University Physics Al No 10 The First Law of Thermodynamics Class Number ame I. Choose the Correct Answer 1. Which of the following processes must violate the first law of thermodynamics?(There may be more than one answer!) A.B. D (A)w>0, 00,00 (C)w>0,00, and AEint 0,0>0, and AEint <0 Solution: Applying the first law of thermodynamics =AE +w 2. In which of the paths between initial state i and final statef in p Fig. I is the work done on the gas the greatest? Solution: The work done by the gas is the area under the path, and B the work done on the gas is the negative of the work done by the Fig 1 3. Consider the following processes that can be done on an ideal gas constant volume, AV=0; constant pressure, AP=0; and constant temperature, AT=O (a) For which process does W=0? (A) (b) For which process does =0 (D) (c)For which process does W+2-0? (d) For which process does AEint =Q (e) For which of these processes does AEint=w? (D) (B)△ (C)△T=0 (D)Non of these Solution: Applying the first law of thermodynamics 4. Which of the following processes is forbidden by the first law of thermodynamics?(There may be more than one correct answer! A)An ice cube is placed in hot coffee; the ice gets colder and the coffee gets hotter ( B)Solid wax is placed in hot metal pan; the wax melts and the metal pan cools (C)Cold water is placed in a cold glass; the glass gets colder and the water gets colder (D)A student builds an automobile engine that converts into work the heat energy released when water changed to ice (E) Dry ice can be made by allowing carbon dioxide gas to expand in a bag. Solution: Only (C) violates the first law of thermodynamics
University Physics AI No. 10 The First Law of Thermodynamics Class Number Name I．Choose the Correct Answer 1. Which of the following processes must violate the first law of thermodynamics? (There may be more than one answer!) （ A,B,D ） (A) W > 0, Q 0, Q 0 (C) W > 0, Q 0, and ∆Eint 0, Q > 0, and ∆Eint < 0 Solution: Applying the first law of thermodynamics Q = ∆E +W . 2. In which of the paths between initial state i and final state f in Fig.1 is the work done on the gas the greatest? ( D ) Solution: The work done by the gas is the area under the path, and the work done on the gas is the negative of the work done by the gas. 3. Consider the following processes that can be done on an ideal gas: constant volume, ∆V =0; constant pressure, ∆P =0; and constant temperature, ∆T =0. (a) For which process does W=0? ( A ) (b) For which process does Q=0? ( D ) (c) For which process does W+Q=0? ( D ) (d) For which process does ∆Eint = Q? ( A ) (e) For which of these processes does ∆Eint = W? ( D ） (A) ∆V =0 (B) ∆P =0 (C) ∆T =0 (D) Non of these Solution: Applying the first law of thermodynamics. 4. Which of the following processes is forbidden by the first law of thermodynamics? (There may be more than one correct answer!) （C） (A) An ice cube is placed in hot coffee; the ice gets colder and the coffee gets hotter. (B) Solid wax is placed in hot metal pan; the wax melts and the metal pan cools. (C) Cold water is placed in a cold glass; the glass gets colder and the water gets colder. (D) A student builds an automobile engine that converts into work the heat energy released when water changed to ice. (E) Dry ice can be made by allowing carbon dioxide gas to expand in a bag. Solution: Only （C）violates the first law of thermodynamics.. i f V P A B C D Fig.1
I. Filling the blanks 1. A.500 kg block of ice at-100C is placed in a punch bowl with 4.00 kg of water(punch)at 20.0C. Assume no heat transfer to the surroundings and neglect the thermal effects of the punch bowl itself. Does all the ice melt? yes. If not, the quantity which is left is final temperature of the punch plus ice system is_281.53 K To warm the ice from 10C to ooC takes a heat transfer to the ice system Q1=maCa△T=0.5×2050×(0+10)=1.025×10°(J) If all the ice melts. it need heat transfer Q2=maLr=0.5×3335×10=16675×10°(J) Let the water(punch) changes from 20.C to oC, the heat transfer is Q3= m C△T=4×4186×(0-20)=-3:3488×10(J) So Q1+02<o3 we can know all the ice will melt As Q1+02+03<O So the system's final temperature will higher than 0oC Assume that the final temperature is T of the punch plus ice system Q,+02+O,=(mater +mice )Cwater At=(mater + mice )C ater .(T-O 7=-9+g2+g (meater +m)c 3.3488×105-1025×104-16675×105 =8.38(C)=281.53(K) 4168×(0.5+4) 2. A typical household water heater has a capacity of about 0.50 m of water. The water is initially at a temperature of 15C and is warmed to 60C for the pleasure of long, hot showers. The quantity of the heat transfer which is necessary to warm the water is 9.42x10'J If the heating element is rated at 5.0 kW. the time it take to warm the water is 5.2 h Solution (a)The quantity of the heat transfer which is necessary to warm the water is Cwa△T= Pote veater c△T =1×10×0.5×4186×(60-15)=942×10(J) (b)∵O=PMt
II. Filling the Blanks 1. A 0.500 kg block of ice at -10.0 °C is placed in a punch bowl with 4.00 kg of water (punch) at 20.0 °C. Assume no heat transfer to the surroundings and neglect the thermal effects of the punch bowl itself. Does all the ice melt? yes . If not, the quantity which is left is . The final temperature of the punch plus ice system is 281.53 K . Solution: To warm the ice from 10°C to 0°C takes a heat transfer to the ice system 0.5 2050 (0 10) 1.025 10 (J) 4 Q1 = miceCice∆T = × × + = × If all the ice melts, it need heat transfer 0.5 3.335 10 1.6675 10 (J) 5 5 Q2 = miceLf = × × = × Let the water (punch) changes from 20°C to 0°C, the heat transfer is 4 4186 (0 20) 3.3488 10 (J) 5 Q3 = mwaterCwater∆T = × × − = − × So Q1 + Q2 < Q3 we can know all the ice will melt. As Q1 + Q2 + Q3 < 0 So the system’s final temperature will higher than 0°C. Assume that the final temperature is T of the punch plus ice system. ( ) ( ) ( 0) Q1 + Q2 + Q3 = mwater + mice Cwater∆T = mwater + mice Cwater ⋅ T − 8.38( C) 281.53(K) 4168 (0.5 4) 3.3488 10 1.025 10 1.6675 10 ( ) 5 4 5 1 2 3 = = × + × − × − × = + + + = − o mwater mice Cwater Q Q Q T 2. A typical household water heater has a capacity of about 0.50 m3 of water. The water is initially at a temperature of 15 °C and is warmed to 60 °C for the pleasure of long, hot showers. The quantity of the heat transfer which is necessary to warm the water is 9.42×107 J . If the heating element is rated at 5.0 kW, the time it take to warm the water is 5.2 h . Solution: (a) The quantity of the heat transfer which is necessary to warm the water is 1 10 0.5 4186 (60 15) 9.42 10 (J) 3 7 = × × × × − = × Q = mwaterCwater∆T = ρ waterVwaterCwater∆T (b) QQ = P∆t
A=g=942×10=18 10°(s)=52(h) 3. Two reservoirs, one at 0.0C and the other at 1000C, are connected by means of two materials as indicated in Figure 2. The relative R-values of the materials are specified. The temperature at the interface between the two materials K Assume that the temperature is T at the interface between the two I Hot reserv materials R For A R Cold reservior Fig So we have T-7,T-7c R 3R r=3Rn+Rrc=3n+c=3×(27315+100+27315 =348.15(K) R+3R 4. A gas is taken from thermal equilibrium state l to P(atm) state 2 on a P-l diagram via the semicircular route indicated in Figure 3. The work done by the gas in going from state I to state 2 is 2.61x10 J 6.00 The work done by the gas if it goes from state 2 ack to state 1, retracting the semicircular path is 2.00 v(liters) 1000 (a) The work is equal to the area under the semicircular route W=6×1013×10×(10-1)×10-2(4×1013×10510-1×0)=261×10(0) (b)W=-W=-261×103(J) P 5. Gas within a chamber passes through the cycle shown in Fig 4. The net heat added to the gas during process CA if OaB-20J, @Bc=0 and WBCA 15J is--5J Solution Fig 4
18.84 10 (s) 5.2(h) 5 10 9.42 10 3 3 7 = × = × × ∴∆ = = P Q t 3. Two reservoirs, one at 0.0 °C and the other at 100.0 °C, are connected by means of two materials as indicated in Figure 2. The relative R-values of the materials are specified. The temperature at the interface between the two materials is 348.15 K . Solution: Assume that the temperature is T at the interface between the two materials. For R T T A t Q R T T A t Q H C 3 ) d d ) , ( d d ( 1 2 − = − = and 1 2 ) d d ) ( d d ( t Q t Q = So we have 348.15(K) 4 3 (273.15 100) 273.15 4 3 3 3 3 = × + + = + = + + ⇒ = − = − H C H C H C T T R R RT RT T R T T A R T T A 4. A gas is taken from thermal equilibrium state 1 to state 2 on a P-V diagram via the semicircular route indicated in Figure 3. The work done by the gas in going from state 1 to state 2 is 2.61×103 J . The work done by the gas if it goes from state 2 back to state 1, retracting the semicircular path is -2.61×103 J . Solution: (a) The work is equal to the area under the semicircular route 10 ) 2.61 10 (J) 2 10 1 (4 1.013 10 2 6 1.013 10 (10 1) 10 5 3 5 3 3 × = × − = × × × − × − × × × − π − W (b) 2.61 10 (J) 3 W ′ = −W = − × 5. Gas within a chamber passes through the cycle shown in Fig.4. The net heat added to the gas during process CA if QAB=20J, QBC = 0 and WBCA= 15J is -5J . Solution: V P A B C Fig.4 0 V(liters) P(atm) 0 Fig.3 1.00 10.00 2.00 6.00 Hot reservior Cold reservior 3R R Fig.2
OB=AUAB, CBC=AUBC+WBC, Cc=AUcA+Wo For the cycle,△U=△UAB+△UBC+△Uc=0 Therefore, QB+ OBC+Oc=y O,,=15-20=-5J 6. Gas within a chamber undergoes the processes shown in the Pl diagram of Fig. 5. The net heat added to the system during one complete cycle is-3.58x10J Solution P( Mpa) The work during one complete cycle is equal to the area of the semicircular route on the diagram W=-2[(30-15)×1013×103× 4-1 Applying the first law of thermodynamics: =AU +w V(L) and AU =0, so the net heat added to the system during Fig 5 one complete cycle is Q=W=-3.58x10J III. Give the Solutions of the following problems 1. When a system is taken from state i to state f along the path iaf in p Fig 6, it is found that 2-50J and W=20J. Along the path ibf, 2-36J (a)What is W along the path ibf?(b)If w=-13J for the curved return path fi, what is Q for this path?(c)Take Eint. i10J. What is Eint,/ ?(d)If Eint. b-22J, find for process ib and prosess bf. Solution (a) Applying the first law of thermodynamics: Q=A+w so A0=U-U,=Q-w, that is Oia -Wia=Oib -Wib Thus Wh=Ou +Wor-O =36+20-50=6) ( b) The same way as(a), we have Oiat-Wiaf=@r-Wa SoQ=Cax-W+Wh=50-20-13=17(J) (c)Applying the first law of thermodynamics: Q=AU+w,so
QAB = ∆UAB QBC = ∆UBC +WBC QCA = ∆UCA +WCA , , For the cycle, ∆U = ∆UAB + ∆UBC + ∆UCA = 0 , Therefore, QAB + QBC + QCA = WBCA = − =15 − 20 = −5J QCA WBCA QAB 6. Gas within a chamber undergoes the processes shown in the PV diagram of Fig. 5. The net heat added to the system during one complete cycle is -3.58×103 J . Solution: The work during one complete cycle is equal to the area of the semicircular route on the diagram: 10 ] 3.58 10 (J) 2 4 1 [(30 15) 1.013 10 2 5 3 3 × = − × − = − − × × × π − W Applying the first law of thermodynamics: Q = ∆U +W and ∆U = 0 , so the net heat added to the system during one complete cycle is 3.58 10 J 3 Q =W = − × III. Give the Solutions of the Following Problems 1. When a system is taken from state i to state f along the path iaf in Fig.6, it is found that Q=50J and W=20J. Along the path ibf, Q=36J. (a) What is W along the path ibf? (b) If W=-13J for the curved return path fi, what is Q for this path? (c) Take Eint,i=10J. What is Eint,f? (d) If Eint,b =22J, find Q for process ib and prosess bf. Solution: (a) Applying the first law of thermodynamics: Q = ∆U +W , so ∆U = U f −Ui = Q −W , that is Qiaf −Wiaf = Qibf −Wibf Thus = + − = 36 + 20 − 50 = 6 (J) Wibf Qibf Wiaf Qiaf (b) The same way as (a), we have Qiaf −Wiaf = Qfi −Wfi So = − + = 50 − 20 −13 =17 (J) Qfi Qiaf Wiaf Wfi (c) Applying the first law of thermodynamics: Q = ∆U +W , so V(L) P（Mpa） 0 Fig.5 1 2 3 4 5 10 20 30 40 V P Fig.6 i f a b
AU =Eint -EintiCiagt-Wia EinOia-Wia+ Einti=50-20+10=40() d)See the diagram, Wih=Wb=6J Applying the first law of thermodynamics, the heat transfer for process ib is Q=AU+w=Eintb- Inti +Wib=22-10+6=18) As Qib = oib Ob, so the heat transfer for process bfis =y-gb=36-18=18(J) 2. Three(3.00)moles of an ideal gas is taken along the excursion on a P-V diagram indicated in Figure 7. The first part of the pati sobaric process and the second part of the path is isothermal a)What is the initial absolute temperature of the gas? P(atm) (b)Calculate the work done by the gas along the isobaric segment of the path (c)What is the temperature of the gas along the isotherm? (d)Is the work done by the gas positive or negative 18.5 B along the isothermal segment of the path? (e) Determine the final volume Vr of the gas so that the v(liters) total work done by the gas is zero. Express your result 4.0 5.0 in lite Fig. 7 What is the final pressure Pr of the gas? Express your result in atmospheres (a) Applying the equation of ideal gas Pv=nRT, the initial absolute temperature of the gas is 7sP_185×1013×105×402230051(K) 8.315×3 (b) The work done by the gas along the isobaric segment of the path Wp=P△=185×1013×103×(5-4)×10-3=187405(J) (c) Applying the equation of ideal gas Pv=nRT, the temperature of the gas along the isotherm 7=P=185×10310×5×10=3564k (d)The work is negative along the isothermal segment of the path e)In order that the total work done by the gas is zero, we have
50 20 10 40 (J) ∆ = int − int = − ⇒ int = − + int = − + = U E f E i Qiaf Wiaf E f Qiaf Wiaf E i (d) See the diagram, = = 6 J Wibf Wib . Applying the first law of thermodynamics, the heat transfer for process ib is 22 10 6 18 (J) Q = ∆U +W = Eint b − Eint i +Wib = − + = As Qibf = Qib + Qbf , so the heat transfer for process bf is = − = 36 −18 =18 (J) Qbf Qibf Qib 2. Three (3.00) moles of an ideal gas is taken along the excursion on a P-V diagram indicated in Figure 7. The first part of the path is an isobaric process and the second part of the path is isothermal. (a) What is the initial absolute temperature of the gas? (b) Calculate the work done by the gas along the isobaric segment of the path. (c) What is the temperature of the gas along the isotherm? (d) Is the work done by the gas positive or negative along the isothermal segment of the path? (e) Determine the final volume Vf of the gas so that the total work done by the gas is zero. Express your result in liters. (f)What is the final pressure Pf of the gas? Express your result in atmospheres. Solution: (a) Applying the equation of ideal gas PV = nRT , the initial absolute temperature of the gas is 300.51(K) 8.315 3 18.5 1.013 10 4 10 5 3 = × × × × × = = − Rn PV T (b) The work done by the gas along the isobaric segment of the path is 18.5 1.013 10 (5 4) 10 1874.05(J) 5 3 = ∆ = × × × − × = − WP Pi V (c) Applying the equation of ideal gas PV = nRT , the temperature of the gas along the isotherm is 375.64(K) 8.315 3 18.5 1.013 10 5 10 5 3 = × × × × × = = − Rn PV T i f (d) The work is negative along the isothermal segment of the path. (e) In order that the total work done by the gas is zero, we have A V(liters) P(atm) Fig.7 4.0 5.00 18.5 Pf B C
Wr=nRT, So the final volume Vf of the gas is 5×103×e815×4=4.09×10-(m3)=409(1) (f Applying the equation of ideal gas Pv=nRT, the final pressure Pr of the gas is 78.315×3×37564 P 409×10-3 2.29×10°(Pa)=2262(atm) 3. Suppose that a sample of gas expands from 2.0 to 8.0 m along the diagonal path in the PI diagram shown in Fig 8. It is then compressed back to 2.0 m along either path I or path 2. Compute the net work done on the gas for the complete cycle in p(kpa) each case Solution Path 1 The net work done on the gas is the area of the cycle 20 on the diagram The area of the cycle along path S=×(20-5)×103×(2-8)=-4.5×104 Path 2 So the net work done on the gas along path I is 246 W=-4.5×104(J The area of the cycle along path 2 is ×(20-5)×103×(8-2)=45×10 So the net work done on the gas along path 2 is ×10(J 4. a gas is taken around a closed cycle on a P-I diagram as indicated in Figure 9 (a)Find the total work W done by the gas (b)What is the change in the internal energy of the 6.00- (c)What is the total heat transfer to the gas in executing this cycle? If the gas is taken around the cycle in the opposite o v(liters) 3.00600 Fig 9
P i f T f W V V W = nRT ln = − So the final volume Vf of the gas is 5 10 4.09 10 (m ) 4.09(l) 8.315 3 375.64 3 3 1874.05 3 = = × × = × = × × − − − − V V e e f P nRT W f i (f) Applying the equation of ideal gas PV = nRT , the final pressure Pf of the gas is 2.29 10 (Pa) 22.62(atm) 4.09 10 8.315 3 375.64 6 3 = × = × × × = = − f f f V nRT P 3. Suppose that a sample of gas expands from 2.0 to 8.0 m3 along the diagonal path in the PV diagram shown in Fig.8. It is then compressed back to 2.0 m3 along either path 1 or path 2. Compute the net work done on the gas for the complete cycle in each case. Solution: The net work done on the gas is the area of the cycle on the diagram. The area of the cycle along path 1 is 3 4 1 (20 5) 10 (2 8) 4.5 10 2 1 S = × − × × − = − × So the net work done on the gas along path 1 is 4.5 10 (J) 4 W1 = − × The area of the cycle along path 2 is 3 4 2 (20 5) 10 (8 2) 4.5 10 2 1 S = × − × × − = × So the net work done on the gas along path 2 is 4.5 10 (J) 4 W1 = × 4. A gas is taken around a closed cycle on a P-V diagram as indicated in Figure 9. (a) Find the total work W done by the gas. (b) What is the change in the internal energy of the gas? (c) What is the total heat transfer to the gas in executing this cycle? If the gas is taken around the cycle in the opposite V(liters) P(atm) 0 Fig.9 3.00 6.00 2.00 6.00 V(m3 ) P（kpa） 0 Fig.8 2 4 6 8 5 10 15 20 Path 1 Path 2
direction: (d) Find the total work done by the gas (e) Find the change in the internal energy of the gas, (f) Find the total heat transfer to the gas in executing the cycle in this direction Solution (a)The total work W done by the gas is W=S4=(6-2)×1.013×105×(6-3)×10-3=6078×10(J) (b) The change in the internal energy of the gas is AU=O0) (c)Applying the first law of thermodynamics: =AU +w and AU=0, the total heat transfer to the gas in executing this cycle is 0 =W=6.078x10-(J) (d)The total work w done by the gas is W'=-W-6.078x10-(J) (e) The change in the internal energy of the gas is AU=O() (f The total heat transfer to the gas in executing the cycle in opposite direction is Q=W=-6078×10(J)
direction: (d) Find the total work done by the gas; (e) Find the change in the internal energy of the gas; (f) Find the total heat transfer to the gas in executing the cycle in this direction. Solution: (a) The total work W done by the gas is (6 2) 1.013 10 (6 3) 10 6.078 10 (J) 2 1 5 3 2 = = − × × × − × = × − W S∆ . (b) The change in the internal energy of the gas is ∆U = 0(J). (c) Applying the first law of thermodynamics: Q = ∆U +W and ∆U = 0 , the total heat transfer to the gas in executing this cycle is 6.078 10 (J) 2 Q = W = × . (d) The total work W done by the gas is 6.078 10 (J) 2 W ′ = −W − × . (e) The change in the internal energy of the gas is ∆U = 0(J). (f) The total heat transfer to the gas in executing the cycle in opposite direction is 6.078 10 (J) 2 Q = W = − ×