2. Calcolare il lavoro compiuto dal gas in questa trasformazione.

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1 <latexit sha1_base64="jyoa07y+mt9uwswjmmxezra0i+u=">aaab6nicbvdlsgnbeoynrxhfuy9ebopgkeykomeaf48rtsiks5id9czdzmaxmvkhlpkelx4u8epv+aeevpkth4htsxokkq66e6kusgn9f1vr7cyura+udwsbwv7o6v9w+ajsk0wwzlrklvipqciuny6a+1qjlzhavjs8mvitb9sgj+rojlimje0rhnngrznuztfolit+1z+cljngtiq14uchons75a9olgzrgwzoma0az+1yu615uzguntjdkaudwkf44qktge+ftumtlxso/eixalljmqvydyko0zych1smohztgbip957czglhovzpzvgykm4esqmz/e16xcozyuqizzq7wwkbuezdemuxajb4svlphlwdfxqcopsoiczinaexakavxada6hdg1g0idheiyxthpqvnmstepozq/gd7/0hadupva==</latexit> <latexit sha1_base64="orz9ok9ib6etihkm+gbfvgnevgu=">aaab6nicbvdlsgnbeoynrxhfuy9ebopgkewgqy8blx4jmgcks5iddjihm7plzkwqlnycfw+kepu/anpvwbj4+djhy0ffxddhdfiedg+v6l1tb9jcym8xdnb9g+kh0cne6eayzfitatibouxghdciuwlwikmhlyjebxu7/5gnrwwnbcykhpapf+5xr66q7a10iyw/7m9avkmwikvq/vmdhgrd4lenf7nuorjmugpagz/ymkpacizwuuikbhpkrnsabucvlwjcbhbqhjw5puf6sxalljmpvycyko0zy8h1smqhztmbiv957dtr8kmqys1qnh8ut8vxmzk+jfpcymirejlgnubivssdvl1qvtcceeyy+vkkalhpjl4nalcqfz5oeetuecariektxaderayacp8awvnvcevffvbd6a8xyzx/ahvspa1+pvq==</latexit> sha1_base64="mdshzsihynbiddbu00qwz+o=">aaab6nicbvdlsgnbeoynrxhfazx6grietechom6mfjrpoazamzk9lkymzsmjmrhcwf4mwdil7fv/apphnzb5w8dppy0fbuddpdfsacaen505ubx1jcyu/xdjzds/ca+ltrnitagixmsihwldnjg4yzttujolienlbc0exub91tpvks78w4oyhaa8kirrcx0qovxtuat4m6bv4i9iuzb//chevzfqpfer49jkqg0hgoto76xmcddyjdc6atqttvnmbnhae1ykrggoshmp07qivx6kiqvlwnqtp09kwgh9vietlngm9tllt8z+ukjroimiat1fbj5ouilcmto+nfqm8ujyaplcfemxsrikosmdenyinwv9+ezu0qxxfq/gno0zmcmpx1ccu/dhhgpwdxvoaiebpmatpdvcexrennd5a85zzbzbhzhvp+nikhm=</latexit> 1 Esercizio (tratto dal Problema 11.1 del Mazzoldi-Nigro-Voci) 0.5 moli di un ideale monoatomico a temperatura T 0 = 8 K sono contenute nella parte inferiore A di un cilindro adiabatico verticale. Un pistone di massa trascurabile e spessore trascurabile divide la parte inferiore A da quella superiore B in cui c è il. Due masse = 6 Kg e m sono appese al pistone tramite un filo che esce dal cilindro. Il sistema è inizialmente in equilibrio termodinamico con il pistone a distanza h = 6.5 cm dal fondo del cilindro. 1. Calcolare il valore della massa m. Si taglia il filo che collega m a. Si osserva che, quando il è di nuovo in equilibrio, esso occupa un volume doppio rispetto a quello iniziale.. Calcolare il lavoro compiuto dal in questa trasformazione. Si riattacca la massa m al filo e si attende che il raggiunga nuovamente l equilibrio.. Calcolare la distanza del pistone dal fondo del cilindro e la temperatura del. B A m

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SOLUZIONE Dati Noti: T 0 = 8 K = 6 Kg h = 0.65 m 1. Iniziamo dalla prima parte del problema in cui entrambe le masse sono attaccate al filo B ps A T 1 T 1 T T m Intuitivamente, la forza che il esercita sull ambiente è data da p S (dove S è la superficie del pistone), mentre la forza che l ambiente esercita sul è data dal peso delle due masse e m (non c è la pressione atmosferica perché dal testo sappiamo che nella parte B del cilindro c è il ). Per vederlo espressamente, consideriamo le forze che agiscono su, m e sul pistone. Essendo tutti in equilibrio si ha ps T 1 = 0 T 1 g T = 0 m g T = 0 (1) Risolvendo il sistema (1) ricaviamo subito che { T = m g T 1 = ( + m )g () e come avevamo intuito. ps = ( + m )g ()

3 Dall equazione dei ideali applicata allo stato di equilibrio iniziale abbiamo che ps }{{} =( +m )g da cui pv A = T 0 VA S }{{} =h = T 0 + m = T 0 gh m = = [moltiplico e divido per la superficie S del pistone] (4) m = T 0 gh (5) 0.5 mol / 8.14 J mol / K/ 8 K/ 9.81 m 0.65 m s [uso J = Kg m s ] = 96 Kg 6 Kg = 1 m/ / s Kg m/ s/ 6 Kg = 6 Kg = = 60 Kg (6). Consideriamo ora la fase in cui la massa m viene staccata. Il si espande liberamente ed il pistone risale. Siccome l espansione è libera, non sappiamo se avvenga quasi-staticamente (ossia passando attraverso stati di equilibrio termodinamico in cui p e T del sono ben definite), e dunque non possiamo calcolare il lavoro compiuto dal come W = pdv. Tuttavia, osserviamo che durante l espansione la forza che il esercita sull ambiente è, istante per istante, uguale ed opposta alla forza che l ambiente esercita sul (principio di azione-reazione). F amb = F amb = g (7) Alla fine dell espansione il pistone è risalito di un altezza h, dato che il testo ci dice che il volume del è raddoppiato. Pertanto il lavoro vale W amb = W amb = W peso m1 = = E fin P E in P = = g(zm fin 1 zm in 1 ) = = gh (8) W amb = 6 Kg 9.81 m s 0.65 m = [uso J = Kg m/s ] = 0.7 J (9)

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Per la legge dei perfetti p S }{{} = g p V = T V S }{{} =h = T T = gh [moltiplico e divido per la superficie S del pistone] = W amb (10) T 0.7 J/ = 0.5 mol/ 8.14 J/ = mol / K = 1.4 K (11). Consideriamo ora la terza fase, in cui la massa m viene riattaccata ed il viene dunque compresso. Indichiamo con z la variazione in altezza del pistone (verso il basso) Alla nuova posizione di equilibrio finale, abbiamo p f = pressione finale T f = temperatura finale V f = S(h z ) volume finale (1) Per la legge dei perfetti V f p f S }{{}}{{} S =( +m )g =h z p f V f = T f = T f [moltiplico e divido per la superficie S del pistone] T f = ( + m )g(h z ) (1)

5 <latexit sha1_base64="dg1lcgxtf7vrizuqprcv59izhgy=">aaab7icbva9swnbejzm8avqiwfzwiqlctciablwea7ioydkipsbeasjxt75+6eei78crslrwz9ozvv4onx+fjj4yelww8y8ibfcg9f9dbywl5zxvnnr+fwnzatws5utcepylhlsyhvi6aabzdyndwibcqkarqiraf9y5fff0cleszvzcbbp6jdyupoqlfsimhphvcbtufoltyxydzxjusynn/+vslacrtwkere7m0qmmyofoptcxfkav4uzgmn9knsau9wkxm5zkgqhs/g9qjklq4jyvlgjjwf09knnj6eawm6kmpe9kfif10xneofnxcapqckmi8juebot0fokwxuyiwawuka4vzwwhlwugrtrobgzb48tqnjc8tetcjtoyiachcajh4me5loekklafbgie4rlenhvnyxl1iatc850zg/+whn/afqtkw4=</latexit> sha1_base64="vbn5borntxgfnwgqzrjv6b0xpkm=">aaab7icbvdlsgnbeoynrxhfuq8evawgwyoexrh0gpcit4jmackszye9yeds7dozk4sqn/diqrgvfoh/4c0f8ogohkcnlggoajqprsrsatxxnu/nczc/mliuny5t7k6tr6r9yq6jhvdcssfrgqb1sj4birhhub9uqhjqkbtedbojx7lfphstr00vqjhh8pazaqxut1ohar6sq1a+4bbdecgs8sakunq5+pp+z9yuw/mpzjtmaytsmeg1bnhuyvw+vyyzgyncm9wyuhzlo9iwvniitd8ftsg+1zpkzbwtqqhi/xrj9gwveiwhzg1ht1tdcu//maqqlp/t6xswpqsvgimbxexgt4pglzhcyiniwukw5vjaxlfwxgrpszixjtl8+s6lhrc4vepujgmbiwi7swqf4cailoicyvicbgad4gmfnznl0xpzxcwvgmcxswx84bz9+vpkv</latexit> <latexit sha1_base64="jyoa07y+mt9uwswjmmxezra0i+u=">aaab6nicbvdlsgnbeoynrxhfuy9ebopgkeykomeaf48rtsiks5id9czdzmaxmvkhlpkelx4u8epv+aeevpkth4htsxokkq66e6kusgn9f1vr7cyura+udwsbwv7o6v9w+ajsk0wwzlrklvipqciuny6a+1qjlzhavjs8mvitb9sgj+rojlimje0rhnngrznuztfolit+1z+cljngtiq14uchons75a9olgzrgwzoma0az+1yu615uzguntjdkaudwkf44qktge+ftumtlxso/eixalljmqvydyko0zych1smohztgbip957czglhovzpzvgykm4esqmz/e16xcozyuqizzq7wwkbuezdemuxajb4svlphlwdfxqcopsoiczinaexakavxada6hdg1g0idheiyxthpqvnmstepozq/gd7/0hadupva==</latexit> <latexit sha1_base64="orz9ok9ib6etihkm+gbfvgnevgu=">aaab6nicbvdlsgnbeoynrxhfuy9ebopgkewgqy8blx4jmgcks5iddjihm7plzkwqlnycfw+kepu/anpvwbj4+djhy0ffxddhdfiedg+v6l1tb9jcym8xdnb9g+kh0cne6eayzfitatibouxghdciuwlwikmhlyjebxu7/5gnrwwnbcykhpapf+5xr66q7a10iyw/7m9avkmwikvq/vmdhgrd4lenf7nuorjmugpagz/ymkpacizwuuikbhpkrnsabucvlwjcbhbqhjw5puf6sxalljmpvycyko0zy8h1smqhztmbiv957dtr8kmqys1qnh8ut8vxmzk+jfpcymirejlgnubivssdvl1qvtcceeyy+vkkalhpjl4nalcqfz5oeetuecariektxaderayacp8awvnvcevffvbd6a8xyzx/ahvspa1+pvq==</latexit> sha1_base64="mdshzsihynbiddbu00qwz+o=">aaab6nicbvdlsgnbeoynrxhfazx6grietechom6mfjrpoazamzk9lkymzsmjmrhcwf4mwdil7fv/apphnzb5w8dppy0fbuddpdfsacaen505ubx1jcyu/xdjzds/ca+ltrnitagixmsihwldnjg4yzttujolienlbc0exub91tpvks78w4oyhaa8kirrcx0qovxtuat4m6bv4i9iuzb//chevzfqpfer49jkqg0hgoto76xmcddyjdc6atqttvnmbnhae1ykrggoshmp07qivx6kiqvlwnqtp09kwgh9vietlngm9tllt8z+ukjroimiat1fbj5ouilcmto+nfqm8ujyaplcfemxsrikosmdenyinwv9+ezu0qxxfq/gno0zmcmpx1ccu/dhhgpwdxvoaiebpmatpdvcexrennd5a85zzbzbhzhvp+nikhm=</latexit> <latexit sha1_base64="sce6bcwpzygag0ol80ddv1jg=">aaab7icbzdlsgmxfibp1futt6pln8eiucoziujogi5cvraxaevjpgfa0exmtdjchfosbqqo0qp4gu481mlwtt/shw8f/nkhoohwuujet+o5mv1bx1jexmbmt7zcvv9q1vgigfzyjcjv96lgwsvwddcc67fcgvoca7/eplxhlfphsl7m4ixfdku5afn1firrxbysh5auclbtgdiiydn4fc1edonaaacjv/1exelalrgiao1gpju0rpcpwjncyayyay8r6tisni5kgqfvpdn4hobfohwsrsk8amnv/d6q01hoq+ryypkanf7oj+v/wsexwuq5jbodks0+chjbteqmy5mov8imgfigthe7ke9qigz9kq5ewrvcevlqj4vpbfobmf0jnmliujoizt8oacsnalzagaawhp8apvzopz4rw741lpxpnhmifor8/v7wsyg==</latexit> sha1_base64="fp1kp6aqtjfdcjxhuex5jinqgey=">aaab7icbzdlsgmxfibpek1vnxpjlgevvgbn1z0ixlcvycbsmznnogzjjjckaoq1/cjxhbpoivoy78zm4w/hd4+p9zydnhj6uw6lrfzsli0vlkam4tv76xubvdnmtmsjrjfdzjcpd8knhuiherygsnlnaehlxvchf1lev+paiejd4ddm7zdlageoitruuss6tkvlmouivldiphsk559pmz4rncjxqxuxjoqkmatgnd0xnzknqom+sjfsgypkrvqhm9avdtkppo5xrq+t0srbp+xsssfu7i6whmcpqt5uhxb6zztlzv6yzyhdwtowke+sktt4kekkwitnypcs0zyihfijtws5kwj9qytcekg+p4mupa+145lnlrxrt1g+gylysa8hcaqeneizrqacvwag4qfe4nw5dr6dn+d9urrgthv4i+cjx+08pqn</latexit> 5 B f p f S A f z m D altra parte per il primo principio della termodinamica la variazione di energia interna del è data da U = Q W amb (14) Ricordando che - il calore Q scambiato è nullo perché il contenitore è adiabatico; - la variazione di energia interna per un perfetto è data da U = nc V T (dove c V = R perché il è monoatomico); abbiamo U = 0 W amb nc V (T f T ) = + W amb }{{} W peso (m1 + m ) n R (T f T ) = (E fin P E in P ) (T f T ) = ( + m )g z (15) Le due equazioni (1) e (15) sono due equazioni per le due incognite T f e z. T f = ( + m )g(h z ) (T f T ) = ( + m )g z (16)

6 6 Per risolvere il sistema sostituiamo la prima nella seconda ( ) (m1 + m )g(h z ) T = ( + m )g z ( + m )g(h z ) T = ( + m )g z ( + m )gh T = 5 ( + m )g z 6 5 h T 5 ( + m )g = z [divido per 5 ( + m )g] z = m 0.5 mol/ K/ mol / K/ 5 (6 + 60) Kg 9.81 m = s J = 0.75 m 0.8 Kg m = [uso J = Kg m s ] s = 0.75 m 0.8 m = = 0.47 m (18) Pertanto la nuova altezza del pistone dal fondo del cinindro è h f = h z = 0.65 m 0.47 m = 0.78 m (19) Dalla prima delle equazioni (16) ricaviamo T f = (6 + 60) Kg 9.81 m 0.78 m s = J 0.5 mol / 8.14 mol / K J/ (17) m m Kg s = K = J = 5.4 K (0)