Chapter 21 - The Kinetic Theory of Gases

Transcription

1 hapter 1 - he Knetc heory o Gases 1. Δv 8.sn 4. 8.sn 4. m s F Nm. 1 kg.94 N Δ t. s F A 1.7 N m 1.7 a N mv 1.6 Use the equaton descrbng the knetc-theory account or pressure:. hen mv Kav where N nna NA N ( 8. atm ( a atm (. 1 m Kav N mol 6. 1 molecules mol K av A 1. 1 J molecule 1.7 ( K N E rom the knetc-theory account or pressure. ( KE A ( 1. 1 ( 4. 1 ( N molecules 6. 1 molecules mol N n N 4. 1 molecules. mol 1.9 (a Nk : 4 ( J K ( 9 K a π.1 m N k.4 1 atoms J J K k (c For helum, the atomc mass s 4. g mol m 6. 1 molecules mol g molecule 7 m kg molecule 1 mv k k: vrms m 1. km s J K 4 K J 1.11 (a K k 1 1 K m v rms J

2 so For helum, v rms J (1 m 4. g mol m 6. 1 molecules mol Smlarly or argon, g molecule 7 m kg molecule 9.9 g mol m 6. 1 molecules mol Substtutng n (1 above, we nd or helum, and or argon, g molecule 6 m kg molecule v 1.6 km s rms v rms 14 m s 1.1 n 1. mol, K Snce constant, W (a Δ Ent Q+ W 9 J+ 9 J (c Δ Ent n Δ n R Δ so 9 ( J ( ΔE nr 1. mol 8.14 J mol K nt Δ 16.8 K +Δ K K 17 K 1.1 We use the tabulated values or and (a Q n Δ 1. mol 8.8 J mol K 4 K.46 kj E n Δ nt Δ 1. mol.4 J mol K 1 K.4 kj (c W Q+Δ Ent.46 kj+.4 kj 1.1 kj 1. mol 8.14 J mol K 719 J kg K.719 kj kg K.89 kg 1.16 (a R m Mn M R

3 1 a. m m (.8 9 kg mol.811 kg ( 8.14 J mol K ( K (c (d We consder a constant volume process where no work s done. Q m Δ.811 kg.719 kj kg K 7 K K kj We now consder a constant pressure process where the nternal energy o the gas s ncreased and work s done. 7R 7 Q mδ m( + R Δ m Δ m Δ 7 Q.811 kg (.719 kj kg K ( 4 K 7 kj 1.17 Q ( n Δ + ( n Δ sobarc sovolumetrc In the sobarc process, doubles so must double, to. In the sovolumetrc process, trples so changes rom to 6. 7 Q n R ( + n R ( 6 1.nR (a so ( (c Snce the process s adabatc, Q R+ Snce 1.4, Δ. 1. Rand Δ Ent n Δ (.16 mol ( 8.14 J mol K 1.( K 1 J and W Q+Δ Ent + 1 J + 1 J (a atm 1.9 atm.

4 ( (. mol ( 8.14 J mol K ( (. mol ( 8.14 J mol K a 1. 1 m nr 66 K a. 1 m nr K (c he process s adabatc: Q R+ 1.4, nt R Δ Ent n Δ. mol ( 8.14 J mol K ( K 66 K 4.66 kj W ΔE Q 4.66 kj 4.66 kj m. m m π he quantty o ar we nd rom nr n R 4 ( a(.4 1 m ( 8.14 J mol K ( K n mol Adabatc compresson: 11. ka + 8 ka 91. ka (a m m 7 nr ( ( K 6 K 91. (c he work put nto the gas n compressng t s Δ Ent n Δ W mol 8.14 J mol K 6 K W.9 J Now magne ths energy beng shared wth the nner wall as the gas s held at constant volume. he pump wall has outer dameter. mm +. mm +. mm 9. mm, and volume π( m π( 1. 1 m m m

5 6 and mass ρ kg m m. g he overall warmng process s descrbed by.9 J n Δ + mcδ.9 J mol 8.14 J mol K K ( (. 1 kg( 448 J kg K ( K (.9 J.7 J K +.9 J K K K.4 K + 1. We suppose the ar plus burnt gasolne behaves lke a datomc deal gas. We nd ts nal absolute pressure: Now Q atm. cm 4 cm 1 1. atm 1.14 atm 8 and W Δ Ent n ( W nr nr FIG N m W 1 atm 1.14 atm 4 cm 1. atm. cm 1 m cm W 1 J 6 he output work s W + 1 J he tme or ths stroke s 1 1 mn 6 s 4 1 mn 6. 1 s W 1 J Δ t 6. 1 s. kw 1.7 he sample's total heat capacty at constant volume s n. An deal gas o datomc molecules has three degrees o reedom or translaton n the x, y, and z drectons. I we take the y axs along the axs o a molecule, then outsde orces cannot excte rotaton about ths axs, snce they have no lever arms. ollsons wll set the molecule spnnng only about the x and z axes.

6 (a I the molecules do not vbrate, they have ve degrees o reedom. Random collsons put equal amounts o energy 1 k nto all ve knds o moton. he average energy o one molecule s k. he nternal energy o the two-mole sample s N k nn A k n R n he molar heat capacty s R and the sample s heat capacty s n n R mol ( 8.14 J mol K n 41.6 J K For the heat capacty at constant pressure we have 7 7 n n( + R n R+ R nr mol ( 8.14 J mol K n 8. J K In vbraton wth the center o mass xed, both atoms are always movng n opposte drectons wth equal speeds. braton adds two more degrees o reedom or two more terms n the molecular energy, or knetc and or elastc potental energy. We have 7 n n R 8. J K and 9 n n R 74.8 J K dn 1. In the Maxwell oltzmann speed dstrbuton uncton take v to nd dv m m v m v 4π N exp v π k k k and solve or v to nd the most probable speed. Reject as solutons v and v hey descrbe mnmally probable speeds. Retan only hen mv k v mp k m *1.4 (a 1 k. So W d k d k 1 For k we can substtute and also to have

7 W de nt dq + dw and dq or an adabatc process. hereore, W +Δ Ent n ( o show consstency between these two equatons, consder that and R. hereore, 1 1 Usng ths, the result ound n part (a becomes R W R Also, or an deal gas n R so that W n (, as ound n part. 1.4 Let the subscrpts 1 and reer to the hot and cold compartments, respectvely. he pressure s hgher n the hot compartment, thereore the hot compartment expands and the cold compartment contracts. he work done by the adabatcally expandng gas s equal and opposte to the work done by the adabatcally compressed gas. nr nr ( 1 1 ( 1 1 hereore K (1 onsder the adabatc changes o the gases. and , snce 1 and 1 nr nr , usng the deal gas law nr nr 1 1, snce 1 and K 1.76 K (

8 Solvng equatons (1 and ( smultaneously gves 1 1 K, 9 K 1.46 he net work done by the gas on the bullet becomes the bullet's knetc energy: mv kg ( 1 m s 7.9 J he ar n ront o the bullet does work Δ (1.1 1 N/m (. m(. 1 4 m.1 J he hot gas behnd the bullet then must do output work +W n +W.1 J 7.9 J W 8.7 J. he nput work on the hot gas s 8.7 J 1 ( 8.7 J 1 Also So J.4 And 1 cm + cm. cm 1. cm 8.7 J(.41 6 cm m hen 1. cm 1/ cm a 7.7 atm 1.8 We want to evaluate d or the uncton mpled by d nr constant, and also or the derent uncton mpled by constant. We can use mplct derentaton: d d From constant + d d d d sotherm From constant d + 1 d d d adabat hereore, d d d d adabat sotherm he theorem s proved. *1.9 (a ( a(. 1 m n R ( 8.14 J mol K ( K. mol. A K 9 K A 1.

9 9 K 9 A. L 1. L A FIG. 1.9 nt, A A. mol 8.14 J mol K K 76 J Ent, Ent, nr (. mol ( 8.14 J mol K ( 9 K.8 kj (c E nr (d (atm (L (K E nt (kj A (e For the process A, lock the pston n place and put the cylnder nto an oven at 9 K. For, keep the sample n the oven whle gradually lettng the gas expand to lt a load on the pston as ar as t can. For A, carry the cylnder back nto the room at K and let the gas cool wthout touchng the pston. ( For A: W Δ Ent Ent, Ent, A.8.76 kj 1. kj QΔE W nt 1. kj For : Δ Ent, W nr ln W. mol 8.14 J mol K 9 K ln kj Q ΔE W nt 1.67 kj For A: E E E nt Δ nt nt, A nt,.76.8 kj 1. kj W Δ nrδ. mol 8.14 J mol K 6 K 1.1 kj Q ΔE W 1. kj 1.1 kj. kj (g We add the amounts o energy or each process to nd them or the whole cycle. Q W AA AA ( E nt + 1. kj kj. kj.66 kj 1.67 kj+ 1.1 kj.66 kj Δ + 1. kj+ 1. kj AA