<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 ()
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 ] 0.5 8.14 8 9.81 0.65 = 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)
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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 = 6 5 0.65 m 0.5 mol/ 8.14 1.4 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 96 9.81 0.78 m Kg s = K = 0.5 8.14 J = 5.4 K (0)