University Physics AI No. 12 The Second Law of Thermodynamics Class Number Name I．Choose the Correct Answer 1. A real engine has an efficiency of 33%. The engine has a work output of 24J per cycle. How much heat energy is extracted from the high- temperature reservoir per cycle? ( D ) (A) 8J (B) 16J (C) 48J (D) 72J (E) The question can be answered only of the engine is a Carnot engine. Solution: Using the definition of the efficiency H Q W =ε , we have )J(7.72 33.0 24 === ε W Q H . 2. A real engine has an efficiency of 33%. The engine has a work output of 24J per cycle. For this engine T L =27°C. What can be concluded about T H ? （ C ） (A) T H =450°C. (B) T H =177°C (C) T H >177°C (D) T H <177°C. (E) 177°C <T H <450°C. Solution: Since real heat engines have efficiencies less than a Carnot heat engine operating between the same two temperatures. The efficiency of the Carnot engine is H C T T ?=1ε , so we have o 175 %67 27327 %331 %331 = + = ? >?>?= C H H C T T T T ε 3. Ten identical particles are to be divided up into two different containers. How many different configurations are possible? （ B ） (A) 1 (B) 11 (C) 120 (D) 1024 (E) 3628800 Solution: The possible configurations are configuration I Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ Ⅶ Ⅷ Ⅸ Ⅹ ⅩI N 1 0 1 2 3 4 5 6 7 8 9 10 N 2 10 9 8 7 6 5 4 3 2 1 0 4. Ten identical particles are to be divided up into two containers. Which configuration has the largest number of microstates? （ D ） (A) 0, 10 (B) 3, 7 (C) 4, 6 (D) 5, 5 Solution: The most probable state is the ten particles divided into two equal parts, in this situation, the number of microstates is largest, 252 !5!5 !10 5 10 == ? ? ? ? ? ? ? ? ，therefore the answer is (D). 5. For which of the following processes is the entropy change zero? （ C ） (A) Isobaric (B) Isothermal (C) Adiabatic (D) Constant volume (E) None of these, since ?S>0 for all processes. Solution: If these processes are quasi-static reversible, according to the definition of entropy ∫ =? T Q S d , the answer is (C). 6. One mole of an ideal gas is originally at P 0 , V 0 , and T 0 . The gas is heated at constant volume to 2T 0 , then allowed to expand at constant temperature to 2V 0 , and finally it is allowed to cool at constant pressure to T 0 . The net entropy change for this ideal gas is （ E ） (A) ?S=(5R/2)ln2 (B) ?S=5R/2 (C) ?S=Rln2 (D) ?S=3R/2 (E) ?S=0 Solution: For the entropy is a state variable of a thermodynamic system, the net entropy change of the ideal gas is (E). 7. A block of aluminum originally at 80°C is places into an insulated container of water originally at 25°C. After a while the system reaches an equilibrium temperature of 31°C. During this process （ C ） (A) ?S aluminum >0. (B) ?S aluminum =0. (C) ?S aluminum <0. Solution: The total entropy change of the block of aluminum is J0 d K304 K353 min ∫ <=? T Tmc S umalu , the answer should be (C). 8. Which of the following is a consequence of the second law of thermodynamics? ( C ) (A) Heat can flow only from high temperature to low temperature. (B) Objects in contact will tend toward having the same temperature. (C) Any system that produces order from disorder must have an external influence. Solution: The second law of thermodynamics tells that if there is no external effect, an isolated system will always change from order to disorder. The answer is (C) II. Filling the Blanks 1. An ideal gas undergoes a reversible isothermal expansion at 132°C. The entropy of the gas increases by 46.2 J/K. The heat absorbed is 18711 J . Solution )(18711)132273(2.46 JSTQ T Q S =+×=?=?=? 2. In Fig. 1, suppose that the change in entropy of the system in passing from state a to state b along path 1 is +0.60 J/K. The entropy change in passing from state a to b along path 2 is 0.6 J/K , and the entropy change in passing from state b to a along path 2 is -0.6 J/K . Solution: The entropy is a state variable of a thermodynamic system. It is settled by the initial and final states. 3. For the Carnot cycle shown in Fig.2, the heat that enters is 60 J , the work done on the system is 22.5 J . Solution: According to the entropy change T Q S =? , We have STQ ?= . So the heat enters is )J(160400)2.06.0( =×?== H QQ According to the first law of thermodynamics, the work done on the system is J60160) 400 250 1( =×?== QW ε 4. A Carnot heat engine has an efficiency of 0.300 operating between a high-temperature reservoir at temperature T H and a low-temperature reservoir at 20 °C. To increase the efficiency of the Carnot heat engine to 0.400, the temperature of the hot reservoir be increased is 69.76 K (in Kelvin). Solution: The efficiency of the Carnot heat engine is H C T T ?=1ε )K(57.418 3.01 20273 1 1 1 1 1 1 = ? + = ? =??= ε ε C H H C T T T T )K(33.488 4.01 20273 1 1 2 2 2 2 = ? + = ? =??= ε ε C H H C T T T T 1 P a b Fig.1 2 0 S(J/K) T(K) Fig.2 0.2 0.4 0.6 100 200 300 400 Then we have )K(76.6957.41833.488 12 =?=? HH TT 5. One (1.00) kilogram of steam at 100 °C condenses into water at 100 °C. The entropy change of the mass is -0.61×10 4 J/K . Does your result violate the second law of thermodynamics? Explain your answer. No. For a system which is not isolated, the entropy change of the system could be less than zero. According to the entropy change ?S=Q/T , since the heat is negative, the entropy is negative too. Solution: The entropy change of the mass is )J/K(1061.0 100273 1057.221 4 5 ×?= + ×× ?==? T mL S 6. Two (2.00) moles of a monatomic ideal gas is warmed slowly from 300 K to 400 K at constant volume. The entropy change of the gas is 7.17 J/K . Solution: The entropy change of the gas is )J/K(17.7 300 400 ln 2 3 2ln dd 1 2 2 1 =×====? ∫∫ R T T nC T TnC T Q S v T T v 7. A heat engine with an efficiency of 0.25 operates between two reservoirs at 1000K and 400K. The power output of the heat engine is 1.00kW. The heat flow to the heat engine from the high-temperature reservoir is 4000 J . Solution: The efficiency of the heat engine is H Q W =ε , so the heat flow to the heat engine from the high-temperature reservoir is )J(4000 25.0 1000 === ε W Q H 8. An ideal gas undergoes an isothermal expansion at 77°C increasing its volume from 1.3 to 3.4 L. The entropy change of the gas is 24 J/K. The quantity of gas present is 3mole (in mole). Solution For isothermal process 1 2 1 2 lnln 1 d 1d V V nR V V nRT T Q TT Q S ====? ∫∫ So the quantity of gas present is )mole(3 3.1 4.3 ln315.8 24 ln 1 2 = × = ? = V V R S n III. Give the Solutions of the Following Problems 1. In Carnot cycle, the isothermal expansion of an ideal gas takes place at 412K and the isothermal compression at 297K. During the expansion, 2090 J of heat energy are transferred to the gas. Determine (a) the work performed by the gas during the isothermal expansion, (b) the heat rejected from the gas during the isothermal compression, and (c) the work done on the gas during the isothermal compression. Solution: (a) According to the first law of thermodynamics, the work performed by the gas during the isothermal expansion is J2090== QW . (b) The efficiency of the Carnot heat engine is H C H C H T T Q Q Q W ?=?== 11ε so the heat rejected from the gas during the isothermal compression is )J(63.1506 412 297 2090 =×=?= H C HC T T QQ According to the first law of thermodynamics, the work done on the gas during the isothermal compression is J63.1506== C QW 2. One (1.00) mole of an ideal diatomic gas (with γ = 1.40) initially at 20.0 °C and 1.00 atm pressure is taken around the following cycle (see Figure 3). Path (1): an isochoric increase in pressure until the temperature of the gas is 150 °C and the pressure is P; Path (2): an isothermal expansion until the pressure returns to 1.00 atm. Path (3): an isobaric compression until the gas reaches its original volume. (a) What is the original volume of the gas at the beginning and end of the cycle? (b) What is the pressure of the gas at the completion of path (1)? the volume of the gas at the completion of path (2)? And the temperature of the gas at the completion of path (2)? (c)Calculate the work done by the gas during each path of the cycle and the total work done by the gas. And the heat transfer to the gas during each path of the cycle and the total heat transfer to the gas over the cycle. (d) Find the efficiency of the cycle. (e) Calculate the maximum efficiency that a heat engine could have if it operated between the hottest and coldest temperature encountered by this gas in this cycle. Solution: A V P 1 2 3 Fig.3 1.00 Pressure (atm) Vo l u me Isothermal process B C Since 4.1 1 2 2 1 = + = + == i i R i R i C C V P γ , so we get i=2.5 (a) Using the equation of state for an ideal gas: nRTPV = We have the original volume of the gas at the beginning and end of the cycle )m(1041.2 013.1101 )2015.273(315.81 32 5 ? ×= ×× +×× == A A A P nRT V (b) For isochoric process: B B A A T P T P = The pressure of the gas at the completion of path (1) is )Pa(1046.1 2015.273 )15015.273(10013.11 5 5 ×= + +××× == A BA B T TP P For isothermal process: CCBB VPVP = The volume of the gas at the completion of path (2) is )m(1047.3 013.1101 1041.21046.1 32 5 25 ? ? ×= ×× ××× == C BB C P VP V The temperature of the gas at the completion of path (2) is K423150273 =+= B T (c) (1) The work Path 1: J0 1 =W Path 2: )J(13.1282 1041.2 1047.3 ln)150273(315.8ln 2 2 2 = × × ×+×== ? ? B C V V RTW Path 3: )J(78.1073)1041.21047.3(10013.11)( 225 3 ?=×?××××?=??= ?? ACA VVPW The total work done by the gas is )J(35.20878.107313.12820 321 =?+=++= WWWW total (2) The heat Path 1: )J(38.2702)293423(315.85.21)( 1 =?×××=?= ABV TTC M Q μ Path 2: )J(13.1282 1041.2 1047.3 ln423315.81ln 2 2 2 = × × ×××== ? ? B C B V V nRTQ Path 3: )J(33.3783)423293(315.85.31)( 3 ?=?×××=?= CAP TTnCQ The total heat transfer to the gas over the cycle is )J(18.20133.378313.128238.2702 321 =?+=++= QQQQ total (d) The efficiency of the cycle is 05.0 59.128238.2702 35.208 21 = + = + == QQ W Q W total H total ε (e) K423(hottest);423;K(coldest)293 === CBA TTT The maximum efficiency that a heat engine is the carnort engine’s efficiency, that is 31.0 423 293 11 =?=?= H C T T ε 3. A gasoline internal combustion engine can be approximated by the cycle shown in Fig.4. Assume an ideal diatomic gas and use a compression ratio of 4:1 (V d =4V a ). Assume that P b =3P a . (a) Determine the pressure and temperature of each of the vertex points of the PV diagram in terms of P a and T a . (b) Calculate the efficiency of the cycle. Solution: For the ideal diatomic gas: i=5, so 2.1 1 = + = i i γ (a) For the isochoric process a→b : P b =3P a aa a a a a b b a b a b TT P P T P P T P P T T 3 3 =?=?=?= For the adiabatic process b→c : V c = V d = 4V a aa d a a c b bcccbb PP V V P V V PPVPVP 57.0) 4 1 (3)(3)( 2.1 ====?= γγγγ aab c b cccbb TTT V V TTVTV 27.23) 4 1 ()( 12.1111 ===?= ???? γγγ For the isochoric process c→d : ad TT = P V Fig.4 P b V a V d a b c Adiabatic Adiabatic Intake Spark d aa a a c c d d c d c d TP T T P T T P P P T T 25.057.0 27.2 =?=?=?= (b) For the isochoric process a→b : aVaaVabV TnCTTnCTTnCQ 2)3()( 1 =?=?= For the isochoric process c→d : aVaaVcdV TnCTTnCTTnCQ 27.1)27.2()( 2 ?=?=?= The efficiency of the cycle is 365.0 2 27.1 11 1 2 = ? ?=?= av av TnC TnC Q Q η 4. One (1.00) mole of an ideal monatomic gas is taken around the cycle shown in Figure 5. (a) What is the work done by the gas in going from point (1) to point (2)? and the work done by the gas in going from point (2) to point (3)? (b)What is the change in the internal energy of the gas in going from point (1) to point (3) along the path (1) → (2) → (3)? (c) What is the entropy change of the gas in going form point (1) to point (3)? (d) What is the change in the internal energy and entropy of the gas in one complete cycle? Solution: (a) The work done by the gas in going from point (1) to point (2) is the area under the curve (1) → (2). That is 000001 )2( VPVVPW =?= The work done by the gas in going from point (2) to point (3) is zero. (b) Using the equation of state for an ideal gas: nRTPV = The temperature at point (1) is R VP R VP T 0011 1 == The temperature at point (3) is R VP R VP R VP T 000033 3 422 = ? == The change in the internal energy of the gas in going from point (1) to point (3) along the path (1) → (2) → (3) is 0000 0000 13 2 3 2 ) 4 ( 2 )( VPVP i R VP R VPiR TTnCE V ==?=?=? 2V 0 2P 0 1 2 3 Fig.5 P 0 P V V 0 (c) Using the equation: T VPTC T Q S V ddd d + == The entropy change of the gas in going form point (1) to point (3) is )J/K(82.22)69.037.15.1(315.8 2ln4ln 2 32 ln 4 ln 2 3 lnln dd d 0 0 00 00 1 3 1 3 3 1 3 1 3 1 =+××= +=+= +=+==? ∫∫∫ R R V V R R VP R VP R V V R T T C V V R T TC SS V V V T T V (d) The entropy change and the internal energy change of the gas in going form point (1) to point (3) are zero, since they are state variables.