University Physics Al No 12 The Second Law of Thermodynamics Class Number ame 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 (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 e-n we have 0-w24 72.7(J) 0.33 2. A real engine has an efficiency of 33%. The engine has a work output of 24J per cycle. For this T1=27 C. What can be concluded about Tp (C) A)TH=450°C.(B)TH=177°C(C)Tn>177°C(DTH3 _27+273 =175 1-33%67% 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|VⅥⅦⅧxxI 4 5 6 7 10 10 6 3 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
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 QH W ε = , we have 72.7(J) 0.33 24 = = = ε W QH . 2. A real engine has an efficiency of 33%. The engine has a work output of 24J per cycle. For this engine TL=27°C. What can be concluded about TH? （ C ） (A) TH =450°C. (B) TH =177°C (C) TH >177°C (D) TH ⇒ > 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 N1 0 1 2 3 4 5 6 7 8 9 10 N2 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, therefore the answer is(D) 5)55 5. For which of the following processes is the entropy change zero? (C) (A)Isobaric B)Isothermal ( C)Adiabatic (D)Constant volume None of these, since AS>0 for all processes Solution If these processes are quasi-static reversible, according to the definition of entropy A5 rd@ 6. One mole of an ideal gas is originally at Po, Vo, and To. The gas is heated at constant volume to 2To, then allowed to expand at constant temperature to 2Vo, and finally it is allowed to cool at constant pressure to To. The net entropy change for this ideal gas is (A)△S=(SR/2)n2(B)△S=5R/2(C)△S=Rln2(D)△S=3R2(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 80C is places into an insulated container of water originally at 25C. After a while the system reaches an equilibrium temperature of 31C. During this process (A)△Saam (B)△ satuminunn (C)△ saluinun0 The total entropy change of the block of aluminum is As 304K mcalumingm dT <0J. the answer 353K should be(C) 8. Which of the following is a consequence of the second law of thermodynamics? 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) I. Filling the blanks
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 P0, V0, and T0. The gas is heated at constant volume to 2T0, then allowed to expand at constant temperature to 2V0, and finally it is allowed to cool at constant pressure to T0. 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) ∆Saluminum>0. (B) ∆Saluminum=0. (C) ∆Saluminum<0. Solution: The total entropy change of the block of aluminum is 0 J 304K d 353K min ∫ ∆ = < T mc T S alu um , 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_ 18711J △S=→Q=△ST=46.2×(273+132)=18711(0) 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 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. T(K) Fig1 It is settled by the initial and final states 3. For the Carnot cycle shown in Fig. 2, the heat that enters 200F is_ J, the work done on the system is_ 22.5 J Solutio 0.6 According to the entropy change 4Ss O 1g2 We have O= AsT. So the heat enters is Q=Qn=(0.6-0.2)×400=160() According to the first law of thermodynamics, the work done on the system is 250 W=a=( 160=60J 4. A Carnot heat engine has an efficiency of 0.300 operating between a high-temperature reservoir at temperature TH and a low-temperature reservoir at 20C. To increase the efficiency of the Carnot heat engine to 0.400, the temperature of the hot reservoir be increased is 69.76K (in Kelvin) Solution The efficiency of the Carnot heat engine is e=l T273+20 =41857(K) 1-0.3 =48833(K) T E21-0.4
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 Q ST 46.2 (273 132) 18711 (J ) 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 haveQ = ∆ST . So the heat enters is = = (0.6 − 0.2) × 400 = 160(J) Q QH According to the first law of thermodynamics, the work done on the system is ) 160 60J 400 250 W = εQ = (1− × = 4. A Carnot heat engine has an efficiency of 0.300 operating between a high-temperature reservoir at temperature TH 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 ε =1− 418.57(K) 1 0.3 273 20 1 1 1 1 1 1 = − + = − = − ⇒ = ε ε C H H C T T T T 488.33(K) 1 0.4 273 20 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 Thn2-Thn1=48833-41857=6976(K) 5. One(1.00)kilogram of steam at 100C condenses into water at 100C. The entropy change of the mass is-061x10J/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 As=O/T, since the heat is negative. the entropy is negative too. Solution The entropy change of the mass is As- mL 1x 22.57x105 -061×104(J/K 273+100 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_ 7.17J/K Solution entropy change △S= do r nC, dT 3R,40 =nC In-2=2xIn=7. 17(J/K) T T 300 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_400 Solution The efficiency of the heat engine is E=/h so the heat flow to the heat engine from the Q W1000 high-temperature reservoir is Q E025=4000 8. An ideal gas undergoes an isothermal expansion at 77C increasing its volume from 1.3 to 3.4 L The entropy of the gas is 24 J/K. The quantity of gas present is 3mole (in mole) Solution For isothermal process 45s rdo 1 ∫= -niTIn=nRln 24 antity of gas present is n 3(mole) RIn 8.315×ln 1.3 II Give the Solutions of the following problems In Carnot cycle, the isothermal expansion of an ideal gas takes place at 4 12K and the isothermal
Then we have 488.33 418.57 69.76 (K) 2 1 TH −TH = − = 5. One (1.00) kilogram of steam at 100 °C condenses into water at 100 °C. The entropy change of the mass is -0.61×104 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 0.61 10 (J/K) 273 100 1 22.57 10 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 7.17(J/K) 300 400 ln 2 3 ln 2 d d 1 2 2 1 ∆ = = = = × = ∫ ∫ R T T nC T nC T 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 QH W ε = , so the heat flow to the heat engine from the high-temperature reservoir is 4000 (J) 0.25 1000 = = = ε W QH 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 ln ln 1 d d 1 V V nR V V nRT T Q T T Q ∆S = = = = ∫ ∫ So the quantity of gas present is 3(mole) 1.3 3.4 8.315 ln 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 sion,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 th isothermal compression (a)According to the first law of thermodynamics, the work performed by the gas during the isothermal expansion is W=0=2090J (b) The efficiency of the Carnot heat engine is a so the heat rejected from the gas during the isothermal compression g|==00207-1506(0 According to the first law of thermodynamics, the work done on the gas during the isothermal compression is w= =150663J 2. One(1.00) mole of an ideal diatomic gas(with r 1.40)initially at 20.0C and 1.00 atm pressure is taken Pressure(atm) around the following cycle(see Figure 3) R Isothermal process Path(1): an isochoric increase in pressure until the temperature of the gas is 150.C and the pressure is 1.00 2 P Path(): an isothermal expansion until the pressure Volume returns to 1. 00 atm Path ( 3): an isobaric compression until the gas Fig 3 ginal 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
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 W = Q = 2090 J. (b) The efficiency of the Carnot heat engine is H C H C H T T Q Q Q W ε = =1− =1− so the heat rejected from the gas during the isothermal compression is 1506.63 (J) 412 297 = ⋅ = 2090× = H C C H T T Q Q According to the first law of thermodynamics, the work done on the gas during the isothermal compression is = =1506.63 J W QC 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) Volume Isothermal process B C
Since y=-= i+r i+1-14, so we get F=2.5 (a) Using the equation of state for an ideal gas: PV=nRT We have the original volume of the gas at the beginning and end of the cycle h=R711×8315×(27315+20)=241×10(m2) P 1×105×1013 b) For isochoric process P Pa The pressure of the gas at the completion of path(1)is P=P2=1×1013027315+150=146×0(a) 273.15+20 For isothermal process: PBB=PVC The volume of the gas at the completion of path(2) P1.46×10°×241×102 1×105×1013 =347×10-(m) P The temperature of the gas at the completion of path(2)is TB=273+150=423K (c)(1)The work Path 1: WI=OJ Path2:W2= RTIn -c=8315×(273+150)×n 3.47×10 2.41×10 2=1282.13(J) Path 3 W3=-PV-V)=-1×1013×103×(347×102-241×10-2)=-107378(J) The total work done by the gas Wu=W+W2+W=0+128213-1073.78=208.35(J) (2)The heat Path1:g1=-C1(T-7A)=1×2.5×8.315×(423-293)=2702.38(J)
Since 1.4 1 2 2 1 = + = + = = i i Ri R i C C V P γ , so we get i=2.5 (a) Using the equation of state for an ideal gas: PV = nRT We have the original volume of the gas at the beginning and end of the cycle 2.41 10 (m ) 1 10 1.013 1 8.315 (273.15 20) 2 3 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 1.46 10 (Pa) 273.15 20 1 1.013 10 (273.15 150) 5 5 = × + × × × + = = A A B B T P T P For isothermal process: PBVB = PCVC The volume of the gas at the completion of path (2) is 3.47 10 (m ) 1 10 1.013 1.46 10 2.41 10 2 3 5 5 2 − − = × × × × × × = = C B B C P P V V The temperature of the gas at the completion of path (2) is TB = 273 +150 = 423K (c) (1) The work Path 1: 0J W1 = Path 2: 1282.13 (J) 2.41 10 3.47 10 ln 8.315 (273 150) ln 2 2 2 = × × = = × + × − − B C V V W RT Path 3: ( ) 1 1.013 10 (3.47 10 2.41 10 ) 1073.78(J) 5 2 2 3 = − − = − × × × × − × = − − − W PA VC VA The total work done by the gas is 0 1282.13 1073.78 208.35 (J) Wtotal = W1 +W2 +W3 = + − = (2) The heat Path 1: ( ) 1 2.5 8.315 (423 293) 2702.38(J) 1 = CV TB − TA = × × × − = M Q µ
347×10-2 Path2:Q2= nRT In=1×8.315×423×ln 241×10-2 282.13(J) Pah3:g3=nCp(T4-Tc)=1×3.5×8.315×(293-423)=-378333J) The total heat transfer to the gas over the cycle is Qoma=Q+Q2+93=270238+128213-378333=201.18(J) (d)The efficiency of the cycle W 208.35 Qn"+Q22702.38+128250005 (e)T,=293K(coldest); T:=423(hottest); TC=423K The maximum efficiency that a heat engine is the carnot engine's efficiency, that is g=1-=1-293=0.3 423 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 4V) Assume that P =3P a. Pbl (a)Determine the pressure and temperature of each of the SparktAdiabatic vertex points of the Pl diagram in terms of Pa and Ta.(b) Calculate the efficiency of the cycle Intake Solution Adiabatic For the ideal diatomic gas: F=5, so r=2*\:1.2 Fig 4 a)For the isochoric process a-b: Pb=3Pa T P 3P T=D Ta T=37 P For the adiabatic process b->c: s=Va=4/ P=P→P=P()=3P()=3P()2=0.57P x-=r-7==(1ny-=()23x=227 For the isochoric process c>d: T=T
Path 2: 1282.13(J) 2.41 10 3.47 10 ln 1 8.315 423 ln 2 2 2 = × × = = × × × − − B C B V V Q nRT Path 3: ( ) 1 3.5 8.315 (293 423) 3783.33(J) Q3 = nCP TA − TC = × × × − = − The total heat transfer to the gas over the cycle is 2702.38 1282.13 3783.33 201.18(J) Qtotal = Q1 + Q2 + Q3 = + − = (d) The efficiency of the cycle is 0.05 2702.38 1282.59 208.35 1 2 = + = + = = Q Q W Q W total H total ε (e) TA = 293K(coldest); TB = 423(hottest); TC = 423K The maximum efficiency that a heat engine is the carnort engine’s efficiency, that is 0.31 423 293 =1− =1− = 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 (Vd=4Va). Assume that Pb=3Pa. (a) Determine the pressure and temperature of each of the vertex points of the PV diagram in terms of Pa and Ta. (b) Calculate the efficiency of the cycle. Solution: For the ideal diatomic gas: i=5, so 1.2 1 = + = i i γ (a) For the isochoric process a→b : Pb=3Pa a a a a a a b b a b a b T T P P T P P T P P T T 3 3 = ⇒ = ⋅ = ⋅ = For the adiabatic process b→c : Vc = Vd = 4Va a a d a a c b b b c c c b P P V V P V V PV PV P P ) 0.57 4 1 ( ) 3 ( ) 3 ( 1.2 = ⇒ = = = = γ γ γ γ b a a c b b b c c c T T T V V V T V T T ) 3 2.27 4 1 ( ) ( 1 1 1 1.2 1 = ⇒ = = = γ − γ − γ − − For the isochoric process c→d : Td = Ta P V Fig.4 Pb Va Vd a b c Adiabatic Adiabatic Intake Spark d
T P P 分≈T.0.57P=0257 2.27T (b) For the isochoric process a->b Q=nCr(T-T)=nCr (3T4-T,)=2nCpT For the isochoric process c->d Q2=nC(T-7)=nCV(T-2277)=-1.27nCT The efficiency of the cycle 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 ()to 2PoF---------. point(2)? and the work done by the gas in going from point(2)to (b)What is the change in the internal energy of the gas in going from PoF---i nt(1) to point3) along the path(1)→(2)→(3)? (c)What is the entropy change of the gas in going form point(1)to 2 point (3)? Fig 5 (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 W=Po(2Vo-V)=PoL 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: PV=nRT The temperature at point(I)is T- Pv, Pvo R R The temperature at point(3)is T, p=2P-2=4PVo R R R The change in the internal energy of the gas in going from point(1)to point (3)along the path(1) →(2)→(3)is AE=nC(T-T)=ir 4P._Po0)=PV=Poo R R 2
a a a a c c d d c d c d P T T T P T T P P P T T 0.57 0.25 2.27 = ⇒ = ⋅ = ⋅ = (b) For the isochoric process a→b : Q nCV Tb Ta nCV Ta Ta 2nCVTa ( ) (3 ) 1 = − = − = For the isochoric process c→d : Q nCV Td Tc nCV Ta Ta 27nCVTa ( ) ( 2.27 ) 1. 2 = − = − = − The efficiency of the cycle is 0.365 2 1.27 1 1 1 2 = − = − = − v a v a nC T nC T 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 1 0 0 0 0 0 W = P (2V −V ) = PV 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: PV = nRT The temperature at point (1) is R PV R PV T 1 1 0 0 1 = = The temperature at point (3) is R PV R P V R PV T 3 3 0 0 0 0 3 2 2 4 = ⋅ = = The change in the internal energy of the gas in going from point (1) to point (3) along the path (1) → (2) → (3) is 0 0 0 0 0 0 0 0 3 1 2 3 2 ) 4 ( 2 ( ) PV PV i R PV R iR PV ∆E = nCV T − T = − = = 2V0 2P0 1 2 3 Fig.5 P0 P V V0
(c)Using the equation: ds= do CdT+Pdv The entropy change of the gas in going form point (1)to point (3)is AS=l dS 厂+y一cm+m 4Po 3R +RIn_3R In 4+ rin 2 R 8315×(1.5×1.37+0.69)=2282(J/K) 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
(c) Using the equation: T C T P V T Q S d V d d d + = = The entropy change of the gas in going form point (1) to point (3) is 8.315 (1.5 1.37 0.69) 22.82(J/K) ln 4 ln 2 2 2 3 ln 4 ln 2 3 ln ln d d d 0 0 0 0 0 0 1 3 1 3 3 1 3 1 3 1 = × × + = = + = + ∆ = = + = + ∫ ∫ ∫ R R V V R R PV R PV R V V R T T C V V R T C T S S 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
