Homework 11. Problems: 20.37, 22.33, 22.41, 22.67

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1 Homework 11 roblems: 0.7,.,.41,.67

2 roblem kg block o alumnum s heated at atmospherc pressure such that ts temperature ncreases rom.0 to Fnd (a) the work done by the alumnum, (b) the energy added to t by heat, and (c) the change n ts nternal energy. a) he macroscopc work done by a system s dened by the change n the system's volume m 1 kg dw d. º; 40 º he stress (hydrostatc pressure) n the block s equal to the external pressure o 1atm. Snce, n ths process, the pressure s constant, t s easy to nd the work (ntegral) ΔW dw process Δ he thermal expanson o the block causes a change n volume. From the denton o the volumetrc coecent o expanson we can relate the change n volume to the change n temperature. ssumng that the change n volume s small when compared wth the ntal volume we can use the approxmate relaton Δ βδ αδ We can express the volume n terms o the mass o the block, usng the denton o densty. For a unorm object ntegraton o the densty leads to a smple relaton

3 m ρ Work done n the process s thereore m ΔW α ρ l Δ 1.01a K 1 1kg kg.7 10 m o o ( 40 ) 48mJ b) he process s sobarc and the specc heat n ths temperature nterval does not depend on temperature. he ntegraton o heat s thereore easy to perorm ΔQ process dq J 1kg 900 kg K mcd mcδ ( 40 ) K 16.kJ c) From the rst law o thermodynamcs, the change n the nternal energy o the block s ΔU ΔQ ΔW 16. kj 0. 48mJ 16. kj

4 roblem. In a cylnder o an automoble engne just ater combuston, the gas s conned to a volume o 0.0 cm and has an ntal pressure o 10 6 a. he pston moves outward to a nal volume o 00 cm, and the gas expands wthout energy loss by heat. (a) I 1.4 or the gas, what s the nal pressure? How much work s done by the gas n expandng? Wth good approxmaton, we can assume that the gas n the engne s deal whch satses the ollowng state equaton (relatng pressure, volume and the temperature o the gas) 1) nr where n s the amount o the gas expressed n moles and R s the gas constant. For an deal gas, the change n nternal energy s related to the change n temperature only ) du n v d where v s the molar heat capacty at constant volume. ccordng to the rst law o thermodynamcs, or an adabatc process (no heat delvered to the gas), the change n nternal energy s opposte to the work perormed by the gas ) du 0 - d he rest s math. Snce the pressure and volume s gven n the problem we want to elmnate the temperature rom the consderaton. From equaton (1), we can express the temperature derental n terms o the volume and pressure derentals d + d nrd Usng () and () we can elmnate temperature rom last equaton d d + d nr n v ecause the derence between the molar heat capacty as the constant pressure and the constant volume or one mole o gas s equal to the gas constant

5 4) p - v R we can rearrange the last equaton and wrte p v 1+ d d v or smply t even urther by substtutng the rato o the molar heat capacty at constant pressure and the molar heat capacty at constant volume p ) v lso dvdng both sdes o the equaton by the product o volume and pressure we can separate the varables (group terms wth volume on one sde o the equaton and terms wth pressure on the other sde o the equaton d Integratng both sde over the consdered process (rom ntal volume and pressure to the nal volume and pressure) process d d process we get ( ln ln ) ( ln ln ) Usng the propertes o logarthmc uncton we can smply the last equaton hereore ln ln d

6 (We can present the last equaton n the tradtonally used orm ) Solvng or the only unknown (the nal pressure) we can nd the answer 10 6 a ( 0cm ) ( 00cm ) omment. In ths problem t s assumed that ntrogen undergoes the process. In the combuston, sgncant amounts o water (H O) and carbon doxde (O ) are produced. hereore, t would be more accurate to assume that 1.< <1.4. (See table 1.) a

7 roblem.41 -L contaner has a center partton that dvdes t nto two equal parts, as shown n Fgure.41. he let-hand sde contans H gas, and the rght-hand sde contans O gas. oth gases are at room temperature and at atmospherc pressure. he partton s removed, and the gasses are allowed to mx. What s the entropy ncrease o the system?.044 mole H.044 mole O a) When the gases are mxed they ll the entre volume o the contaner. he ntal and the nal temperatures are equal, thereore we can choose an sothermal process n order to calculate the change n entropy. (he sobarc process s ncorrect, because although the pressure n the mxture s 1 atm as each gas expands and ts partal pressure decreases.) In the consdered condtons, both gases can be treated as deal gases. her partal pressure, volume and temperature are related by the equaton o the state o an deal gas. 1) nr In the sothermal process or an deal gas, the nternal energy o the gas does not change. From the rst law o thermodynamcs, the heat delvered to the deal gas must be equal to the work done by the gas n an sothermal process. nrd ) dqr dwr d Usng the denton o entropy, the change n entropy o the deal gas undergong an nntesmal expanson s thereore dq ds r nrd he entropy n the entre expanson process o an deal gas changes by

8 ΔSx x x nrd nr ln oth the hydrogen and the oxygen ncrease entropy. he change n the entropy o the system s thereore ΔS H + ΔS O n H R ln H H + n R ln J l J 0.44mol 8.1 ln mol 8.1 ln mol K 1l mol K,, O O O,, x x l 1l 0.07 J K

9 roblem.6 One mole o an deal monatomc gas s taken through the cycle shown n Fgure.67. he process s a reversble sothermal expanson. alculate (a) the net work done by the gas, (b) the (thermal) energy added to the gas, (c) the (thermal) energy expelled by the gas, and (d) the ecency o the cycle. (atm) a) In each process the work depends on the change n volume o the gas and the pressure o the gas durng the process. 1) dw d 1 In order to nd the value o the (lters) 10 0 work, we have to know the explct dependence o the pressure durng the process. In an sothermal process, the pressure o the gas s nversely proportonal to ts volume ) ( ) nr hereore the work done n the sothermal process s W d atm 1atm nr d a ln m 0l ln 10l 811.8J

10 In process, the pressure o the gas s constant. In ths sobarc process the ntegraton wll be much easer W d ( ) a ( m 0 10 m ) 40J In the thrd process, the volume s not changed, thereore the gas does not perorm work. he work done by the gas n the whole cycle s thereore Δ W W + W + W 81J 40J + 0J 4kJ b,c) We have to perorm smlar calculatons or the heat. In the sothermal expanson o an deal gas ts nternal energy does not change. ccordng to the rst law o thermodynamcs, the heat delvered to the gas s equal to the work done by the gas n ths process. ) Q ΔU + W W 8.1kJ In ths process the heat s delvered to the gas. In the sobarc process the temperature o the gas changes. We can express the heat Q delvered to the gas, wth the change n ts temperature and the molar heat capacty (at constant pressure) o a monatomc deal gas 1 p R 1 See chapter 1 or the molar heat capacty o an deal gas.

11 Q n p d n p nr nr ( ) nr ( ) ( a m a 0 10 m ) 10.1kJ In ths process the gas expels the heat. In the sochorc process the temperature o the gas also changes. We can express the heat Q delvered to the gas, n terms o the change n ts temperature and the molar heat capacty v R (at constant volume) o a monatomc deal gas Q n v d n atm 1atm v ( ) nr ( ) a nr m nr a m 6.1kJ Heat s delvered to the gas n ths process. In ths cycle heat was delvered to the gas rom the heat reservor n processes and. he heat delvered n the whole cycle s thereore Δ Qh Q + Q 8.1kJ + 6.1kJ 14.kJ In ths cycle, the gas only n process expels heat. he heat expelled n the whole cycle s thereore Δ Qc Q 10.1kJ (o very our calculatons we can check energy s conserved. In a cycle the nternal energy o the gas does not change thereore the energy delvered must be equal to the sum o the energy used and wasted n the heat snk 14.kJ 4.1kJ kJ.

12 In the soluton, we could use ths act to nd heat delvered to the gas n one o the three processes.) d) From the denton o ecency, we have to compare the energy used n the orm o work perormed by the gas to the energy delvered to the gas engne rom the reservor ΔW e ΔQ h 4.1kJ 14.kJ 8.9%

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