8.2. THE EQUATION OF STATE FOR A PERFECT GAS

Transcription

1 8 del nd Rel Gses 8.. ntroduction. 8.. he eqution of stte for erfect gs surfce of n idel gs nternl energy nd enthly of erfect gs Secific het ccities of n idel gs Rel gses Vn der Wls eqution Viril eqution of stte Bettie-Bridgemn eqution Reduced roerties. 8.. Lw of corresonding sttes. 8.. Comressibility chrt. Highlights Objectie ye Questions heoreticl Questions Unsoled Problems. 8.. NRODUCON An idel gs is defined s gs hing no forces of intermoleculr ttrction. he gses which follow the gs lws t ll rnges of ressures nd temertures re considered s idel gses. Howeer, rel gses follow these lws t low ressures or high temertures or both. his is becuse the forces of ttrction between molecules tend to be ery smll t reduced ressures nd eleted temertures. An idel gs obeys the lw R. he secific het ccities re not constnt but re functions of temerture. A erfect gs obeys the lw R nd hs constnt secific het ccities. A erfect gs is well suited to mthemticl mniultion nd is therefore most useful model to use for nlysis of rcticl mchinery which uses rel gses s working substnce. n relity there is no idel or erfect gs. At ery low ressure nd t ery high temerture, rel gses like hydrogen, oxygen, nitrogen, helium etc. behe nerly the sme wy s erfect gses. hese gses re clled semi-erfect or ermnent gses. he term semi-erfect hs the imliction tht the behiour of the gses re nerly the sme s tht of erfect gs. he term ermnent ws used for these gses by erlier chemists who thought tht these gses did not chnge their hse (i.e., did not condense to liquid stte). Hence they re clled ermnent gses. here is no gs which does not chnge hse, nd there is no ermnent gs in the rel sense. Howeer, these gses cn be chnged into liquid hse only if they re subjected to gret decrese in temerture nd increse in ressure. All gses behe in nerly in similr wy, esecilly t ressures considerbly lower thn the criticl ressure, nd t temertures boe the criticl temerture. he reltion between the indeendent roerties, such s ressure, secific olume nd temerture for ure substnce is known s the eqution of stte. or engineering clcultions, the eqution of stte for erfect gses cn be used for rel gses so long s the ressures re well below their criticl ressure nd the temertures re boe the criticl temerture. 8.. HE EQUAON O SAE OR A PEREC GAS Boyle s lw. t sttes tht olume of gien mss of erfect gs ries inersely s the bsolute ressure when temerture is constnt. 376 C\HERMAL\HER7-

2 DEAL AND REAL GASES 377 f is the bsolute ressure of the gs nd V is the olume occuied by the gs, then V or V Constnt, so long s the temerture is constnt...(8.) ig. 8. shows the grhicl reresenttion of Boyle s lw. he cures re rectngulr hyerbols symtotic to the - xis. Ech cure corresonds to different temerture. or ny two oints on the cure, V V...(8.) (Pressure) < < 3 3 V (Volume) ig V reltion of erfect gs t constnt temerture. Chrle s lw. t sttes tht if ny gs is heted t constnt ressure, its olume chnges directly s its bsolute temerture. n other words, V V or Constnt, so long s ressure is constnt...(8.3) f gs chnges its olume from V to V nd bsolute temerture from to without ny chnge of ressure, then V V...(8.4) ig. 8. gies the grhicl reresenttion of Chrle s lw. (Abs. temerture) º C < < 3 V (Volume) ig reltion of erfect gs constnt ressure. \M-therm\h8-.m5

3 378 ENGNEERNG HERMODYNAMCS o derie the eqution of stte for erfect gs let us consider unit mss of erfect gs to chnge its stte in the following two successie rocesses (ig. 8.3) (i) Process - t constnt ressure, nd (ii) Process - t constnt temerture. Constnt Constnt ig ormultion of eqution of stte of erfect gs. or the rocess -, lying Chrle s lw nd since, we my write or the rocess -, using Boyle s lw...(i) nd since i.e., Substituting the lue of from eqn. (ii) in eqn. (i), we get...(ii) or i.e., constnt...(8.5) \M-therm\h8-.m5

4 DEAL AND REAL GASES 379 he mgnitude of this constnt deends uon the rticulr gs nd it is denoted by R, where R is clled the secific gs constnt. hen R he eqution of the stte for erfect gs is thus gien by the eqution R...(8.6) or for m kg, occuying V m 3, V mr...(8.7) f the mss is chosen to be numericlly equl to the moleculr weight of the gs then mole of the gs hs been considered, i.e., kg mole of oxygen is 3 kg oxygen, or kg mole of hydrogen is kg hydrogen. he eqution my be written s V 0 MR...(8.8) where V 0 Molr olume, nd M Moleculr weight of the gs. Aogdro discoered tht V 0 is the sme for ll gses t the sme ressure nd temerture nd therefore it my be seen tht MR constnt ; R 0 nd thus V 0 R 0...(8.9) R 0 is clled the molr or uniersl gs constnt nd its lue is kj/kg mol K. f there re n moles resent then the idel gs eqution my be written s V nr 0...(8.0) where V is the olume occuied by n moles t ressure nd temerture SURACE O AN DEAL GAS he eqution of stte of n idel gs is reltionshi between the ribles ressure (), olume (V) nd temerture (). On lotting these ribles long three mutully erendiculrs xes, we get surfce which reresents the eqution of stte ( R). Such surfce is clled -- surfce. hese surfces reresent the fundmentl roerties of substnce nd roide tool to study the thermodynmic roerties nd rocesses of tht substnce. ig. 8.4 shows ortion of -- surfce for n idel gs. Ech oint on this surfce reresents n equilibrium stte nd line on the surfce reresents rocess. he ig. 8.4 lso shows the constnt ressure, constnt olume nd constnt temerture lines. ig NERNAL ENERGY AND ENHALPY O A PEREC GAS Joule s Lw. Joule s lw sttes tht the secific internl energy of gs deends only on the temerture of the gs nd is indeendent of both ressure nd olume. i.e., u f() \M-therm\h8-.m5

5 380 ENGNEERNG HERMODYNAMCS Joule concluded this result from series of exeriments conducted with n rtus similr to the one shown in ig wo tnks connected by le were submerged in bth of wter. nitilly one tnk ws ecuted nd the other ws filled with ir under high ressure. A thermometer ws lced in the wter bth. After the tnk nd wter hd ttined the sme temerture, the le between the two tnks ws oened to ss ir slowly from high ressure tnk to the ecuted tnk. ime ws llowed for equilibrium to be ttined. Joule obsered tht there ws no chnge in temerture of wter during or fter the rocess. Since there ws no chnge in the temerture of wter, he concluded tht there ws no het trnsfer between ir nd wter i.e., δq 0. And since there ws no work during the rocess, i.e., δw 0, from the first lw of thermodynmics, δq de + δw, Joule concluded tht chnge in internl energy of the ir is zero, i.e., de 0. hermometer Ecuted tnk Air under high ressure Bth of wter Vle ig Artus for demonstrtion of Joule s lw. Agin, since both ressure nd olume chnged during the rocess, he remrked tht internl energy ws function only of temerture ; since during the rocess temerture did not chnge, the internl energy remined constnt. Lter on when exeriments were conducted with more refined instruments, it ws found tht there ws ery smll chnge in temerture of wter, indicting tht for rel gses internl energy ws not function of temerture lone. Howeer, t low ressure nd high temerture where rel gses behe like semi-erfect gses nd where the eqution of stte for semi-erfect gs, R, is sufficiently ccurte, Joule s lw holds eqully good in tht rnge. rom definition of enthly, h u + Also R h u + R...(8.) Since u is function of temerture only, h is function of temerture, i.e., h f()...(8.) 8.5. SPECC HEA CAPACES O AN DEAL GAS he secific het ccity t constnt olume of ny substnce is defined by c u K J \M-therm\h8-.m5

6 DEAL AND REAL GASES 38 t my be seen tht s Joule s lw for n idel gs sttes u f(), then c du d...(8.3) Since h u +, Boyle s lw, V f() nd Joule s lw u f() together show, h f() nd by similr rgument to the boe it my be seen tht : c dh d...(8.4) urther s h u +, then h u + R nd by differentition dh d du d + R Substitution from eqns. (8.3) nd (8.4) gies, c c + R i.e., c c R...(8.5) f exressed in terms of molr quntities then eqn. (8.5) becomes C C R 0...(8.6) where C nd C re molr secific het ccities. Equtions for secific het ccities of idel gses Since both u nd h re functions of temerture, the equtions to c nd c must lso be functions of temerture. hey re usully exressed in form : c + K + K + K 3...(8.7) c b + K + K + K 3...(8.8) where, b, K, K nd K re constnts. Vlues of secific enthly etc. re then obtined by integrtion REAL GASES t hs been obsered tht when exeriments re erformed t reltiely low ressures nd temertures most of the rel gses obey Boyle s nd Chrle s lws quite closely. But the ctul behiour of rel gses t eleted ressures nd t low temertures deites considerbly. he idel gs eqution R cn be deried nlyticlly using the kinetic theory of gses by mking the following ssumtions : (i) A finite olume of gs contins lrge number of molecules. (ii) he collision of molecules with one nother nd with the wlls of the continer re erfectly elstic. (iii) he molecules re serted by lrge distnces comred to their own dimensions. (i) he molecules do not exert forces on one nother excet when they collide. As long s the boe ssumtions re lid the behiour of rel gs roches closely tht of n idel gs VAN DER WAALS EQUAON Vn der Wls eqution (for rel gs) my be written s : + K J ( b) R...[8.9 ()] he constnts nd b re secific constnts nd deend uon the tye of the fluid considered, reresents the olume er unit mss nd R is the gs constnt. \M-therm\h8-.m5

7 38 ENGNEERNG HERMODYNAMCS mol). f the olume of one mole is considered then the boe eqution cn be written s K J + ( b) R 0...[8.9 (b)] he units of,,, R, nd b re s follows : (N/m ), (m 3 /kg-mol), (K) nd R 834 Nm/kg mol K, [Nm 4 /(kg-mol) ], b (m 3 /kg ble 8.. Constnts of Vn der Wls Eqution S.No. Substnce b Nm 4 /(kg-mol) m 3 /kg-mol. Hydrogen (H ) Oxygen (O ) Crbon dioxide (CO ) Helium (He) Air Wter (H O) our Mercury (Hg) our Vn der Wls eqution ws roosed in 873 for the gseous nd liquid sttes of fluid, nd ccounts qulittiely for mny imortnt roerties, but quntittiely it fils in mny rticulrs. he chrcteristic eqution for erfect gs is obtined by neglecting the finite size of the molecules. f this be tken into ccount it is obious tht the eqution must be modified, for the distnce trelled by molecule between two successie λ encounters will be less thn if the molecules were oint sheres. Let the erge distnce trersed by molecule between two successie encounters be denoted by λ, the men free th. n ig. 8.6 suose L nd M L M to be the two molecules of dimeter d t distnce λ rt. f these molecules were to iminge long the line of centres the th moed oer would be less by n d d mount d thn if the molecules were oint sheres. Now ll the encounters between molecules re not ig. 8.6 direct, so their men free ths will be lessened by n mount kd, where k is frction. ht is, the men free th is diminished in the rtio (λ kd) : λ or f the men free th is lessened in this rtio, the encounters er second will be incresed in the rtio : kd. But the ressure of the gs deends uon the encounters er second with the λ wll of the contining essel. Hence the new ressure is gien by 3 ρ C. kd λ (where ρ is the density nd C is the erge elocity). kd K J λ :....(8.0) \M-therm\h8-.m5

8 DEAL AND REAL GASES 383 he men free th is inersely roortionl to the density of the gs, for if the olume were hled, i.e., the density doubled, there would be twice s mny molecules in the sme sce, nd therefore ny molecule would only he to trel roximtely hlf s fr before encountering nother molecule. Hence writing for ρ nd b for kd λ b K J in eqn. (8.0), we get C 3 R or ( b) R...(8.) Next consider the forces of cohesion which ct between molecule nd those surrounding it. When the molecule is sufficiently fr remoed from the surfce of the gs in ll directions the resultnt of these cohesies forces re eqully robble, s the indiidul forces re rying continuously s the surrounding molecules chnge their ositions. Hence if the resultnt is erged oer sufficient length of time the ggregte force will be nil. his is not true, howeer, when the molecule is ner the surfce. Let the force from ech molecule be resoled into norml nd tngentil comonents. All directions for the resultnt in the tngentil lne re eqully likely, but the resultnt norml comonent is most often directed inwrds. Aerged oer sufficient length of time the totl resultnt force will therefore be norml force lwys directed inwrds. hus the erge effect of the cohesie forces is the sme s if there ws ermnent field of force cting t nd ner the surfce. his field of force cn be regrded s exerting ressure oer the boundry of the gs. he ressure is roortionl to the number of molecules er unit re ner the boundry surfce nd to the norml comonent of the force. Both of these fctors re roortionl to the density, so will be roortionl to the squre of the density. i.e., ρ...(8.) where is constnt. Hence the molecules re not deflected by imct lone on reching the boundry, but s the totl result of their imct nd of the ction of the suosed field of force. ht is, their chnge of momentum my be suosed to be roduced by totl ressure + insted of by the simle ressure. Hence eqn. (8.) now becomes : ( + )( b) R, or K J + ( b) R by substitution from from (8.) nd relcing ρ by Elution of constnts nd b : he generl form of the isothermls for crbon dioxide gien by Vn der Wls eqution is shown in ig hese cures re obtined from the eqution, K J ( 0.003) (8.3) 73 where the unit of ressure is the tmoshere, nd the unit of olume tht of the gs t 0 C under one tmoshere ressure.. \M-therm\h8-.m5

9 384 ENGNEERNG HERMODYNAMCS Vn der Wls eqution being cubic in hs three roots which my be either ll rel, or two imginry nd one rel, s imginry roots lwys occur in irs. n ig. 8.7, the 40 C isotherml corresonds to the first condition, nd the other isothermls to the ltter. here is one isotherml where there re three rel coincident roots t oint of inflexion. All the isothermls for temertures higher thn tht corresonding to the isotherml with three rel coincident roots he no horizontl tngent, nd ll those lower he mximum nd minimum. Consequently this cure is identified with the criticl isotherml. he temerture of the criticl isotherml is obtined in the following mnner. Eqution (8.9) my be written 3 b + R + K J + ( b) R Pressure, tm º C 0º C 0C º 0º C Perfect gs t 0 C / 0 ig Vn der Wls sotherml for CO b. 0...(8.4) b + b R 0 b + b R 0 Multilying both sides by, we get b + 3 b + b R R 0 + b 0 Now t the criticl oint, s the three roots re equl, the eqution must be of the form : ( c ) (8.5) where the suffix c denotes conditions t the criticl oint. or the criticl oint eqution (8.4) becomes 3 R c b + c + b 0...(8.6) c c Equtions (8.5) nd (8.6) re identicl, hence equting coefficients 3 c b + R 3 c c, c c, \M-therm\h8-.m5

10 DEAL AND REAL GASES c b, c nd from these by simle reduction, we he c 3b c 7b 8 c. 7 br U V W...(8.7) rom these equtions it follows tht the criticl olume, ressure, nd temerture re ll comletely determined by the constnts of eqution (8.9). he eqution (8.7) indictes the criticl constnts for rticulr gs nd leds to the following results : he lues of nd b re lso gien by 3 c c 9 8 R c c R c b c 3 R 8 c c nd R 8 c c...(iii) 3 c Using the lues of, b nd R in eqution (8.3), nd substituting in (8.6), we he for crbon dioxide c 6. tmosheres, c K or 3. C. t is frequently ssumed tht the roximte greement between the clculted nd exerimentl lues of the criticl temerture for crbon dioxide is sufficient erifiction of Vn der Wls theory, but the constnt b cnnot be clculted with the required degree of ccurcy from Regnult s exeriments to mke this n dequte test of the theory. Also from equtions (8.7), we he c c 3 Rc wheres exeriment shows tht bout 0.7 s the erge lue of this rtio, rying considerbly, howeer, from gs to gs. he Reduced Eqution : When the ressure, olume nd temerture of the fluid re exressed s frctions of the criticl ressure, olume nd temerture the reduced form of Vn der Wls eqution is obtined. hus, writing e e c 7b, n c 3nb, m c 8 7. m br nd substituting these lues in eqn. (8.9), this reduces to c...(i)...(ii) \M-therm\h8-.m5

11 386 ENGNEERNG HERMODYNAMCS K J 3 e + (3n ) 8m n n this reduced eqution the three constnts which chrcterised rticulr fluid he disered. he eqution is ccordingly true of ny substnce which stisfies n eqution of the Vn der Wls tye, nd the form of the cures connecting e, n nd m is the sme for ll these substnces. hus we see tht two substnces, the behiour of ech of which is reresented by Vn der Wls eqution, will be in corresonding sttes when the ressure, olume nd temerture re the sme multiles of their criticl lues. his theorem of corresonding sttes, enuncited by Vn der Wls, ws tested by Amgt nd found to be roximtely true for lrge number of fluids. he theorem of corresonding sttes is not unique to the eqution of Vn der Wls. Any eqution of stte giing criticl oint nd hing not more thn three constnts will sere eqully well to gie reduced eqution, in which the constnts eculir to ny one fluid diser, nd therefore become the bsis of the theorem of corresonding sttes. t must be remembered in lying the theorem tht the ccurcy of results deduced by its id cnnot be greter thn the ccurcy with which the originl eqution reresents the behiour of the fluids under considertion. Amgt s Exeriments As er Amgt s exeriments Vn der Wls eqution ccounts for the rition of the roduct with incresing ressure s follows. Writing eqution (8.9) in the form R, b nd differentiting with resect to, keeing constnt, we he d ( ) d R S Rb ( b) U V W Since the condition for minimum on ny isotherml is d( ) 0, d the right-hnd side of eqution (8.8) must nish t this oint. Now he s the condition for minimum : d...(8.8) d b K J d is neer zero, so we d Rb ( b) or R. b...(8.9) his eqution shows tht the olume t which the minimum lue of occurs on ny isotherml grdully increses s the temerture is rised. o find the locus of minim the temerture must be eliminted from eqution (8.9) by substitution from the originl eqution. hus from eqution (8.9) R + ( b), nd substituting this in eqution (8.8), we he K J \M-therm\h8-.m5

12 DEAL AND REAL GASES 387 which reduces to b b K J + ( b) + Multily ech side of this eqution by, nd ut y nd x, nd we obtin y b y b K J, + x or y( by) bx he boe exression gies the locus of minim nd is rbol with xis rllel to the x-xis s shown in ig Consider the isotherml which goes through the oint A. Here x 0 nd y b. Writing Vn der Wls eqution in terms of x nd y, we he x + y (y bx) R, nd substituting the coordintes of the oint A R b ig. 8.8 or...(8.30) br or temertures boe tht gien by eqution (8.30) the minim lie in the region of negtie ressure, so n Amgt isotherml for temerture equl to or greter thn will sloe br uwrds long its whole length for incresing lues of, but for temerture less thn isothermls first di to minimum nd then rise. Using the result from eqution (8.7) c 8 7 br, we see tht the limiting temerture for n isotherml to show minimum is br the 7 8. c he reson for Amgt finding no di in the isothermls for hydrogen is now rent. he criticl temerture is 35 K, nd therefore the limiting temerture boe which minim do not occur is K or 55 C, nd ll Amgt s exeriments were conducted between 0 C 8 nd 00 C. he Cooling effect : he most gses show n inersion of the cooling effect t certin temerture. he eqution of Vn der Wls indictes t wht temerture this occurs. \M-therm\h8-.m5

13 388 ENGNEERNG HERMODYNAMCS or We he + K J ( b) R [rom eqn. (8.5)] Keeing constnt nd differentiting with resect to, we get Substituting this lue of of cooling effect), we get b d + 3 RS UVW d K J R d R d K J b + 3 RS UVW d K J in the eqution c µ d R c µ b + 3 RS UVW nd substituting for R from eqution (8.9) this reduces to c µ 3b b + b + 3 d d he denomintor of this exression is lwys ositie, since it is R cooling effect, µ, is ositie if nd negtie if nd inersion occurs when or b b < 3 b b > 3 b b 3 b K J (where µ is mesure d d K J. Hence the...(8.3)...(8.3) 3b K J...(8.33) n order to get the temerture of inersion this eqution must be combined with the originl eqution. hus b K J R...(8.34) b Since is necessrily lwys greter thn b, it will be seen tht s increses so lso does the temerture of inersion. \M-therm\h8-.m5

14 DEAL AND REAL GASES 389 he form of cure gien by eqution (8.33) is shown in br nd br 3b Heting Cooling ig he ressure is zero when 3 b, or infinity. hese lues of determine the limiting lues of the temerture of inersion, s it is only between these limits tht is ositie. Substituting these limits of in eqution (8.34) the limiting lues of the temerture of inersion re, or from 9 (8.7), c nd 4 c. ig. 8.9 he eqution (8.33) being qudrtic there re two lues of for constnt lue of t which inersion occurs, s my lso be seen by reference to ig Consequently by eqution (8.34) there re two temertures for constnt lue of t which inersion occurs. As the temerture increses through the lower of these lues the chnge is from heting to cooling effect, nd s it increses through the higher of these lues the chnge is from cooling to heting effect. he inersion will occur when the mximum lue of is, when 3b. or ny lue 3b of less thn this there is cooling effect roided the condition of the substnce is reresented by oint inside the re enclosed by the cure nd the xis of olume, ig. 8.9, nd for ny greter lue of there is heting effect s indicted by equtions (8.3) nd (8.3) resectiely. Let us tke the cse of hydrogen. n the exeriments of Joule nd homson the ressure used ws 4.7 tmosheres. he criticl temerture nd ressure re 35 K nd 5 tmosheres. rom eqution (8.33) we cn find the lues of b corresonding to the ressure used by Joule nd homson, nd by substitution in eqution (8.34) find the two temertures t which inersion occurs t this ressure. Eqution (8.33) cn be written s : L NM H G K J b b 7 c 3 O QP Hence b ± c or by substitution of the boe lues for nd c. Writing eqution (8.34) in the form 7 4 c b K J, we he by substitution for b : 33.5 K or 7. K \M-therm\h8-.m5

15 390 ENGNEERNG HERMODYNAMCS tht is, below 45.9 C there would be heting effect, between 45.9 C nd 39.6 C cooling effect, nd boe 39.6 C heting effect. hus Vn der Wls eqution qulittiely ccounts for the heting effect obsered t ordinry temertures. Limittions of Vn der Wls Eqution Vn der Wls eqution under ctul condition becomes inlid s discussed below : he lues of nd b (which re ssumed to be constnt) re found to ry with temerture. hus the results obtined from the eqution re incorrect when the rition of nd b is lrge with resect to temerture. he eqution is not ccurte enough in the criticl region nd it is lso obious from its derition VRAL EQUAON O SAE he iril ( Ltin word used for force which refers to interction forces between molecules) eqution of stte my be exressed s follows : R A 0 + A + A + A (8.35) or R B 0 + B B + + B (8.36) where A 0, A,... nd B 0, B,... re clled the iril co-efficients which re functions of temerture only. he iril eqution cn be used only for gses t low nd medium densities. he dntge of iril eqution is tht the iril co-efficients cn be determined from exerimentl -- dt BEAE-BRDGEMAN EQUAON Bettie-Bridgemn eqution is exressed s follows : R 0 ( e ) A ( + B) ( ) ( )...(8.37) where ressure nd e A A 0 B B 0 c 3 K J b K J he fctors A 0,, B 0, b nd c re constnts whose lues for different gses re gien in ble 8.. his eqution is normlly used for substnces t ressures less thn criticl ressure. he eqution is ccurte enough when the olumes inoled re greter thn twice the criticl olume. he eqution fits the dt of fourteen gses down to the criticl oint nd oer wide rnge of ressure within ± 0.5% error. Howeer, it is inccurte ner criticl oint. \M-therm\h8-.m5

16 DEAL AND REAL GASES 39 ble 8.. Constnts of the Bettie-Bridgemn Eqution of Stte Gs A 0 B 0 b c 0 4 Hydrogen (H ) Oxygen (O ) Crbon dioxide (CO ) Helium (He) Air Nitrogen Argon REDUCED PROPERES he rtios of ressure, temerture nd secific olume of rel gs to the corresonding criticl lues re clled the reduced roerties. r c, r c, r c...(8.38) ble 8.3. Criticl Constnts Substnce Pressure ( c ) br emerture ( c ) K Air Argon Crbon dioxide Crbon monoxide Helium Hydrogen Nitrogen Oxygen Wter Ethne Ethylene Methne Prone ble 8.4. Proerties of Gses Gs Moleculr c c R c c γ c c weight (kj/kg K) (kj/kg K) (kj/kg K) (M) Zc c R 0 c Air Oxygen Nitrogen Hydrogen Crbon monoxide \M-therm\h8-.m5

17 39 ENGNEERNG HERMODYNAMCS Crbon dioxide Wter Methne Sulhur dioxide Ammoni LAW O CORRESPONDNG SAES f ny two gses he equl lues of reduced ressure nd reduced temerture, then they he sme lues of reduced olume ; i.e., R f( r, r ) for ll gses nd the function is the sme. his lw is most ccurte in the icinity of the criticl oint. 8.. COMPRESSBLY CHAR he comressibility fctor (Z) of ny gs is function of only two roerties, usully temerture nd ressure, so tht Z f( r, r ) excet ner the criticl oint. he lue of Z for ny rel gs my be less or more thn unity, deending on ressure nd temerture conditions of the gs. he generl comressibility chrt is lotted with Z ersus r for rious lues of r. his is constructed by lotting the known dt of one or more gses nd cn be used for ny gs. Such chrt is shown in ig his chrt gies best results for the regions well remoed from the criticl stte for ll gses. Z /R r Reduced ressure r ig Generlised comressibility chrt. DEAL GASES Exmle 8.. he olume of high ltitude chmber is 40 m 3. t is ut into oertion by reducing ressure from br to 0.4 br nd temerture from 5 C to 5 C. How mny kg of ir must be remoed from the chmber during the rocess? Exress this mss s olume mesured t br nd 5 C. ke R 87 J/kg K for ir. Solution. V 40 m 3 V 40 m 3 br 0.4 br K K kg of ir to be remoed : Assuming nitrogen to be erfect gs, \M-therm\h8-.m5

18 DEAL AND REAL GASES 393 V m R or m V R V m R or m Mss of ir remoed during the rocess (m m ) kg (m m ) V R R V R V V 87 L NM V R 6.7 kg. (Ans.) Volume of this mss of gs t br nd 5 C is gien by V mr 5 5 ( 0 ) 40 ( ) m (Ans.) Exmle 8.. A steel flsk of 0.04 m 3 ccity is to be used to store nitrogen t 0 br, 0 C. he flsk is to be rotected ginst excessie ressure by fusible lug which will melt nd llow the gs to esce if the temerture rises too high. (i) How mny kg of nitrogen will the flsk hold t the designed conditions? (ii) At wht temerture must the fusible lug melt in order to limit the ressure of full flsk to mximum of 50 br? Solution. Ccity of the steel flsk, V 0.04 m 3 Pressure, 0 br emerture, K (i) kg of nitrogen the flsk cn hold : Now, R for nitrogen (moleculr weight, M 8) R J/kg K M 8 Assuming nitrogen to be erfect gs, we get Mss of nitrogen in the flsk t designed condition m V kg. (Ans.) R (ii) emerture t which fusible lug should melt, t : When the fusible lug is bout to melt 50 br ; V 0.04 m 3 ; m 5.5 kg herefore, temerture t t which fusible lug must melt is gien by V mr K t C. (Ans.) Exmle 8.3. A blloon of shericl she 6 m in dimeter is filled with hydrogen gs t ressure of br bs. nd 0 C. At lter time, the ressure of gs is 94 er cent of its originl ressure t the sme temerture : 5 5 O QP \M-therm\h8-.m5

19 394 ENGNEERNG HERMODYNAMCS (i) Wht mss of originl gs must he esced if the dimensions of the blloon is not chnged? (ii) ind the mount of het to be remoed to cuse the sme dro in ressure t constnt olume. Solution. Dimeter of the shericl blloon 6 m Pressure of hydrogen gs, br bs. emerture of hydrogen gs, 0 C or 93 K At lter time ressure of the gs, 0.94 t 93 K. (i) Mss of originl gs esced : m m m [where m nd m re the initil nd finl msses of the gs] or V R V R V R V ( R 0.94 ) V ( 0.94) R %ge mss esced m 00 m V ( 094. ) R 6%. (Ans.) V R (ii) Amount of het to be remoed : Using the gs eqution, V V 094. ( ) [Q V V, nd 0.94 ] K or.4 C he het to be remoed is gien by Q mc ( ) (Q V V nd 0.94 ) Q MR 834 where m V π R 9.8 kg R s M for H c 0400 J/kg K for H Q (het to be remoed) ( ).69 MJ. (Ans.) Exmle 8.4. A essel of ccity 3 m 3 contins kg mole of N t 90 C. (i) Clculte ressure nd the secific olume of the gs. (ii) f the rtio of secific hets is.4, elute the lues of c nd c. (iii) Subsequently, the gs cools to the tmosheric temerture of 0 C ; elute the finl ressure of gs. L N M O Q P \M-therm\h8-.m5

20 DEAL AND REAL GASES 395 (i) Elute the increse in secific internl energy, the increse in secific enthly, increse in secific entroy nd mgnitude nd sign of het trnsfer. Solution. Mss of N, m kg mole i.e., 8 kg Ccity of the essel, V 3 m 3 emerture, K (i) Pressure ( ) nd secific olume ( ) of the gs : Using the reltion V mr 834 L 0 834O Q P or K J 363 Q R NM R M J/m or 0.06 br. (Ans.) Secific olume, V 3 m m3 /kg. (Ans.) (ii) c?, c? c c.4 (gien)...(i) But c c R Soling for c nd c between (i) nd (ii) c.039 kj/kg K ; c 0.74 kj/kg K. (iii) inl ressure of the gs fter cooling to 0 C : nitilly After cooling 0.06 br? V 3 m 3 V 3 m 3 Now, 363 K V V K 8. br. (Ans.) (Ans.) (i) u, h, s, Q : or erfect gs, ncrese in secific internl energy u c ( ) 0.74(93 363) 5.94 kj/kg. ncrese in secific enthly, h c ( ).039(93 363) 7.73 kj/kg. ncrese in secific entroy, s c log e + R log e...(ii) (s V V ) (Ans.) (Ans.) \M-therm\h8-.m5

21 396 ENGNEERNG HERMODYNAMCS But s c log e 0.74 log 93 e 363 K J kj/kg K. (Ans.) Now, Q u + W Here W 0 s chnge in olume is zero Q u Het trnsfer, Q 5.94 kj/kg kj. (Ans.) Exmle 8.5. () kg of ir t ressure of 8 br nd temerture of 00 C undergoes reersible olytroic rocess following the lw. constnt. f the finl ressure is.8 br determine : (i) he finl secific olume, temerture nd increse in entroy ; (ii) he work done nd the het trnsfer. Assume R 0.87 kj/kg K nd γ.4. (b) Reet () ssuming the rocess to be irreersible nd dibtic between end sttes. Solution. () Mss of ir, m kg Pressure, 8 br emerture, K he lw followed :. constnt inl ressure,.8 br Chrcteristic gs constnt, R 0.87 kj/kg K Rtio of secific hets, γ.4 (i), nd s : Assuming ir to be erfect gs, R or R Also,.. or ( ) /. / i.e., inl secific olume, m 3 /kg. (Ans.) Agin, R m 3 /kg / K J m 3 /kg K R ( ) i.e., inl temerture, t C. (Ans.) ncrese in entroy s is gien by, s c log e 5 + R log e \M-therm\h8-.m5

22 DEAL AND REAL GASES 397 But γ c c.4 (gien)...(i) nd c c R ( 0.87 kj/kg K for ir)...(ii) Soling for c between (i) nd (ii), c 0.77 kj/kg K 90 8 s 0.77 log e K J log e K J kj/kg K i.e., ncrese in entroy, s 0.78 kj/kg K. (Ans.) (ii) Work done nd het trnsfer : he work done in olytroic rocess is gien by, W n R ( ) n ( ) (. ) 7.96 kj/kg i.e., Work done 7.96 kj/kg. (Ans.) Het trnsfer, Q u + W where u c ( ) 0.77 ( ) kj/kg Q kj/kg Hence het trnsfer 59.0 kj/kg. (Ans.) (b) (i) hough the rocess is ssumed now to be irreersible nd dibtic, the end sttes re gien to be the sme s in (). herefore, ll the roerties t the end of the rocess re the sme s in (). (Ans.) (ii) As the rocess is dibtic, Q (het trnsfer) 0. (Ans.) u u in () Alying first lw for this rocess Q u + W 0 u + W or W u ( 58.94) Work done kj/kg. (Ans.) Exmle 8.6. wo sheres ech.5 m in dimeter re connected to ech other by ie with le s shown in ig. 8.. One shere contins 6 kg of ir nd other 8 kg of ir when the le is closed. he temerture of ir in both shere is 5 C. he le is oened nd the whole system is llowed to come to equilibrium conditions. Assuming there is no loss or gin of energy, determine the ressure in the sheres when the system ttins equilibrium. Neglect the olume of the ie. Solution. Volume of ech shere 4 3 πr3 4 3 π 5. 3 K J 8.8 m 3 \M-therm\h8-.m5

23 398 ENGNEERNG HERMODYNAMCS Sheres.5 m.5 m ig. 8. he temerture in both sheres is sme (5 C) i.e., K As no energy exchnge occurs, the temerture reched fter equilibrium is 98 K. Mss of ir in shere, m 6 kg Mss of ir in shere, m 8 kg After oening the le otl olume, V m 3 otl mss, m m + m kg Now using chrcteristic gs eqution V mr mr N/m or.55 br V Hence ressure in the sheres when the system ttins equilibrium.55 br. (Ans.) Exmle 8.7. CO flows t ressure of 0 br nd 80 C into turbine, locted in chemicl lnt, nd there it exnds reersibly nd dibticlly to finl ressure of.05 br. Clculte the finl secific olume, temerture nd increse in entroy. Neglect chnges in elocity nd eletion. f the mss flow rte is 6.5 kg/min. elute the het trnsfer rte from the gs nd the ower deliered by the turbine. Assume CO to be erfect gs nd c kj/kg K. Solution. At entry to turbine At exit of turbine Pressure, 0 br Pressure,.05 br emerture, K Since the exnsion is reersible nd dibtic, therefore, the eqution γ constnt is licble. γ γ...(i) Eliminting nd using the erfect gs eqution We cn write eqution (i) s R ( γ )/ γ \M-therm\h8-.m5

24 DEAL AND REAL GASES K J ( γ )/ γ c kj/kg K (gien) R R (Moleculr weight of CO M 44 44) kj/kg K Also c c R c c.059 kj/kg K γ c c Substituting for γ in eqution (ii) K J (. 3 )/ K inl temerture C. (Ans.) R ( ) 97 ( ) m 3 /kg i.e., inl secific olume m 3 /kg. (Ans.) As the rocess is reersible nd dibtic s 0 i.e., ncrese in entroy 0. (Ans.) Since the rocess is dibtic, therefore, het trnsfer rte from turbine 0. Alying stedy flow energy eqution (S..E.E.) on unit time bsis, L NM O P Q C m& h Z Q& + + P + C &m h + + Z W M P + By dt chnges in elocity nd eletion re negligible, nd Q 0. S..E.E. reduces to i.e., W &m (h h ) L M N O P Q L NM (Ans.) O, ( ) Q P &m c ( ) s dh d c h h c (453 97) 7.34 kw 60 Hence ower deliered by the turbine 7.34 kw. (Ans.) Exmle 8.8. A certin quntity of ir initilly t ressure of 8 br nd 80 C hs olume of m 3. t undergoes the following rocesses in the following sequence in cycle : () Exnds t constnt ressure to 0. m 3, \M-therm\h8-.m5

25 400 ENGNEERNG HERMODYNAMCS (b) ollows olytroic rocess with n.4, nd (c) A constnt temerture rocess (which comletes the cycle). Elute the following : (i) he het receied in the cycle ; (ii) he het rejected in the cycle ; (iii) Efficiency of the cycle. Solution. ig. 8. shows the cycle on -V nd -s lnes. (Pressure) (em.) Constnt n V Const. Const. n V Const. V Const. 3 V(Volume) Const. 3 s (Entroy) nd ig. 8. Pressure, 8 br Volume, V m 3 emerture, K Pressure, 8 br ( ) Volume, V 0. m 3 ndex, n.4 o find mss of ir, use the reltion V mr m V R rom V mr V mr Also, V.4 3 V (. 4 )/. 4 But 3 s nd 3 re on n isotherml line / kg 580 K \M-therm\h8-.m5

26 DEAL AND REAL GASES (. 857) / (. 857) br i.e., Now, 3 V 3 mr V V m 3 (i) he het receied in the cycle : Alying first lw to the constnt ressure rocess -, Q U + W W z dv (V V ) ( ) 5000 J or 5 kj (work done by ir) Q m c ( ) ( ) kj Het receied 80.6 kj Alying first lw to reersible olytroic rocess -3 Q U + W But W V V n 3 3 mr ( 3) n (s the rocess is reersible) ( ) 9.98 kj (work done by ir) 4. Q mc ( 3 ) ( ) kj (het receied) otl het receied in the cycle kj. (Ans.) (ii) he het rejected in the cycle : Alying first lw to reersible isotherml rocess 3-, Q U + W W 3 V 3 log e V V log e K J kj (work done on the ir) Q mc ( 3 ) + W kj (Q 3 ) Hence het rejected in the cycle 03.9 kj. (Ans.) \M-therm\h8-.m5

27 40 ENGNEERNG HERMODYNAMCS (ii) Efficiency of the cycle, η cycle : Het receied Het rejected η cycle Het receied or 43.3% i.e., Efficiency of the cycle 43.3%. (Ans.) nd REAL GASES Exmle 8.9. One kg of CO hs olume of m 3 t 00 C. Comute the ressure by (i) Vn der Wls eqution (ii) Perfect gs eqution. Solution. (i) Using Vn der Wls eqution : Molr secific olume, m 3 /kg-mol (Q M for CO 44) emerture, K he lues of nd b for CO (from ble 8.) Nm 4 /(kg-mol) b m 3 /kg-mol R Nm/kg-mol K Vn der Wls eqution is written s + K J ( b) R 0 or R 0 b Substituting the lues in the boe eqution, we get N/m or br. (Ans.) (ii) Using erfect gs eqution : R 0 R N/m 44 or br. (Ans.) Exmle 8.0. A continer of 3 m 3 ccity contins 0 kg of CO t 7 C. Estimte the ressure exerted by CO by using : (i) Perfect gs eqution (ii) Vn der Wls eqution (iii) Bettie Bridgemn eqution. Solution. Ccity of the continer, V 3 m 3 Mss of CO, m 0 kg emerture of CO, K Pressure exerted by CO, : \M-therm\h8-.m5

28 DEAL AND REAL GASES 403 (i) Using erfect gs eqution : Chrcteristic gs constnt, R R Nm/kg K (for CO M 44 ) Using erfect gs eqution V mr mr V N/m or.889 br. (Ans.) (ii) Using Vn der Wls eqution : + K J ( b) R 0 R 0 b rom ble 8. or CO : Nm 4 /(kg-mol) b m 3 /(kg-mol) Molr secific olume m 0 3 /kg-mol Now substituting the lues in the boe eqution, we get (. ) N/m or.875 br. (Ans.) (iii) Using Bettie Bridgemn eqution : where ressure, A A 0 K J R 0 ( e ) ( ), B B 0 b K J A ( + B) nd e rom ble 8. A , B , b C A B C ( 300) 3 c 3 K J K J Now substituting the rious lues in the boe eqution, we get ( ) ( 3. ) ( ) ( 3. ) ~_ N/m.9 br. (Ans.) \M-therm\h8-.m5

29 404 ENGNEERNG HERMODYNAMCS Exmle 8.. One kg-mol of oxygen undergoes reersible non-flow isotherml comression nd the olume decreses from 0. m 3 /kg to 0.08 m 3 /kg nd the initil temerture is 60 C. f the gs obeys Vn der Wls eqution find : (i) he work done during the rocess (ii) he finl ressure. Solution. he Vn der Wls eqution is written s K J + ( b) R 0 where ressure of the gs ;, b constnts ; molr olume ; R 0 uniersl gs constnt rom ble 8. or O : 3950 Nm 4 /(kg-mol) b m 3 /kg-mol nd R Nm/kg-mol K m 3 /kg-mol m 3 /kg-mol. (i) Work done during the rocess : he work done er kg mole of O is gien by z z L R 0 W. d M O d b P NM O QP bo b P Q L N M R log ( b) 0 R 0 L NM L NM log e e + LO NM QP L NM e QP P ( ) log P Nm/kg-mol. (Ans.) (ii) he finl ressure, : R 0 b O QP O Q P L N M ( 56. ) N/m or 0.73 br. (Ans.) Exmle 8.. f the lues for reduced ressure nd comressibility fctor for ethylene re 0 nd.5 resectiely, comute the temerture. Solution. Reduced ressure, r 0 Comressibility fctor, Z.5 emerture,? rom the generlised comressibility chrt on Z r co-ordintes corresonding to r 0 nd Z.5, r 8.0. Now, since c r [rom ble 8.3, c 8.4 K] 59. K. (Ans.) K J O Q P \M-therm\h8-.m5

30 DEAL AND REAL GASES 405 ~ Exmle 8.3. Clculte the density of N t 60 br nd 5 C by using the comressibility chrt. Solution. Pressure, 60 br emerture, K Density, ρ? or N (from ble 8.3) : c br c 6. K Now r c nd r 88.8 c 6. rom the comressibility chrt for r 7.6 nd r.8, Z ~.08 Also Z R, where ρ stnds for density ρr or ρ ZR kg/m 3. (Ans.) Exmle 8.4. Wht should be the temerture of.3 kg of CO gs in continer t ressure of 00 br to behe s n idel? Solution. Pressure, 00 br emerture,? or CO (from ble 8.3) c br c 304. K As the gs behes like n idel gs, Z r c rom comressibility chrt for Z, r.7 r.48 r c K. (Ans.) Exmle 8.5. A shericl shed blloon of m dimeter contins H t 30 C nd. br. ind the mss of H in the blloon using rel gs eqution. Solution. Dimeter of shericl blloon m Volume, V 4/3 π (6) m 3 emerture, K Pressure,. br Mss of H in the blloon, m : or H (from ble 8.3) c.97 br c 33.3 K Now, r c. 97 r 303 c \M-therm\h8-.m5

31 406 ENGNEERNG HERMODYNAMCS rom comressibility chrge, corresonding to r nd r 9. Z ~ (his indictes tht the gs hing higher criticl ressure nd lower criticl temerture behes like n idel gs t norml ressure nd temerture conditions.) Also, V ZmR 5 V or m ZR H G 86.9 kg. (Ans.) K J Exmle 8.6. Determine the lue of comressibility fctor t criticl oint (Z c ) for the Vn der Wls gs. Solution. Refer ig c C.P. C.P. Criticl oint sotherms c ig. 8.3 rom the isotherms lotted on - digrm in ig. 8.3 it cn be seen tht the criticl isotherms hs n inflection oint, whose tngent is horizontl t the criticl oint. c K J 0 nd c 0 c he Vn der Wl s eqution t the criticl oint is As c is constnt c R 0 c b c c c c K c R ( b) c + 0 c 3 c c...(i) 0...(ii) J R0 c (iii) ( b) ( ) c c \M-therm\h8-.m5

32 DEAL AND REAL GASES 407 or 3 c (ii) + (iii) gies 3R 0 ( b) c c + R 0 3 ( b) c 0 3 c ( b) c or c 3b Substituting for b in (ii), we get R 0 c [ ( / 3) ] c c + 3 c ( ) 0 Substituting for nd b in (i), we get But c R 0 c R R 0 c c c c c c c c ( / 3) Z c Z c 3 R 0 c ( / 3) c ( 9/ 8) (Ans.) ( 9/ 8) R0 c c c HGHLGHS. An idel gs is defined s gs hing no forces of intermoleculr ttrction. t obeys the lw R. he secific het ccities re not constnt but re functions of temerture. A erfect gs obeys the lw R nd hs constnt secific het ccities.. he reltion between the indeendent roerties, such s ressure, secific olume nd temerture for ure substnce is known s eqution of stte. 3. Ech oint on -- surfce reresents n equilibrium stte nd line on the surfce reresents rocess. 4. Joule s lw sttes tht the secific internl energy of gs deends only on the temerture of the gs nd is indeendent of both ressure nd olume. 5. Vn der Wls eqution my be written s + ( b) R where nd b re constnts for the rticulr fluid nd R is the gs constnt. \M-therm\h8-.m5

33 408 ENGNEERNG HERMODYNAMCS OBJECVE YPE QUESONS Choose the Correct Answer :. () A erfect gs does not obey the lw R (b) A erfect gs obeys the lw R nd hs constnt secific het (c) A erfect gs obeys the lw R but he rible secific het ccities.. Boyle s lw sttes tht, when temerture is constnt, the olume of gien mss of erfect gs () ries directly s the bsolute ressure (b) ries inersely s the bsolute ressure (c) ries s squre of the bsolute ressure (d) does not ry with the bsolute ressure. 3. Chrle s lw sttes tht if ny gs is heted t constnt ressure, its olume () chnges directly s it bsolute temerture (b) chnges inersely s its bsolute temerture (c) chnges s squre of the bsolute temerture (d) does not chnge with bsolute temerture. 4. he eqution of the stte er kg of erfect gs is gien by () R (b) R (c) R (d) R. where,, R nd re the ressure, olume, chrcteristic gs constnt nd temerture of the gs resectiely. 5. he eqution of stte of n idel gs is reltionshi between the ribles : () ressure nd olume (b) ressure nd temerture (c) ressure, olume nd temerture (d) none of the boe. 6. Joule s lw sttes tht the secific internl energy of gs deends only on () the ressure of the gs (b) the olume of the gs (c) the temerture of the gs (d) none of the boe. 7. Eqution for secific het t constnt ressure of n idel gs is gien by () c + K + K + K 3 (b) c + K + K 3 + K 4 (c) c + K + K 4 + K (d) c + K + K 3 + K. where, K, K nd K re constnts. 8. Vn der Wls eqution my be written s () (c) + K J ( b) R (b) + ( b) R (d) Answers + ( b) R + ( b) R.. (b). (b) 3. () 4. (b) 5. (c) 6. (c) 7. () 8. (b). HEORECAL QUESONS. Wht is n idel gs?. Wht is the difference between n idel nd erfect gs? 3. Wht re semi-erfect or ermnent gses? 4. Define Eqution of stte. 5. Stte Boyle s nd Chrle s lws nd derie n eqution of the stte for erfect gs. 6. Wht is -- surfce? Drw ortion of such surfce. 7. Derie the reltionshi between the two rincil secific hets nd chrcteristic gs constnt for erfect gs. 8. Write short note on Vn der Wls eqution. \M-therm\h8-.m5

34 DEAL AND REAL GASES 409 UNSOLVED PROBLEMS DEAL GASES. A essel of 0.03 m33 ccity contins gs t 3.5 br ressure nd 35 C temerture. Determine the mss of the gs in the essel. f the ressure of this gs is incresed to 0.5 br while the olume remins constnt, wht will be the temerture of the gs? or the gs tke R 90 J/kg K. [Ans. 0.8 kg, 650 C]. he tyre of n utomobile contins certin olume of ir t guge ressure of br nd 0 C. he brometer reds 75 cm of Hg. he temerture of ir in the tyre rises to 80 C due to running of utomobile for two hours. ind the new ressure in the tyre. Assume tht the ir is n idel gs nd tyre does not stretch due to heting. [Ans..6 br] 3. A tnk mde of metl is designed to ber n internl guge ressure of 7 br. he tnk is filled with gs t ressure of 5.5 br bs., nd 5 C. he temerture in the tnk my rech to 50 C when the tnk stnds in the sun. (i) f the tnk does not exnd with temerture, will the design ressure be exceeded on dy when tmosheric ressure is br nd ir in the tnk reches 50 C when exosed to hot sun? (ii) Wht temerture would he to be reched to rise the ir ressure to the design limit? [Ans. (i) 6.6 br, (ii) 47 C] 4. A essel of shericl she is.5 m in dimeter nd contins ir t 40 C. t is ecuted till the cuum inside the essel is 735 mm of mercury. Determine : (i) he mss of ir umed out ; (ii) f the tnk is then cooled to 0 C wht is the ressure in the tnk? he brometer reds 760 mm of mercury. Assume tht during ecution, there is no chnge in temerture of ir. [Ans. (i).9 kg, (ii) 3 kp] 5. A blloon of shericl she is 8 m in dimeter nd is filled with hydrogen t ressure of br bs. nd 5 C. At lter time, the ressure of gs is 95 er cent of its originl ressure t the sme temerture. (i) Wht mss of originl gs must he esced if the dimensions of the blloon re not chnged? (ii) ind the mount of het to be remoed to cuse the sme dro in ressure t constnt olume. [Ans. (i) 5 er cent, (ii) 3.6 MJ] 6. ind the moleculr weight nd gs constnt for the gs whose secific hets re s follows : c.967 kj/kg K, c.507 kj/kg K. [Ans kj/kg K] 7. A constnt olume chmber of 0.3 m33 ccity contins kg of ir t 0 C. Het is trnsferred to the ir until its temerture is 00 C. ind : (i) Het trnsferred ; (ii) Chnge in entroy nd enthly. [Ans. (i) 8.09 kj, (ii) kj/kg K, 80.8 kj] 8. kg of ir t 0 C occuying olume of 0.3 m333 undergoes reersible constnt ressure rocess. Het is trnsferred to the ir until its temerture is 00 C. Determine : (i) he work nd het trnsferred. (ii) he chnge in internl energy, enthly nd entroy. [Ans. (i) 5.5 kj, 80.8 kj ; (ii) 8.09 kj, 80.8 kj, kj/kg K] 9. A blloon of shericl she, 0 m in dimeter is filled with hydrogen t 0 C nd tmosheric ressure. he surrounding ir is t 5 C nd brometer reds 75 mm of Hg. Determine the lod lifting ccity of the blloon. [Ans kg] 0. Air exnds in cylinder in reersible dibtic rocess from 3.73 br to.96 br. f the finl temerture is to be 7 C, wht would be the initil temerture? Also clculte the chnge in secific enthly, het nd work trnsfers er kg of ir. [Ans. 54 K, 4.79 kj/kg, zero, kj/kg]. kg mole of N is contined in essel of olume.5 m 3 t 00 C. (i) ind the mss, the ressure nd the secific olume of the gs. (ii) f the rtio of the secific hets is.4, elute the lues of c nd c. \M-therm\h8-.m5

35 40 ENGNEERNG HERMODYNAMCS (iii) Subsequently, the gs cools to the tmosheric temerture of 30 C, elute the finl ressure of the gs. (i) Elute the increse in secific internl energy, the increse in secific enthly, increse in secific entroy nd mgnitude nd sign of het trnsfer. [Ans. (i) 8 kg,.37 br, m 3 /kg ; (ii) c.038 kj/kg K, c kj/kg K ; (iii) 0. br ; (i) 5.6 kj/kg, 7.67 kj/kg, kj/kg K, 465. kj]. he ressure nd olume of gs, during rocess, chnge from br bsolute nd m resectiely to 33 6 br bsolute nd 0.4 m resectiely. During the rocess the increse in the enthly of the gs is 00 kj. 33 king 0.4 kj/kg K, determine c c, R nd U. [Ans. 3 kj/kg K,.6 kj/kg K, 60 kj] 3. kg of ir t 7 C is heted reersibly t constnt ressure until the olume is doubled nd then heted reersibly t constnt olume until the ressure is doubled. or the totl th find : (i) he work ; (ii) Het trnsfer ; (iii) Chnge of entroy. [Ans. (i) 86.4 kj, (ii) kj, (iii).86 kj/kg K] 4. A mss of ir initilly t 60 C nd ressure of 6.86 br hs olume of 0.03 m 3. he ir is exnded t constnt ressure to 0.09 m 3, olytroic rocess with n.5 is then crried out, followed by constnt temerture rocess which comletes the cycle. All rocesses re reersible. ind (i) he het receied nd rejected in the cycle, (ii) he efficiency of the cycle. Show the cycle on - nd -s lnes. [Ans. (i) kj, 0.3 kj ; (ii) 38.4%] REAL GASES 5. One kg-mol of oxygen undergoes reersible non-flow isotherml comression nd the olume decreses from 0.5 m 3 /kg to 0.06 m 3 /kg nd the initil temerture is 50 C. he gs obeys Vn der Wls eqution during the comression. ind : (i) he work done during the rocess ; (ii) he finl ressure. [Ans. (i) Nm/kg-mol, (ii) 3.85 br] 6. Determine the comressibility fctor for O t (i) 00 br 70 C nd (ii) t 5 br nd 30 C. [Ans. (i) 0.7, (ii) 0.98] 7. Determine the ressure of ir t 05 C hing secific olume of m 3 /kg by mens of : (i) del gs eqution. (ii) Vn der Wls eqution ; (iii) Bettie-Bridgemn eqution. [Ans. (i) br, (ii) br, (iii) 55.8 br] \M-therm\h8-.m5