Physcs 4C Solutons to Chater 9 HW Chater 9: Concetual Questons: 6, 8, 0 Problems:,, 4,,, 48,, 6, 6, 78, 87 Queston 9-6 (a) 0 (b) 0 (c) negate (d) oste Queston 9-8 (a) 0 (b) 0 (c) negate (d) oste Queston 9-0 (a) same (b) ncreases (c) decreases (d) ncreases Problem 9- In solng the deal-gas law equaton = nr or n, we rst conert the temerature to the Keln scale: = (40.0 + 7.) K =. K, and the olume to SI unts: V = 000 cm = 0 m (a) he number o moles o oxygen resent s n. R (.0 0 Pa)(.000 0 m ) = = =.88 0 mol. ( 8.J/mol K)(.K) (b) Smlarly, the deal gas law = nr leads to (.06 0 Pa)(.00 0 m ) = = = 49K, nr (.88 0 mol)( 8.J/mol K) whch may be exressed n degrees Celsus as 0 C. Note that the nal temerature can also be V calculated by notng that =, or V = = = V.0 0 Pa 000 cm.06 0 Pa 00 cm (. K) 49 K
Problem 9- (a) At ont a, we know enough normaton to comute n: ( 00Pa)(.0m ) n = = =.mol. R ( 8. J/mol K) ( 00 K) (b) We can use the answer to art (a) wth the new alues o ressure and olume, and sole the deal gas law or the new temerature, or we could set u the gas law n terms o ratos (note: n a = n b and cancels out): 7.kPa.0m = b = ( 00K).kPa.0m b b b a a a whch yelds an absolute temerature at b o b =.8 0 K. (c) As n the reous art, we choose to aroach ths usng the gas law n rato orm:.kpa.0m = c = ( 00K).kPa.0m c c c a a a whch yelds an absolute temerature at c o c = 6.0 0 K. (d) he net energy added to the gas (as heat) s equal to the net work that s done as t rogresses through the cycle (reresented as a rght trangle n the dagram shown n Fg. 9-0). hs, n turn, s related to ± area nsde that trangle (wth area = (base)(heght) ), where we choose the lus sgn because the olume change at the largest ressure s an ncrease. hus, Q net net (.0m ) (.0 0 Pa ).0 0 = W = = J. Problem 9-4 We can exress the deal gas law n terms o densty usng n = M sam /M: = M R = M M R sam ρ. We can also use ths to wrte the rms seed ormula n terms o densty: rms R ( M / ρ) = = = M M ρ.
(a) We conert to SI unts: ρ =.4 0 kg/m and =.0 0 Pa. he rms seed s (00) / 0.04 = 494 m/s. (b) We nd M rom ρ = M/R wth = 7 K. M ρr (0.04 kg/m ) ( 8.J/mol K )(7K) = = = 0.079 kg/mol = 7.9 g/mol..0 0 Pa (c) From able 9., we denty the gas to be N. Problem 9- (a) We use the deal gas law = nr = Nk, where s the ressure, V s the olume, s the temerature, n s the number o moles, and N s the number o molecules. he substtutons N = nn A and k = R/N A were made. Snce cm o mercury = Pa, the ressure s = (0 7 )( Pa) =. 0 4 Pa. hus, N V = = = = k 4. 0 Pa (.8 0 J/K)(9 K) 6 0.7 0 molecules/m.7 0 molecules/cm. (b) he molecular dameter s d =.00 0 0 m, so, accordng to Eq. 9-, the mean ree ath s λ= 7 m. 0 6 πd N / V = π (.00 0 m) (.7 0 m ) = Problem 9- (a) he aerage seed s ag N = = [4(00 m/s) + (00 m/s) + 4(600 m/s)] = 40 m/s. 0 N = (b) he rms seed s rms N [4(00 m/s) (00 m/s) 4(600 m/s) ] 48 m/s N = 0 = = + + = (c) Yes, rms > ag.
Problem 9-48 (a) Accordng to the rst law o thermodynamcs Q = ΔE nt + W. When the ressure s a constant W = ΔV. So 6 0 m Δ Ent = Q Δ V = 0.9 J (.0 0 Pa)( 00 cm 0 cm ) =.9 J. cm (b) he molar secc heat at constant ressure s C ( ) ( 8. J/mol K)( 0.9 J) Q Q R Q = = = = = 4.4J mol K. nδ n ΔV / nr Δ V.0 0 Pa 0 0 m (c) Usng Eq. 9-49, C V = C R = 6. J/mol K. 6 ( )( ) Problem 9- (a) Snce the rocess s at constant ressure, energy transerred as heat to the gas s gen by Q = nc Δ, where n s the number o moles n the gas, C s the molar secc heat at constant 7 ressure, and Δ s the ncrease n temerature. For a datomc deal gas C = R hus,. Q = 7 nrδ = 7 ( 4.00mol )( 8.J/mol K )( 60.0 K ) = 6.98 0 J. (b) he change n the nternal energy s gen by ΔE nt = nc V Δ, where C V s the secc heat at constant olume. For a datomc deal gas C = R, so V Δ Ent = nrδ = ( 4.00mol )( 8.J/mol.K )( 60.0 K ) = 4.99 0 J. (c) Accordng to the rst law o thermodynamcs, ΔE nt = Q W, so W = Q Δ E nt = 6.98 0 J 4.99 0 J =.99 0 J. (d) he change n the total translatonal knetc energy s Δ K = nrδ = ( 4.00mol )( 8.J/mol K )( 60.0 K ) =.99 0 J.
Problem 9-6 In the ollowng, C = R s the molar secc heat at constant olume, C = R s the molar V secc heat at constant ressure, Δ s the temerature change, and n s the number o moles. he rocess takes lace at constant olume. (a) he heat added s Q= ncv Δ = nrδ = (.00mol )( 8.J/mol K )( 600K 00K ) =.74 0 J. (b) Snce the rocess takes lace at constant olume, the work W done by the gas s zero, and the rst law o thermodynamcs tells us that the change n the nternal energy s Δ E = = nt Q.74 0 J. (c) he work W done by the gas s zero. he rocess s adabatc. (d) he heat added s zero. (e) he change n the nternal energy s Δ Ent = ncv Δ = nrδ = (.00 mol )( 8.J/mol K )( 4K 600 K ) =.8 0 J. () Accordng to the rst law o thermodynamcs the work done by the gas s W = Q Δ E nt =+.8 0 J. he rocess takes lace at constant ressure. (g) he heat added s (.00 mol)(8.j/mol K)(00K 4K). 0 J. Q= ncδ = nrδ = = (h) he change n the nternal energy s (.00 mol)(8.j/mol K)(00 K 4K).9 0 J. Δ Ent = ncv Δ = nrδ = = () Accordng to the rst law o thermodynamcs the work done by the gas s
W = Q Δ E nt =. 0 J +.9 0 J =.9 0 J. (j) For the entre rocess the heat added s (k) he change n the nternal energy s (l) he work done by the gas s Q =.74 0 J + 0. 0 J = 0 J. Δ E nt =.74 0 J.8 0 J.9 0 J = 0. W = 0 +.8 0 J.9 0 J = 0 J. (m) We rst nd the ntal olume. Use the deal gas law V = nr to obtan nr (.00mol)(8.J / mol K)(00 K) V = = = (.0 0 Pa).46 0 m. (n) Snce s a constant olume rocess, V = V =.46 0 m. he ressure or state s nr (.00 mol)(8. J / mol K)(600 K) = = = V.46 0 m.0 0 Pa. hs s aroxmately equal to.00 atm. (o) s a constant ressure rocess. he olume or state s V nr (.00mol)(8.J / mol K)(4K) = = =.0 0 Pa.7 0 m. () he ressure or state s the same as the ressure or state : = =.0 0 Pa (.00 atm) Problem 9-6 γ γ (a) We use = V to comute γ: ( ) ( V V) ( ) 6 ( ) ln ln.0atm.0 0 atm γ = = =. ln ln.0 0 L.0 0 L
hereore the gas s monatomc. (b) Usng the gas law n rato orm, the nal temerature s ( ) ( )(.0 0 atm.0 0 L ) 6 (.0atm)(.0 0 L) 4 = = 7K =.7 0 K. (c) he number o moles o gas resent s n R (.0 0 Pa)(.0 0 cm ) 4 = = = 4. 0 mol. ( 8. J/mol K)( 7K) (d) he total translatonal energy er mole beore the comresson s K = R = ( 8. J/mol K )( 7K ) =.4 0 J. (e) Ater the comresson, K 8. J/mol K.7 0 4 K.4 0 R J. ( )( ) = = = () Snce rms, we hae rms, 7K 4 rms, = = = 0.00..7 0 K Problem 9-78 (a) In the ree exanson rom state 0 to state we hae Q = W = 0, so ΔE nt = 0, whch means that the temerature o the deal gas has to reman unchanged. hus the nal ressure s 0 0 0 0 = = = 0 = = 0.. V.00V.00.00 0 0 (b) For the adabatc rocess rom state to we hae V γ = V γ, that s, (.00V ) γ = (.00).00 0 0 0V γ 0 whch ges γ = 4/. he gas s thereore olyatomc.
(c) From = /nr we get K = = = (.00) / =.44. K Problem 9-87 For conenence, the nt subscrt or the nternal energy wll be omtted n ths soluton. Recallng Eq. 9-8, we note that E = 0, whch ges cycle Δ E +Δ E +Δ E +Δ E +Δ E = 0. A B B C C D D E E A Snce a gas s noled (assumed to be deal), then the nternal energy does not change when the temerature does not change, so Δ E =Δ E = 0. A B D E Now, wth ΔE E A = 8.0 J gen n the roblem statement, we hae Δ E +Δ E + 8.0 J = 0. B C C D In an adabatc rocess, ΔE = W, whch leads to and we obtan ΔE C D =.0 J..0 J +Δ + 8.0 J = 0, EC D