CHAPTER 20: Second Law of Thermodynamics

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "CHAPTER 20: Second Law of Thermodynamics"

Transcription

1 CHAER 0: Second Lw of hermodynmics Responses to Questions 3. kg of liquid iron will hve greter entropy, since it is less ordered thn solid iron nd its molecules hve more therml motion. In ddition, het must e dded to solid iron to melt it; the ddition of het will increse the entropy of the iron. 5. he mchine is clerly doing work to remove het from some of the ir in the room. he wste het is dumped ck into the room, nd the het generted in the process of doing work is lso dumped into the room. he net result is the ddition of het into the room y the mchine. 6. Some processes tht would oey the first lw of thermodynmics ut not the second, if they ctully occurred, include: cup of te wrming itself y gining therml energy from the cooler ir molecules round it, ll sitting on soccer field gthering energy from the grss nd eginning to roll, nd owl of popcorn plced in the refrigertor nd unpopping s it cools. 7. No. While you hve reduced the entropy of the ppers, you hve incresed your own entropy y doing work, for which your muscles hve consumed energy. he entropy of the universe hs incresed s result of your ctions. 9. No. Even if the powdered milk is dded very slowly, it cnnot e re-extrcted from the wter without very lrge investments of energy. his is not reversile process. 0. Entropy is stte vrile, so the chnge in entropy for the system will e the sme for the two different irreversile processes tht tke the system from stte to stte. However, the chnge in entropy for the environment will not necessrily e the sme. he totl chnge in entropy (system plus environment) will e positive for irreversile processes, ut the mount my e different for different irreversile processes.

2 Solutions to rolems In solving these prolems, the uthors did not lwys follow the rules of significnt figures rigidly. We tended to tke quoted tempertures s correct to the numer of digits shown, especilly where other vlues might indicte tht. 3. Het energy is tken wy from the wter, so the chnge in entropy will e negtive. he het trnsfer is the mss of the stem times the ltent het of vporiztion. 5 Q ml 0.5 kg.6 0 J kg vp S 500J K K 33. Energy hs een mde unville in the frictionl stopping of the sliding ox. We tke tht lost kinetic energy s the het term of the entropy clcultion. i S Q mv 7.5 kg 4.0m s 93 K 0.0J K Since this is decrese in vilility, the entropy of the universe hs incresed. 35. Becuse the temperture chnge is smll, we cn pproximte ny entropy integrls y S Q. here re three terms of entropy to consider. First, there is loss of entropy from vg the wter for the freezing process, S. Second, there is loss of entropy from tht newly-formed ice s it cools to 0 o C, S. ht process hs n verge temperture of 5 o C. Finlly, there is gin of entropy y the gret del of ice, S, s the het lost from the originl mss of wter in steps nd 3 goes into tht gret del of ice. Since it is lrge quntity of ice, we ssume tht its temperture does not chnge during the processes.

3 Q ml kg3.330 J kg fusion 6 S.98 0 J K Q mc 3 o o.00 0 kg00j kggc 0C ice 4 S J K 3 fusion ice K Q Q Q ml mc S K o o 0C.00 0 kg J kg 00J kggc 0 73 K J K S S S S J K 5 0 J K J K J K J K 37. he sme mount of het tht leves the high temperture het source enters the low temperture ody of wter. he tempertures of the het source nd ody of wter re constnt, so the entropy is clculted without integrtion. Q Q S S S Q high low low high S Q 4.86 J 9.50cl s t t cl low high 73 K 5 73 K JK s 39. Becuse the process hppens t constnt temperture, we hve S Q. he het flow cn e found from the first lw of thermodynmics, the work for expnsion t constnt temperture, nd the idel gs eqution E Q W 0 Q W nr ln ln int Q m 7.5tm S 40 K.0tm ln ln 9.3J K

4 4. Since the process is t constnt volume, dq ncd. For ditomic gs in the temperture 5 rnge of this prolem, C R. dq nc d 5 5 nr g K S ln.0mol8.34j mol Kln 4.0J K 73 5 K 44. Entropy is stte vrile, nd so the entropy difference etween two sttes is the sme for ny pth. Since we re told tht sttes nd hve the sme temperture, we my find the entropy chnge y clculting the chnge in entropy for n isotherml process connecting the sme two sttes. We lso use the first lw of thermodynmics. E nc 0 Q W Q W nr ln int dq Q nr ln S nr ln 45. () he figure shows two processes tht strt t the sme stte. he top process is ditic, nd the ottom process is isothermic. We see from the figure tht t volume of /, the pressure is greter for the ditic process. We lso prove it nlyticlly. Isotherml: / Aditic: we see tht. he rtio is ditic isothermic Since, ditic isothermic () For the ditic process: No het is trnsferred to or from the gs, so S dq 0. ditic. For the isotherml process: 0 ln E Q W nr int isotherml isotherml isotherml

5 dq isotherml isotherml S dq = isotherml isotherml nr ln nr ln nr ln Q nr ln (c) Since ech process is reversile, the energy chnge of the universe is 0, nd so S For the ditic process, S 0. For the isotherml process, surroundings S. surroundings system S nr ln. surroundings 46. ()he equilirium temperture is found using clorimetry, from Chpter 9. he het lost y the wter is equl to the het gined y the luminum. f f m c m c HO HO HO Al Al Al f m c m c m c m c Al Al Al HO HO HO Al Al HO HO 0.50 kg900j kggc 5C 0.5kg486J kggc 00C 0.50 kg900j kggc 0.5kg486J kggc 88.9C 89C finl finl finl finl dq dq Al HO d d () S S S Al HO m c m c Al Al HO HO finl m c ln m c Al Al HO HO Al Al H O Al HO ln finl HO 0.50 kg900j kggk ln g K K K 0.5kg 486J kg K ln 3.7 J K K 48. ()he gses do not interct since they re idel, nd so ech gs expnds to twice its volume with no chnge in temperture. Even though the ctul process is not reversile, the entropy chnge cn e clculted for reversile process tht hs the sme initil nd finl sttes. his is discussed in Exmple 0-7.

6 S S nr nr N Ar totl N Ar ln ln g S S S nr ln.00 mol 8.34 J mol K ln.5j K () Becuse the continers re insulted, no het is trnsferred to or from the environment. hus dq S surroundings 0. (c) Let us ssume tht the rgon continer is twice the size of the nitrogen continer. hen the finl nitrogen volume is 3 times the originl volume, nd the finl rgon volume is.5 times the originl volume. S nr ln nr ln 3 ; S nr ln nr ln.5 N Ar totl N Ar N Ar g S S S nr ln 3 nr ln.5 nr ln mol 8.34 J mol K ln 4.5.5J K 50. We ssume tht the process is reversile, so tht the entropy chnge is given y Eq he het trnsfer is given y dq nc d. 3 dq nc d n d 3 3 S n d n 4 0.5mol.08 mj molgk.0 K 3.0 K.57 mj molgk.0 K 3.0 K 3 4.0mJ K All of the processes re either constnt pressure or constnt volume, nd so the het input nd output cn e clculted with specific hets t constnt pressure or constnt volume. his tells us tht het is input when the temperture increses, nd het is exhusted when the temperture decreses. he lowest temperture will e the temperture t point. We use the idel gs lw to find the tempertures.

7 nr nr 3 3,, 3, 6 nr nr nr nr c d W ; Q 0 () process : process c: W 0 ; Q nc nr nr nr 3 nr c c c 0 0 process cd: W 3 ; c Q nc nr nr 3 nr 3 nr cd d c 0 0 process d: W 0 ; Q 0 d d e W rectngle 5 Q 3 H () e 6 e H L rectngle ; 0.3 Crnot 6 e H Crnot 5 7. We hve montomic gs, so Also the pressure, volume, nd temperture for stte re 3. known. We use the idel gs lw, the ditic reltionship, nd the first lw of thermodynmics. () Use the idel gs eqution to relte sttes nd. Use the ditic reltionship to relte sttes nd c..4 L 73K.00 tm tm 56.0 L 73K c c c.4 L.00tm 0.7 tm 0.7 tm 56.0 L c () Use the idel gs eqution to clculte the temperture t c. 5/ 3

8 c c c c 0.7 tm c 73K 48 K 0.400tm c (c) rocess : E nc int 0 ; Q W nr S J 080 J Q g ln.00 mol 8.34 J mol K 73K ln J 73K 7.6 J K rocess c: W 0 ; c int c c 559 J 560J 3.00 mol 8.34 J molgk 48 K 73K E Q nc c c dq nc d c 3 nc c 48 K S ln.00 mol 8.34 J mol Kln g 73K 7.64 J K rocess c: Q S c 0 ; 0 ditic ; c int int int c c int c E W E E 0 560J E 560 J ; W 560 J c (d) W 080J 560J e Q 080J input ke the energy trnsfer to use s the initil kinetic energy of the crs, ecuse this energy ecomes unusle fter the collision it is trnsferred to the environment. m s 00 kg 75km h mv Q i 3.6 km h S 5 73 K 700J K

9 78. Since two of the processes re ditic, no het trnsfer occurs in those processes. hus the het trnsfer must occur long the isoric processes. c Aditic exp nsion Aditic compression d ; nc d d Q Q nc Q Q nc e H c c L d d Q L Q nc H c c Use the idel gs reltionship, which sys tht nr. d d nr nr e nr nr d d d c c c c c d c Becuse process is ditic, we hve ditic, we hve expression. / c d d c /.. Becuse process cd is Sustitute these into the efficiency

10 / / / c d c e c c c