Homework 3 Answer (Due 02/04/2011 Friday) P2.3, P2.7, P2.8, P2.11, P2.16, P2.19, P2.20, P2.21, P2.23, P2.25, P2.26

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1 Homework Answer (Due 0/04/0 Frday) P., P.7, P.8, P., P.6, P.9, P.0, P., P., P., P.6 P.).00 moles o an deal gas are comressed sothermally rom 60.0 to 0.0 L usng a constant external ressure o.00 atm. Calculate q, w, U, and H. he work aganst a constant ressure: w external 4 0 Pa 0 0 m 60 0 m.00 U 0 and H 0 snce 0 and H U P q w.00 4 P.7) For.00 mol o an deal gas, P external = P = Pa. he temerature s changed rom 00.0 C to.0 C, and C,m = /R. Calculate q, w, U, and H. U n C mol K mol 98 K - 7 K 9, m H n C, m,m - - n C R.0 mol K mol 98 K - 7 K 9 q P H 9 w U - q P ( 9 ) ( 9 ) 64 P.) Calculate H and U or the transormaton o.00 mol o an deal gas rom 7.0 C and.00 atm to 7 C and 7.0 atm C P,m K n unts o K mol For an deal gas, H s gven by:

2 ΔH n C d n 600 K K, m d 00 K 600 K K 00K K ΔH ΔU ΔH Δ ΔH n R Δ mol mol K 00 K 9.40 P.) Calculate w or the adabatc exanson o.00 mol o an deal gas at an ntal ressure o.00 bar rom an ntal temerature o 40. K to a nal temerature o 00. K. Wrte an exresson or the work done n the sothermal reversble exanson o the gas at 00. K rom an ntal ressure o.00 bar. What value o the nal ressure would gve the same value o w as the rst art o ths roblem? Assume that C P,m = /R. w ad U n C R - - mol K mol 0 K.870,m w reversble n R ln and ln w n R reversble mol8.447 K mol 00 K w reversble ln n R bar P.) An deal gas undergoes an exanson rom the ntal state descrbed by P,, to a nal state descrbed by P,, n (a) a rocess at the constant external ressure P and (b) n a reversble rocess. Derve exressons or the largest mass that can be lted through a heght h n the surroundngs n these rocesses.

3 wmgh m gh w mgh nr ln nr m ln gh Alternate answer: mg ext A mg ext Ahmgh A m ext gh I w s the average weght to be lted m g w Ah m gh nr ln A m nr ln gh I m s the nal weght mg Ah A m A g gh P.9).0 moles o an deal gas are exanded rom 40. K and an ntal ressure o.00 bar to a nal ressure o.00 bar, and C P,m = /R. Calculate w or the ollowng two cases: a. he exanson s sothermal and reversble. b. he exanson s adabatc and reversble.

4 Wthout resortng to equatons, exlan why the result or art (b) s greater than or less than the result or art (a). a) Calculatng the ntal and nal volumna: n R n R.0 mol K mol 40 K.000 Pa.0 mol K mol 40 K.000 Pa w or an sothermal, reversble rocess s then gven by: nal n R ln ntal w k m.0 mol K mol 40 K ln 0.06 m w b) For an adabatc, reversble rocess: nal ln, where γ C P, m/c, m equaton.4 ntal ln nal ntal nal - ln ntal hereore: nal - ln ntal nal ntal nal ntal - ln - ln ln ln ntal ntal nal nal ntal γ - ntal nal ntalex ln γ nal Wth C P, m R, and C,m R, the nal temerature s: bar nal 40K Ex 0.4 ln 6K bar And nally w or an adabatc rocess and or. moles o gas: w q nc (.mol) (8.447K mol ) (6K 40K) k Less work s done on the surroundngs n art b) because n the adabatc exanson, the temerature alls and thereore the nal volume s less that that n art a). nal m m 4

5 P.0) An deal gas descrbed by = 00. K, P =.00 bar, and = 0.0 L s heated at constant volume untl P = 0.0 bar. It then undergoes a reversble sothermal exanson untl P =.00 bar. It s then restored to ts orgnal state by the extracton o heat at constant ressure. Dect ths closed-cycle rocess n a P dagram. Calculate w or each ste and or the total rocess. What values or w would you calculate the cycle were traversed n the ooste drecton? 0 [bar] [L] Frst we calculate the number o moles: R.00 bar 0.0 L n L bar K mol 00 K 0.40 mol he rocess n the dagram above s descrbed by the stes: ste :,,,, ste :,,,, ste :,,,, In ste (,,,, ) w = 0 snce stays constant In ste (,,,, ) we rst calculate :

6 0.0 bar 00 K bar K hen the work s: w n R ln n R ln 0.0 bar.00 bar mol K mol 000 K ln.0 0 In ste (,,,, ) we rst calculate : and 0 00L And the work: 0 Pa.00 bar 0 m 0 L 00 L w externmal bar L And or the entre crcle: w cycle w w w I the cycle were traversed n the ooste drecton, work o each ste has the same value wth ooste sgn. P.) A ellet o Zn o mass 0.0 g s droed nto a lask contanng dlute H SO 4 at a ressure o P =.00 bar and temerature o = 98 K. What s the reacton that occurs? Calculate w or the rocess. he chemcal equaton or the rocess s: Zn (s) H SO 4 (aq) Zn (aq) SO 4 (aq) H (g) - Frst we calculate the volume o H that s roduced: 6

7 n H H n Zn nh R P 0g 6.9gmol - 0.9mol 0.9mol 8.447K 0 Pa mol 98K.79 0 m Assumng that H w ext, the work s: - 0 Pa.79 0 m - 79 P.) One mole o an deal gas, or whch C,m = /R, ntally at 0.0 C and Pa undergoes a two-stage transormaton. For each o the stages descrbed n the ollowng lst, calculate the nal ressure, as well as q, w, U, and H. Also calculate q, w, U, and H or the comlete rocess. a. he gas s exanded sothermally and reversbly untl the volume doubles. b. Begnnng at the end o the rst stage, the temerature s rased to 80.0 C at constant volume. 6 a) Pa w n R ln mol K mol 9. K ln.69 0 U H 0 because 0 q w.69 0 b) constant volume, then 6, thereore Pa K K Pa U n C, m mol K mol K - 9 K 748 7

8 w = 0 because = 0 q = U = 748 H n C, m n,m R For the overall rocess: - - C.0 mol K mol K - 9 K. 0 q w U H P.6) One mole o an deal gas, or whch C,m = /R, ntally at 98 K and.00 0 Pa undergoes a reversble adabatc comresson. At the end o the rocess, the ressure s Pa. Calculate the nal temerature o the gas. Calculate q, w, U, and H or ths rocess Pa Pa

9 . 98 K 749 K q = 0 or an adabatc rocess. w U n C, m mol K mol 749 K - 98 K H U U R K mol 749 K-98 K P.6) One mole o an deal gas, or whch C,m = /R, ntally at 98 K and.00 0 Pa undergoes a reversble adabatc comresson. At the end o the rocess, the ressure s Pa. Calculate the nal temerature o the gas. Calculate q, w, U, and H or ths rocess Pa Pa. 98 K 749 K q = 0 or an adabatc rocess. w U n C, m mol K mol 749 K - 98 K H U U R K mol 749 K-98 K P.9) A cylndrcal vessel wth rgd adabatc walls s searated nto two arts by a rctonless 9

10 0 adabatc ston. Each art contans 0.0 L o an deal monatomc gas wth C,m = /R. Intally, = 98 K and P =.00 bar n each art. Heat s slowly ntroduced nto the let art usng an electrcal heater untl the ston has moved sucently to the rght to result n a nal ressure P = 7.0 bar n the rght art. Consder the comresson o the gas n the rght art to be a reversble rocess. a. Calculate the work done on the rght art n ths rocess and the nal temerature n the rght art. b. Calculate the nal temerature n the let art and the amount o heat that lowed nto ths art. he number o moles n each art s gven by: mol.0 98 K mol K L bar L.00 bar R n - a) We rst calculate the nal temerature n the rght sde: bar.00 bar.4 98 K 667 K

11 w ΔU n C Δ.0 mol K mol 667 K - 98 K b) Frst we calculate the volume o the rght art: r n R -.0 mol bar L K mol 667 K r r 7.0 bar 4.9 L hereore 000 L 4.9 L 8. L l, and 7.0 bar8. L 800 K n R.0 mol bar L K mol l l l - ΔU n C Δ.0 mol K mol 800 K - 98 K From art a) w = q = U w = P.4) One mole o N n a state dened by = 00. K and =.0 L undergoes an sothermal reversble exanson untl =.0 L. Calculate w assumng (a) that the gas s descrbed by the deal gas law and (b) that the gas s descrbed by the van der Waals equaton o state. What s the ercent error n usng the deal gas law nstead o the van der Waals equaton? he van der Waals arameters or N are lsted n able.. a) For an deal gas: w reversble n R ln.0 L.0 L mol K mol 00 K ln.4 0 b) For a van der Waals gas:

12 w external d R a m b m d R d m b he rst ntegral can be solved by substtutng y = m - b: a m d R d m b y y R dy R ln y hereore, the work s gven by: b ln b w reversble n R ln mol K mol 00 K 0 Pa.66 L bar bar b a b 0 6 m 6 L ln.0 L L.0 L L m.0 0 m.0 And the ercent error s: ercent error %

13 . Calculate dy/dx or the ollowng y: (a) y =ax+b; (b) y =x ; (c) y = x + (x+a) - (d) y = ln(x/); (e) y = C ex(-ax); () Acos(x) (g) Asn(bx) +Ccos(x/b) (h) y = x ex(-ax ) a) dy d(ax+b) a dx dx b) dy d(x ) x 4x dx dx c) ( ) - - dy d(x xa ) d x d( xa ) x ( ) xa x 6 x a dx dx dx dx d) x x d ln( ) d ln( ) dy a a a dx dx x a x x ad a e) dy d[ C ex(- ax)] Cd[ex(- ax)] ac ex( ax) dx dx d (- ax ) a ) dy d[ Acos( x)] Ad[cos( x)] Asn( x) dx dx d( x) g) dy d[ Asn( bx) C cos( x / b)] d[ Asn( bx)] d[ C cos( x / b)] Abcos( bx) C/ bsn( x/ b) dx dx dx dx h) dy d[ x ex(- ax )] ( ) [ex(- )] [ex( d x d ax d ax )] ex( ax ) x ex( ax ) x x dx dx dx dx d( ax ) ax xex( ax ) ax ex( ax ) xex( ax )( ax ) { d( uv) udv vdu }. Calculate dy/dx rom the ollowng equatons (a) ln(x) + y = xy (b) = cos(x)y

14 a) ln( x) ln( x) y xy y x dy dx ln(x) d d[ln(x)] (x -) - ln(x) x - dx dx (x -) d(x -) dx ( x ) ln( x) ln( ) x x x x ( x ) x( x ) { u vdu udv d( ) } v v b) cos( x) y y cos( x) d cos(x) dy cos(x) 0 - (-snx) sn( x) dx dx (cosx) (cos x) mc. P ( C) 4 C ex / Z k. (a) Calculate dp(c)/dc. (b) Show there are two solutons o dp(c)/dc = 0 (C 0) and one soluton gves the most robable seed, C =(R/M) /, where m s the weght o a gas molecule, M s ts molar mass, and k s the Boltzmann constant. a) 4 P( C) C Z mc ex k mc d ex( ) dp( C) 4 k mc d( C ) 4 C ex C dc Z dc k dc Z 4 mc mc mc ex C ex Z k k k 4 mc mc C ex Z k k b) dp( C) 4 mc 0 ex mc C dc Z k k 0 mc d ex( ) k mc mc d( C ) ex mc k k dc d k 4

15 C 0 or mc k Nk A R R 0C C k m N Am M M R C C 0 M / 4. Calculate / x and / y or the ollowng (x, y, z): (a) = xy - (b) = sn(xy) (c) = ln(x y) (d) = yx 4 + ex(-xy ) a) x y y x x x x y x y ) x y y y y b) snxy snxy y cos(xy) x x xy y snxy snxy x cos(xy) y y xy x c) ln xy ln xy y x x xy xy x y ln xy ln xy xy y y xy xy y xy d) 4 4 yx yex( xy) yx y ex( xy) 4yx x x x x yexxy yx y ex( xy) yx y ex( xy) x y y ex xy xy ex xy y y y y 4. Calculate C v = ( U/ ) and C = { (U + P)/ } P or n mole o deal gas assumng that U = nr/ and n s a constant. What knd o molecular roertes do they deend on? U nr

16 C nr U nr C P 7 nr nr nr U P 7 nr P C and C P deend on moles n. P P 6