VI. Entropy. VI. Entropy
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1 A. Introduction ( is an alternative way of looking at entropy). Observation shows that isolated systems spontaneously change to a state of equilibrium. a. wo blocks of iron with A > B. Bring the blocks together and allow the system to come to equilibrium. A B he system is A + B. At equilibrium, A = B. he first law says nothing about the direction of change. A. Introduction. Observation shows that isolated systems spontaneously change to a state of equilibrium. b. wo chambers connected by a valve. Open valve and allow system to reach equilibrium. Air, P A, A Ideal gas A B X Vacuum P B =0 he system is A + B. At equilibrium, A = B and P A = P B. he first law is silent on the direction of change.
2 A. Introduction. he magic vortex tube will it work? he first law says yes, but will it work? Cold air kg at atm 73 K Magic Vortex ube Compressed air kg at 4 atm, 300 K Hot air kg at atm, 333 K We need a quantitative answer. his suggests that we are looking for a property (like, P, u, or volume). How can we find such a property? A. Introduction 3. Entropy (use the symbol S) a. A property that in an isolated system either increases or is constant. he entropy is a maximum at equilibrium. b. Entropy is a measure of the disorder of a system on a microscopic scale. here is only one form of entropy internal entropy. entropy (S) of isolated system ds isolated dt 0 S he entropy is a maximum at equilibrium. time
3 . Four steps to a quantitative definition of S a. Adding thermal energy, Q, to a closed system will increase its disorder. b. Adding Q to a closed system at a low temperature will cause a larger increase in disorder than adding the same Q to a system at a high temperature. c. Performing reversible work, under adiabatic conditions, on a closed system will not change its entropy. d. o avoid the complexities associated with irreversibilities, use Q rev to define S.. Definition of S for a closed system ds Units on S are kj/k δq = rev Qrev absolute temperature, kelvin δ S = Q rev ds dt integrated form = rate form 3
4 a. wo blocks of iron with A > B. What is S once equilibrium is reached? A B he system is A + B. 0.5 kg, A = 0.5 kg, B = 373 K 93 K A = B at equilibrium. Iron is incompressible. From first law: du = δq + δw or δq = du δw For a reversible process with only P-V work: δ qrev = cvd + Pdv a. wo blocks of iron. What is S once equilibrium is reached? Change in entropy: ds δq cd Pdv rev v = = + Integrate and recall c = c v = c p for solids: s = ds = c d cln = 0 since iron is incompressible How to find? 4
5 a. wo blocks of iron. Find S. Find from first law. 0 0 U = Q+ W ( ) ( ) U = U U + U U = A A B B 0 Rewrite in terms of specific heat and temperature. ( ) ( ) U= mc + mc = Solve for. = + A B A B 0 a. wo blocks of iron. Find S. Entropy balance. ( ) ( ) S = S S + S S = S A A B B A + A B+ B S = mc ln + ln = mcln 4 A B A B J (373) J S = 0.5 kg 450 ln = 3.7 kgk 4(373)(93) K J S = S = 3.7 K S > 0 as advertised 5
6 b. Unrestrained expansion of ideal gas. Open valve. Find S. Air, m A = kg, ideal gas, V A = 0. m 3, A =300K A B X Energy balance. Ideal gas law. Change in entropy. Vacuum P B = 0, V B = 0. m 3 he system is A + B. 0 0 U = Q+ W = = 300 K P VA P mr = and = P V V A ds 0 δq cd Pdv rev = = + = is also observed experimentally. b. Unrestrained expansion of ideal gas. Open valve. Find S. Air, m A = kg, ideal gas, V A = 0. m 3, A =300K A B X Vacuum P B = 0, V B = 0. m 3 he system is A + B. δqrev PdV dv Change in entropy. ds = = = mr V Integrate. S = mr dv = mrln V V V kj 8.34 V kmol K J S = S = mrln = kg ln = 99 S > 0 as V kg 9 K kmol advertised 6
7 c. Conclusion Our definition of entropy is consistent with all experiments that have been performed to test it. For a closed system, our entropy balance equations are δq S = S S = + S sys dssys Qj = + S dt j j where S S S > 0 irreversible process = 0 reversible process < 0 impossible process 7
VI. Entropy. VI. Entropy
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