality for this cycle. What do you conclude about these two integrals? P 2 B A 1 A cycle composed of two reversible processes. V Apply the Clausius inequality for the cycle made of two internally reversible processes: 7
8 Entropy You should find: Since the quantity (δq net /T) int rev is independent of the path and must be a property, we call this property the entropy S. The entropy change occurring during a process is related to the heat transfer and the temperature of the system. The entropy change during a reversible process, sometimes called an internally reversible process, is defined as 8
9 Entropy Consider the cycle 1A2B1, shown below, where process A is arbitrary that is, it can be either reversible or irreversible, and process B is internally reversible. P 2 B A 1 A cycle composed of reversible and irreversible processes. V The integral along the internally reversible path, process B, is the entropy change S 1 S 2. Therefore, 9
10 Entropy or In general the entropy change during a process is defined as where ds Q δ T = holds for the internally reversible process > holds for the irreversible process Consider the effect of heat transfer on entropy for the internally reversible case. net ds Q = δ T Which temperature T is this one? If then ds > net δ Q net > 0 0 δ Q net = 0 then ds = 0 δ < 0 then ds < 0 Q net 10
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