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.
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
. 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
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
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 + (373)93 + 93 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
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
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