hy 212: General hysics II 1 hapter 20 orksheet (2 nd Law of hermodynamics & eat Engines) Entropy: 1. A sample of 10.0 moles of a monatomic ideal gas, held at constant temperature (1000), is expanded from 0.10 m to 0.20 m. alculate the entropy change during this process. V f nr ln V = = = 10.0 mol 8.1 ln 2 = 7.6 ( )( ) ( ) i mol 2. A sample of 10.0 moles of a monatomic ideal gas, held at constant pressure (1. atm or 1.2x10 a), is compressed from 0.10 m to 0.0 m. alculate the entropy change during this process. f f f Vf f Vf f d 2nRd pdv 2nRd pdv 2nRd f 2 i i i Vi i Vi i i = d = = + = + = = nrln 1 ( )( ) ( ) = 10.0 mol 8.1 ln = -1. 2 mol 2. A sample of 10.0 moles of a monatomic ideal gas, held at constant volume (1.0 m ), is heated from 00 to 00. alculate the entropy change during this process. d 2nRd f 00 2 2 mol 00 i i i f f ( )( ) ( ) = = = nrln = 10.0 mol 8.1 ln =.9. A sample of 1. moles of a thermally insulated (adiabatic) monatomic ideal gas is expanded from 0.10 m to 0.20 m. uring the expansion, the pressure of the gas decreases from.09x10 a to 2.026x10 a. alculate the entropy change during this process. f i f d = d = = 0 i
hy 212: General hysics II 2 hapter 20 orksheet (2 nd Law of hermodynamics & eat Engines) eat Engines:. A arnot heat engine consists of 2 isothermal and 2 adiabatic processes. teps A and are isothermic and steps and A are adiabatic. he temperature at ( ) is 100 and the temperature at A ( ) 00. For state A, =290 a & V =.0 m, and arnot Engine performs.x10 of work. Note: the working fluid is moles of ideal diatomic gas. A) ketch the -V diagram for this heat engine. ) hat is the efficiency of this heat engine? e = 1 - = 0.80 ) ow much heat energy ( ) is absorbed during the isothermic phase?. = = =.6 e 0.80 ) ow much heat energy ( ) is released during the isothermic phase A? = - + = -.6 +. = -1.1 E) hat is the pressure at each point A-? alculate using the ideal gas law. V Get V from : = A = nrln = -1.1 VA A nr ( ) mol -1.1 (mol)(8.1 )(00) V = V e =.0m e = 0.66m (mol)(8.1 mol )(00) = 0.66m = 1.1 nr nr Get from V = V : = = 7 2 00 6 = 1.1 =.16 (100)(1.1 ) 1 1.6 nr (mol)(8.1 nr nr )(100) mol Get V from : V = e = e = 0.01m (mol)(8.1 mol )(100) = 0.01m = 7.0 F) ow much work is performed by the gas during each step? = = -1.1 and = nr = -.9 A 2 = =.6 and = nr =.9 A 2
hy 212: General hysics II hapter 20 orksheet (2 nd Law of hermodynamics & eat Engines) ) hat is the change in entropy ( A- ) during A? -1.1 A = = = -7.7 00 I) hat is the change in entropy ( - ) during?.6 = = = 7.7 100 ) hat is the net change in entropy ( universe ) between the steps in and I? = + = 0 universe A A 6. onsider the following heat engine, where steps A and are isobaric and steps and A are adiabatic (yes, the adiabatic steps should have a slight curvature). Note: the working fluid in this engine is a monatomic ideal gas (10 moles)..0e+0.00e+0 A = =1.0x10 a 2.0E+0 = =.0x10 a ressure (a) 2.00E+0 1.0E+0 V A =0.0m V =0.20m 1.00E+0.00E+0 A V =0.10m V =0.207m A) hat is the temperature at each point A-? 0.00E+00 0 0.1 0.2 0. 0. 0. Volume (m^) V = A=81, =21, =72, and =77 nr ) ow much work is performed by the gas during each step? = p V = -2.00 and = nr = -1.6 A 2 = p V =.12 and = nr =. A 2 ) hat is the net work during the complete cycle? Net= A + + + A=-2.00-1.6 +.12 +. Net=2.79
hy 212: General hysics II hapter 20 orksheet (2 nd Law of hermodynamics & eat Engines) ) ow much heat ( ) is absorbed by the gas? in = = ncp = 7.80 E) ow much heat ( ) is discarded by the gas? out = = ncp = -.00 F) hat is the efficiency of the heat engine? 2.79 e = = = 0. 7.80 G) hat is the arnot efficiency for the engine? e = 1 - = 1 - =0.68 21 77 7. onsider the following heat engine (the tirling Engine). teps A and are isothermic and steps and A are isochoric. he temperature at ( ) is 200 and the temperature at A ( ) 00. Note: the working fluid in this 20000 engine is an ideal monatomic gas (10 moles). A) hat is the pressure at each point A-? alculate using the ideal gas law. nr = V A=8.1 =8.8 =.2 =.16 ) ow much work is performed by the gas during each step? ( )( )( ) 2 A = nr ln = 10.0 mol 8.1 mol 00 ln = -8.0 VA 0.0 ( )( )( ) = nr ln = 10.0 mol 8.1 mol 200 ln =.20 VA 0.9 = = 0 A V 0.9 V 0.0 ) hat is the net work during the complete cycle? 2 net = A + = -8.0 +.20 =.6 ) ow much heat ( in = + ) is absorbed by the gas? 200000 10000 100000 0000 in = + A=.20 + 2.9 = 2. ressure (a) 0 0.88 0.9 0.92 0.9 0.96 0.98 0. 0.02 Volum e (m ^) A
hy 212: General hysics II hapter 20 orksheet (2 nd Law of hermodynamics & eat Engines) E) ow much heat ( out = A + ) is discarded by the gas? 2 out = A + = -8.0-2.9 = -2.0 F) hat is the efficiency of this tirling engine?.6 e = = = 0.01 2. in G) hat is the arnot efficiency for this engine? e = 1 - = 1 - =0.80 00 200 ) hat is the change in entropy ( - ) during? = + = 2nRln +.20 = 10.0 mol 8.1 ln + = 202. 200 200 2 ( )( mol ) ( 00 ) I) hat is the change in entropy ( - ) during? A = A+ = + 2nRln 2-8.0 = + ( 10.0 mol)( 8.1 ) ln ( ) = -202. 00 00 2 mol 200 ) hat is the net change in entropy ( universe ) between the steps in and I? = =0 system universe tatistical Interpretation of Entropy: 8. onsider a system of 2 identical coins, with equal probability of heads or tails, respectively. A) ow many possible unique configurations are possible in this system? tot = X N - 1 = 2 2 1 = ) hat is the multiplicity of configurations () that both coins will and on tails? N! 2! = = = 1, n 1="heads" & n 2="tails" n!n! 0!2! 1 2 ) hat is the entropy of the state when they are tossed and both land on tails? -2 ( ) = k ln = 1.8 10 ln 1 = 0 ) hat does your answer in () imply about that state of the 2 dice system? e as specific as possible. All tails represents a perfectly ordered state and this corresponds to =0, the lowest possible entropy state.
hy 212: General hysics II 6 hapter 20 orksheet (2 nd Law of hermodynamics & eat Engines) 9. onsider a system of identical dice, with equal probability of landing on any side, respectively. A) ow many possible unique configurations are possible in this system? tot = 6 (yahtzee) + 0 (full house) + 0 (-of-a-kind) + 120 (-of-a-kind) + 120 (2 pairs) and 60 (1 pair) = 667 unique combinations (states) ) hat is the multiplicity of configurations () that all dice will land on the same value (i.e. a Yahtzee)? for any particular Yahtzee value: N!! = = = 1 n!n!n!n!n!n!!0!0!0!0!0! 1 2 6 ) hat is the entropy of a Yahtzee state (all the same value) for the dice system? -2 ( ) = k ln = 1.8 10 ln 1 = 0 10. In information theory, entropy (measured in bits) is used to define a lower boundary on the necessary number of numerical bit digits required to uniquely specify all possible states of an information system (e.g. unique letters of an alphabet). A) hat is the entropy of a 6 character alphabet letter (incl. 10 numbers) system? Assume that all letters and numbers are equally probable. N 6 1 1 in bits=- 1. pi ln p i =- 1. ln = 1. ln 6 =.2 bits i=1 i=1 6 6 Ans: ( ) ( ) ( ) ( ) ( ) A minimum of 6 bits is necessary to uniquely express this character system. ) hat is the entropy of this 6 character alphabet letter system, when numbers are times more probable that letters? Ans: Note, the probability of each character is: 26p L + 10 # = 26p L + 10. p L = 1 herefore: p L = 1/76 & p # = /76. 10 26 =- 1. p ln p + p ln p ( ) ( ) ( ) in bits # # L L i=1 i=1 0 26 1 in bits=-( 1.) ln + ln =.7 bits 76 76 76 76 A minimum of bits is necessary to uniquely express this character system.