Prole 3: Crnot Cyle of n Idel Gs In this prole, the strting pressure P nd volue of n idel gs in stte, re given he rtio R = / > of the volues of the sttes nd is given Finlly onstnt γ = 5/3 is given You do not know how ny oles of the gs re present Red over steps (- (4 elow nd sketh the pth of the yle on P plot on the grph elow Lel ll pproprite points ( In the first of four steps, to, n idel gs is opressed fro to while no het is llowed to flow into or out of the syste he opression of the gs rises the teperture fro n initil teperture nd to finl teperture During this proess the quntity γ = onstnt, where γ = 5/3 Wht is the pressure P nd volue of the stte of the gs fter the opression is finished? Wht is the hnge in internl energy of the gs during this hnge of stte? Wht is the work done y the gs during this opression? ( he gs is now llowed to expnd isotherlly fro to, fro volue to volue d Express the work done y the gs in this proess W nd the ount of het Q tht ust e dded fro the het soure t in ters of P,,,, nd
Is this het positive or negtive? Explin whether it is dded to the syste or reoved e Wht is the pressure P of the gs fter the expnsion is finished? (3 When the gs hs rehed point is expnds fro to d while no het is llowed to flow into or out of the syste he expnsion of the gs lowers the teperture nd pressure fro n initil teperture to finl teperture During this proess the quntity γ = onstnt f Wht is the pressure P nd volue d of the stte d of the gs fter the d expnsion is finished? g Wht is the hnge in internl energy of the gs during this hnge of stte? h Wht is the work done y the gs during this expnsion? (4 he gs is now opressed isotherlly fro d to t onstnt fro volue d k to i Find the work done y the syste on the surroundings W d nd the ount of het Qd tht flows etween the syste nd the surroundings Are these quntities positive or negtive? Explin whether het is dded to the syste or reoved fro the het soure t otl Cyle: j Wht is the totl work W yle done y the gs during this yle? k Wht is the totl het Q yle ( fro drwn fro the higher teperture het soure during this yle? l Wht is the effiieny of this yle ε x = W yle / Q yle ( fro? Prole 4 Het pup A reversile het engine n e run in the other diretion, in whih se it does negtive work W yle on the world while puping het Q yle (into into reservoir t n upper teperture,, fro lower teperture, he het gin of this yle, defined to e
g Q yle (into /W yle = (/ ε x where ε = ( x / is the xiu therodyni effiieny of het engine he refrigertor perforne is defined to e K Q yle ( fro / W yle = /( Consider tht you hve lrge swiing pool nd pln to het your house with het pup tht pups het fro the pool into your house A lrge plte in the wter will rein t 0 o C due to the fortion of ie You pik to e 50 o C, whih will e the teperture of the (lrge rditors used to het your house Assue tht your het pup hs the xiu effiieny llowed y therodynis Wht is the het gin nd the refrigertor perforne for this yle? Be reful to use units of Kelvin for teperture If your house forerly urned 00 gllons of oil in winter (t $00/gllon, how uh will the eletriity ost (t $00 per kilowtt-hour to reple this het using 8 the het pup? A gllon of oil hs ss 34 kg nd ontins 4 0 J gl - he ie ue tht ppers in your pool over the winter will e how ny eters on 6 eh side? (It tkes 335 0 J to elt one kg of ie; it tkes up this uh het when 3 freezing he density of ie is 093 0 kg -3 his would e gret for ooling your house in the suer even if the pool wred up enough to swi in it, you ould still ool your house y running the het pup in reverse s n ir onditioner! More prtilly, you ight e le to use ground wter (nd the dirt round it s the het sink
Prole 3: Crnot Cyle of n Idel Gs In this prole, the strting pressure P nd volue of n idel gs in stte, re given he rtio R = / > of the volues of the sttes nd is given Finlly onstnt γ = 5/3 is given You do not know how ny oles of the gs re present Red over steps (- (4 elow nd sketh the pth of the yle on P plot on the grph elow Lel ll pproprite points ( In the first of four steps, to, n idel gs is opressed fro to while no het is llowed to flow into or out of the syste he opression of the gs rises the teperture fro n initil teperture nd to finl teperture During this proess the quntity γ = onstnt, where γ = 5/3 Wht is the pressure P nd volue of the gs fter the opression is finished? Answer: Aording to the idel gs lw, = n R nd = n R so So the pressure = P = P he opression stisfies γ = γ so using the ove result for pressure P, we get 6
γ = P γ = γ his eoes using γ = 5/3 / 3 = /3 he volue is then 3/ = hus the rtio of the volues is 3/ = So the pressure P is 5/ P = P Wht is the hnge in internl energy of the gs during this hnge of stte? Answer: he hnge in internl energy is 3 3 ( U U = n R = Wht is the work done y the gs during this opression? Answer: Sine no het is exhnged Q = 0 U U = W + Q = W = 3 ( So 3 ( W = < 0 7
he surroundings do work opressing the gs ( he gs is now llowed to expnd isotherlly fro to, fro volue to volue d Express the work done y the gs in this proess W nd the ount of het Q tht ust e dded fro the het soure t in ters of P,,,, nd Is this het positive or negtive? Explin whether it is dded to the syste or reoved Answer: his is n isotherl expnsion so the teperture does not hnge the internl energy is onstnt, =0 hus 3 U U = n R = 0 he gs does work on the surroundings euse it is expnding he pressure is not onstnt during this expnsion Sine the gs is expnding y n isotherl proess, the Idel Gs Lw reltes the pressure nd volue vrition ording to nr P = herefore the work done y the gs on the surroundings is the integrl d W = n R = n R ln( / Using the result for the volue fro prt the work is 3/ =, = n R d = n R ln( 3/ W / Rell tht the volues re relted ording to R = / > 0 nd nr = P / so the work done is positive nd given y 8
W = n R ln( / = ln( 3/ R > 0 Fro he First Lw of herodynis, 0 = U U = W + Q, hus the het tht flows into the syste fro the het soure t teperture is equl to the work done y the expnding gs Q = W = ln( 3/ R > 0, Note tht this het flow ust flow fro the higher teperture het soure into the syste euse s the gs expnds it should lose internl energy nd would derese its teperture unless het flows into the syste keeping the internl energy nd hene the teperture onstnt e Wht is the pressure P of the gs fter the expnsion is finished? Answer: = n R = hus P = = P R (3 When the gs hs rehed point is expnds fro to d while no het is llowed to flow into or out of the syste he expnsion of the gs lowers the teperture nd pressure fro n initil teperture to finl teperture During this proess the quntity γ = onstnt f Wht is the pressure P d nd the volue d of the stte d of the gs fter the expnsion is finished? Answer: his lultion is identil to prt, with stte d repling stte, nd stte repling stte So the volue is then 9
3/ = d hus the rtio of the volues is So the pressure P is hene d 3/ = 5/ P = P d 5/ P d = P g Wht is the hnge in internl energy of the gs during this hnge of stte? Answer: he derese in the internl energy is due to the teperture derese of the idel gs during expnsion 3 ( U d U = P h Wht is the work done y the gs during this expnsion? Answer: Sine no het is exhnged Q d = 0 U d U = W d + Q = W d = 3 ( d P So 3 ( W d = P > 0 he gs does work on the surroundings sine the gs is expnding (4 he gs is now opressed isotherlly fro d to t onstnt fro volue d k to i Find the work done y the syste on the surroundings W d nd the ount of het Q tht flows etween the syste nd the surroundings Are these quntities d 0
positive or negtive? Explin whether het is dded to the syste or reoved fro the het soure t Answer: When the gs undergoes opression it will inrese its internl energy ut het flows out of the syste intining onstnt internl energy, U = 0 nd hene the opression is isotherl he lultion of the work nd het is siilr to step ( exept the teperture is held t he work done y the syste on the surroundings is negtive nd is given y the integrl W d d = n R = n R ln( / = P ln( / = P ln(r d d d 3/ Fro prt f the volue d = so the work done is d 3/ 3/ W d = n R = n R ln( / = P ln( / = P ln( R d d Aording to the First Lw this is equl to the het tht flows into the syste whih is lso negtive whih ens tht it tully flows out of the syste into the surroundings t teperture, 3/ Q d = W d = P ln( R otl Cyle: j Wht is the totl work W yle done y the gs during this yle? Answer: he work done y the het engine on the surroundings during the yle is positive nd given y W yle = P ln( 3/ R P 3/ 3/ ln( R = P ln( R k Wht is the totl het Q yle ( fro drwn fro the higher teperture het soure during this yle? Answer: he het tht flowed fro the higher teperture het soure ourred during step ( isotherl expnsion,
totl Q tken fro het soure t = P ln( 3/ R l Wht is the effiieny of this yle ε x = W yle / Q yle ( fro? Answer: he effiieny is given y rtio of the work done divided y the eht flowing into the syste fro the higher teperture het soure 3/ 3/ ε x = W yle / Q yle ( fro = P ln( R / P ln( R ε = x / = = le : Sury of Het Engine Proess U f U i W f,i Q f,i 3 ( 0 diti 3 ( opression isotherl expnsion d diti expnsion d isotherl opression otl positive 0 3 ( negtive 0 0 negtive 3/ ln( R positive 3 ( positive 3/ ln( R negtive 3/ R 3/ ln( R ln( positive 3/ R ln( positive fro 0 3/ ln( R negtive, into 3/ R 3/ ln( R ln( positive, fro into