LEEL ROBLEMS rblem Set Slutins. MI ressr Gerbrand Ceder Fall rblem. Gas is heating in a rigid cntainer rm 4 C t 5 C U U( ) U( ) W + Q (First Law) (a) W Since nly wrk is pssible & since the cntainer is rigid. Q U (. + C ) (. + C) Q U. ( ) 46.5 k kg (b) Entrpy change: since W r,. d ds. d S. ln ds Q Q du. d. k kgk (r reversible heating)
rblem. Calculate entrpy change by ging rm t accrding t: ( ) A reversible isthermal cmpressin ( ) Fllwed by an isbaric heating Entrpy change Entrpy change U ( U U( ) ) nr Q W r Q pd d Q nr ds d S nrln nr ln 6. 5 mle K Q c d Cp n R Q cd p ds nr d S nrln 5. 59 mle K p (isbaric heating) Nw r the ttal prcess tt S S + S. 94 mle K
rblem. (a) 9 C C H c d + H + c d C mlek j mlek 9C HFe 4 (9 C C) 9 + 8 ( mle K mle mle K C 9 C ) (b) System Fe+ice water is adiabatic H H ( m H ), m mass ice transrmed m Fe H H p, H H F e Fe H F e ice melting ice ice Fe melting icewater 495 g 648 ice -4.95k m g p, ml 56g ml 8g rblem.4 Diatmic ideal gas c R and c R p v 5
Assume the prcess is reversible. here are tw methds ne can use t ind the wrk dne alng this path. Fr bth appraches it will be useul t ind p,, r the starting and ending pints. We are given the starting pressure, vlume and temperature and the inal pressure. Frm this we can calculate the inal vlume and temperature. Methd I Start with cp cv p p 6. 99m he inal temperature can be und with a similar relatin c c cv R c c p p S, d w pd p ) w ( ( ) ) ( ) ( ) ( ) Inserting in the values r,, we get w. M Methd II 5 p p (with ) Start with the knwn act the the internal energy an ideal gas is nly a unctin temperature. U Q+ W nc d w nc ( ) v (with c ) K It will als be useul t calculate the number mles rm p nr giving n 65 mles. In summary the cnditins at each pint are with p ( a) 595 ( m ) 6. 99 ( K) 98 p Hwever, since this is an adiabatic path, Q and the wrk dne is simple nc d p w. M v v cp cv 5 5 4
his prcess is at cnstant pressure, s the wrk is given by w pd p d w p ( ) We just have t ine. hat can be dne easily by using the equatin state an ideal gas alng an istherm S, w a 5 6. 99m 5m w 89k he ttal wrk dne is then w tt + 89-849k rblem.5 Functin : dy ( x + y ) dx + x( x + y ) dy () Integrate alng y x ( dy dx) x( x + x ) dx+ x( x + x ) dx 4 x (4 x + x ) dx x dx 4 () Integrate alng y x ( dy xdx) 4 4 x ( x + x ) dx + x( x + x )xdx 6 4 5 x (5 x + 5 x ) dx x + 5 Functin : dy ( x + y) dx + x( x + y) dy () Integrate alng y x ( dy dx) x( x + y) dx+ x( x + x) dx 4 x x (4x + x) dx 4 4 + () Integrate alng y x ( dy xdx) x ( x + x ) dx + x( x + x )xdx 4 x xdx 5 5 4 Since unctin is path independent it is an exact dierential. 5
rblem.6 du w + Q his is an ideal gas s du alng an isthermal path. (a) R w pd, p R w ln ln d R p R p w R ln(5) 654 (b) (c) du Q w Q 654 H U + H U + Fr an ideal gas at cnstant temperature ( ) U U Als, when herere H H (d) Q pd nr ds d S Rln R ln S. 5 K 6
rblem. (a) Wrk dne in the prcess is: Deine water as the system S we write w w inal initial vapr liq vapr w p d (Since p is cnstant) liq 5 pd w p( ) vap p a w p vap vapr. 59 his is abut.5% the heat evaprtatin (b) g, Initial State Liquid Final State apr kg m pd Since the speciic vlume a vapr is much larger than r a liquid liq U H p U 6 6 94 g g 654 6 kg H U + p H U + ( p ) U + p Cnclusin: he heat ne has t transer t water t evaprate it is partly used r increasing the internal energy water (94 breaking bnds) but asl r the wrk required by the vapr expansin. g g rblem.8
Given: N p atm m liter. m liters. m K Need t knw hw many mles gas: (isthermal) (isbaric) w p n. mles R N p p p atm 565 m nr pd d nr ln w nr ln. 9 p p cp 5 with c r mnatmic ideal gas v p. 5 liters. 5m p N w p 565. 5. m m w 4. (reversible adiabatic) w pd w p p p p [ ] p p d w. S the ttal wrk dne is ttal w w + w + w. 9 + 4. +. ttal w 9. 5 p p rblem.9 (i) he amunt wrk dne by the gas is zer, since the gas des n wrk n the surruindings utside the chamber. he expressin w pd cannt be applied since the prcess is nt reversible. (ii) he walls the chamber are insulating, thus Q and rm (i) w. hus U Q+ W (iii) Fr an ideal gas U is nly a unctin temperature. Since r a ree expansin in an insulated chamber U initital inal (iv) Fr a nn-ideal gas where U (, ), will change ater a ree expansin in an insulated chamber since changes and U U (, ) U (, ) inal inal initial initital 8
(Mst gasses at lw pressures can be well apprximated as being ideal gasses) (v) I nw the walls the chambers cnduct heat, r an ideal gas U U( ) is still true. he initial and inal states are in equilibrium with the envirnment. hus initial inal envirnment (vi) Fr nn-ideal gases U since U U(,) and changes. LEEL ROBLEMS rblem. (a) S (b) S> (c) S> rblem. Given inrmatin i i m? i 5 atm atm 98 K 98K (a) Find the inal vlume ater the expansin i i i i 5 5. m Finding the number mles, n, will be useul r parts b & c L atm R. 85, m L K mle 5 n R. 85 98 n 6. 5 mles (b) Find the wrk dne i the prcess is isthermal w pd. w nr ln 8. 4 6. 5 98 ln 5 i w 6 k (c) Find the wrk in the multi-step prcess 9
w parts: (I) Adiabatic expansin rm t (II) Heating at cnstant rm t ( wi + wii)+( QI + QII) Usy s Since state and are at the same temperature and U r an ideal gas is nly a unctin temperature, Usys and ( wi + wii) ( QI + QII) S we can calculate w s r Q s. Since QI (adiabatic prcess) let s calculat QII QII ncp ncp( ) n 6. 5 mles, cp R 9 mle K Need t ind R cp 98 68K 5 Nw r QII Q II 6. mles 9 (98 68 K) mle K Q k S, II w ( Q + Q ) ( + k) ttal I II w ttal k rblem. U Ap ( A psitive cnstant) du w + Q Alng a reverisble path, W pd when nly p- wrk is pssible. Alng a reversible adiabatic path, Q and therere du w (First Law)
Ging back t equatin Ater intergratin U U du dp + d p p U Ap p U Ap p ( Ap ) dp + ( Ap ) d pd A dp ( Ap +) d dp d ( Ap +) A ln( Ap + ) ln + cnstant [ A ] A ln ( Ap + ) new cnstant () ( Ap +) even newer cnstant rblem.4 (a) Using the irst law Usys Q(5 C) + Q( C) + W + Q(5 C). 5 MW +. 5MW MW + Q(5 C) Q(5 C) MW (b) Chack t see i the secnd law is satisied ds machine Q But ds machine has t be zer since we are perating at steady state Q(5 C) Q C Q C + ( ) + (5 ) 5 98 5. MW. MW MW + 5 54. 6 K 5K 98K K s Ks 54. 6
rblem.5 (a) Since the prcess is adiabatic, Q (b) U Q+ w U ncv w 5 ( mles ) 8. 4 8 mle K 85K 88K S S +S S since this prcess is reversible and adiabatic S S isat cnstant pressure, temperatre varies p What is thugh? he temperatre that wuld be reached i expansin was reversible. R cp p 5K p 88 S c d p 88 8. 4 ln 5 5 S. 4 mle K rblem.6 In case (b) extra wrk is dne n the gas s that its internal energy will remain higher than in (a) > b a rblem. (a) tal amunt energy required Necessary heat input is the ttal enthalpy change the material (since p is cnstat) Determined by the inal and initial state
Enthalpy change is nly due t the L (the 5 L remains in exactly the same state) dh nc d g cal H L 5K L g K H 5 kcal 4k H. 8kWh (b) Entrpy change Again,entrpy change is sley determined by the inal and initial state ds p nc d p g cal S L L g K S 999 ln 5 8 mlk 8. kcal K rblem.8 d w pd R Rln 4 w R ln he secnd step where Q gives du dw and the urth step hus and Bd pd R d B ln R ln B ln R ln 4 r 4 4
w 4 w R ln R ln w( ) w( ) Q ( ) Q( ) w w w w + w rblem.9 Q By drawing the heat Q rm a heat suce at ne may ths prduce mechanical wrk in the amunt i the rest the energy, can be dispsed as heat t a heat sink at Q( ) Suniverse Ssys +Senvirnment here is n heat lw int the envirnment, s Senvirnment But r the system (the gas) he gas ges rm state at i, i, i t a state at,, ( i) with < i cmpute the Sgas we need t hw the entrpy changes with pressure (r vlume) at cnstant S (Maxwell Relatin) and r an ideal gas R Sgas R Rln i i since i > i R ln > Since S > the prcess is irreversible universe rblem. calculate U(,) r any (,) given a value U(,) at sme (,) it is sitable r this system t llw the path sketched belw. 4
First g rm,, A cnstant clume, i nly - wrk is pssible du Q U U(, ) U(, ) Q A( ) is unknwn, but we d knw because the path we chse rm (,) (,) that (, ) and (,) are cnnected by a reversible adiabat. herere, r S, (i.e. cnstant vlume) Secnd g rm,, Alng the reversible adiabat Q and therere U U(,) U(, ) d U, ( ) U, ( ) Ar ( ) (with r ) with and representing values alng the adiabat ging thrugh (, ). calaculate this integral use S, Cmbining the irst and secnd step U d U ( ( ) ) ( ) ( ) ( ) U U(,) U(, ) r (with r ) ( ) U U(,) U(, ) A( r )+ r (with r ) tt ( ) LEEL ROBLEMS rblem. he reactin at K is nr NH N +H, and are cnstant but nincreases by a actr in the reactin. herere atm i (b) Heat lw he system is undercnstant vlume ( d ) s W and U Q We can cmpute U rm H because U H 5
U H ( ) H ( nr ) H R n k H 8 and n mles mle U 8 () 8. 4 ( K) mle mle K (c) System is adiabtic and under cnstant vlume U S, use c c R p v a U + ( c, + c, ) d reactin a U Q 646 a v N v H U c, +c, vn vh 646 4 4. 686 a 5K mle reactin mle mlek he system cls dwn which is expected given that the reactin is endthermic. rblem. We can deine the system as the gas in the tank ( n mles) and the gas that will be squeezed int the tank ( n mles). It will helpul t deine sme variables c he initial pressure in the tank he external pressure pushing gas in he initial temperature the gas he inal temperature the gas (what we want t slve r) he vlume the tank he vlume the gas pushed in We can write the llwing relatinships n R n + nc R c nr c Since this is an ideal gas we knw that the internal energy change is nly a unctin temperature, given by Givem that the prcess is adiabatic (islated), U ( n + n ) c ( ) c v But r this case we knw that U Q+ W W ( Q ) W n R c c 6
r the external pressure (which is cnstant) times the vlume gas pushed in. Nw we can just wrk thrugh eliminating variables t ind the inal temperature. nr ( n + n) c( ) c c v nr c v ( ) R c R v ( ) R R R c R R c ( ) v ( R+ cv) ( R+ cv) ( R+ c ) R v + c v reating air as a diatmic ideal gas, we can use cv R and slve r 98 ( R + R ) 98 (. 5 R+ cv) 6 4. 6K 5 rblem. Assume that the vlume change resulting rm the reactin A + B AB is and the heat reactin is Q. During the reactin, the system will perrm wrk against the envirnment which has a cnstant pressure. Since the pressure the envirnment is cnstant, the external wrk dne by the system is w Using the irst law U Q Q U + In the initial and inal states the reactin (bth equilibrium states), the pressure the gas has t equal (therwise they wuldn t be equilibrium states). herere i is the pressure the gas r rblem.4 Q U +( ) Q H where H is the dierence enthalpies in the initial and inal states. Clsed system slutin System is gas lwing int the tank Fr an ideal gas, U U Q+ w w ( Q (adiabatic)) U U U U nc ( ) v nr
S, Open system slutin System is the tank U nc ( ) nr v c ( R+ c) c c v v p p c v Hm U U U m initial i i i because m m in m U U initial H U U + i in in in c ( ) R cp c i i in in v i i v rblem.5 (a) Gas in tank I is underging an adiabatic reversible expansin R cp R 5 R 46K (b) ry t get II rm the act that Utt UI + UII because the ttal system is isthermal. UI is the internal energy change gas that remains in I and UII is the internal energy change gas that ends up in tank II. U n c ( ) need t ind n I nw rm U + U I II II II v II U n c ( ) I I v I n R ni I atm K 5 6. n I atm 46K n n n (. 6). 9n II I UII UI. 9 n c ( ). 6 n c (46 ) v II v II 84K (c) Entrpy change S varies with and 8
S S ds d + d S cp S R and cp ds d R d Fr the gas remaining in tank I dsi as it is underging an adibatic reversible expansin. Fr the gas ending up in tank II II II S nii cp ln R ln atm, K LEEL 4 ROBLEMS n 9. (per mle gas initial in tank I) II II 84K niirii nii II II 5atm n 9 ln 84 5 S. n R Rln S 5. 5n Kmle rblem 4. his is a very tricky questin which is why it is alne in Level 4. First let s determine the equilibrium cnditins. Assume that the system has sme hw reached equilibirum with inal values,, r cmpartment and,, r cmpartment. We can ssk what criteria must there variable satisy r the system t be in equilibrium. Well we knw that r any reversible change arund equilibrium du du + du i.e. U,S ( ) must be minimal at equilibrium, and ds ds + ds i.e. SU,S ( ) must be maximal at equilibrium. Since there is n heat lw between cmpartments and between the chamber and the envirnment (since all walls are adiabatic) ds ds d S at equilibrium, the nly reversible changes we can make are in the vlume the cmpartments, i.e. hen we can write (r a reversible change d du d + d du d 9
d du du+ du ( ) d since this shuld hld r arbitrary d, i.e. equilibrium is characterized by mechanical equilibrium. NOE: Since ds and ds, the equilibrium cnditin that du des nt yield any inrmatin abut the inal temperatures and. his means that the inal temperatures this system depend n the path llwed by the system t attain equilibrium. he questin remains: can we btain equilibrium rm the initial state,,, and n n by llwing a reverible path? he answer is: prbably nt. One candidate reversible path is t allw the pistn t very slw mve until the equilibrium pressures are attained (i.e. ). his path, hwever, cannt prceed reversible r the llwing reasn: Assume that is can prceed reversibly. hen an ininitesimal change in the vlume cmpartment by d will result in an increase the internal energy cmpartment ne by du he change in internal energy cmpartment is du d d But since bth cmpartments are adiabatically sealed rm the envirnment, du du du r d d r But this is nt true except in the inal equilibrium state. In the initial state by assumptin. herere the suggested path cannt prceed reversibly. he suggest path must be accmpanied by dissipatin (thrugh rictin between the pistn and the walls r example). his dissipatin will be accmpanied by a prductin heat. We cannt predict a-priri hw much that heat evlves t cmpartment r cmpartment thugh.