The statement's free expansion, the reversible isotherm that undoes it, and what changes when the expansion is made to lift a load.
The real gas under study expands freely, in thermal contact with a heat reservoir at $T = 300\ \mathrm{K}$. In this process it absorbs $Q = 50\ \mathrm{J}$. After this first process the gas is compressed isothermally and reversibly until it recovers its initial state. In this second process the gas gives off $300\ \mathrm{J}$.
Obtain: 1. the entropy change of the gas in the first process; 2. the work done by the gas in the second process.
Aztertu beharreko gas erreala era askean zabaldu da, $T = 300\ \mathrm{K}$-eko bero-iturriarekin ukipen termikoan. Prozesu horretan xurgatu duen beroa da $Q = 50\ \mathrm{J}$. Lehenengo prozesu horren ondoren, gasa konprimitu dugu, era isotermo itzulgarrian; hasierako egoera berreskuratu arte. Bigarren prozesu horretan, gasak $300\ \mathrm{J}$ kanporatu du. Lortu honako hauek: gasaren entropia-aldaketa, lehenengo prozesuan; gasak egindako lana, bigarren prozesuan.
The two states, $\mathrm{i}$ and $\mathrm{f}$, are fixed, and so is everything the gas owns: $\Delta U = +50\ \mathrm{J}$ and $\Delta S = +1\ \mathrm{J\,K^{-1}}$ from i to f. What the expansion does to its surroundings is not fixed. With no load on the piston (the statement), nothing is lifted. With a load of mass $M$, the gas lifts it, the reservoir pays for that work as extra heat, and the entropy of the universe rises by less.
A cylinder in the reservoir, a piston with vacuum behind it, a cord over a pulley to the load.
Equilibrium states only. The irreversible process 1 has none between i and f.
Hidden. Show it when the class is ready for the balance.
| 1 · irreversible | 2 · reversible | cycle | |
|---|---|---|---|
| $\Delta U$ of the gas | $+50.0\ \mathrm{J}$ | $-50.0\ \mathrm{J}$ | $0\ \mathrm{J}$ |
| Work stored in the load, $Mg\,\Delta h$ | $-250.0\ \mathrm{J}$ | ||
| Net heat from the reservoir | $-300.0\ \mathrm{J}$ | ||
| $\Delta S$ of the gas | $+1.000\ \mathrm{J\,K^{-1}}$ | $-1.000\ \mathrm{J\,K^{-1}}$ | $0\ \mathrm{J\,K^{-1}}$ |
| $\Delta S$ of the reservoir | $+1.000\ \mathrm{J\,K^{-1}}$ | ||
| $\Delta S$ of the universe | $0\ \mathrm{J\,K^{-1}}$ | ||
| Work lost, $T\,\Delta S_\mathrm{univ}$ | $0\ \mathrm{J}$ |
Enuntziatuko hedapen askea, hura desegiten duen isoterma itzulgarria, eta zer aldatzen den hedapenak zama bat altxatzen duenean.
Aztertu beharreko gas erreala era askean zabaldu da, $T = 300\ \mathrm{K}$-eko bero-iturriarekin ukipen termikoan. Prozesu horretan xurgatu duen beroa da $Q = 50\ \mathrm{J}$. Lehenengo prozesu horren ondoren, gasa konprimitu dugu, era isotermo itzulgarrian; hasierako egoera berreskuratu arte. Bigarren prozesu horretan, gasak $300\ \mathrm{J}$ kanporatu du. Lortu honako hauek: gasaren entropia-aldaketa, lehenengo prozesuan; gasak egindako lana, bigarren prozesuan.
Bi egoerak, $\mathrm{i}$ eta $\mathrm{f}$, finkoak dira, eta gasari dagokion guztia ere bai: $\Delta U = +50\ \mathrm{J}$ eta $\Delta S = +1\ \mathrm{J\,K^{-1}}$, i-tik f-ra. Hedapenak ingurunean eragiten duena, ordea, ez dago finkatuta. Pistoiaren gainean zamarik ez badago (enuntziatuko kasua), ez da ezer altxatzen. $M$ masako zama bat badago, gasak altxatu egiten du; bero-iturriak ordaintzen du lan hori, bero gehigarri gisa, eta unibertsoaren entropia gutxiago handitzen da.
Zilindro bat bero-iturrian, pistoi bat atzean hutsa duela, eta soka bat, txirrika baten gainetik, zamaraino.
Oreka-egoerak baino ez. 1. prozesu itzulezinak ez du halakorik i-ren eta f-ren artean.
Ezkutatuta. Erakutsi ezazu klasea balantzerako prest dagoenean.
| 1 · itzulezina | 2 · itzulgarria | zikloa | |
|---|---|---|---|
| Gasaren $\Delta U$ | $+50.0\ \mathrm{J}$ | $-50.0\ \mathrm{J}$ | $0\ \mathrm{J}$ |
| Zaman metatutako lana, $Mg\,\Delta h$ | $-250.0\ \mathrm{J}$ | ||
| Bero-iturriak emandako bero garbia | $-300.0\ \mathrm{J}$ | ||
| Gasaren $\Delta S$ | $+1.000\ \mathrm{J\,K^{-1}}$ | $-1.000\ \mathrm{J\,K^{-1}}$ | $0\ \mathrm{J\,K^{-1}}$ |
| Bero-iturriaren $\Delta S$ | $+1.000\ \mathrm{J\,K^{-1}}$ | ||
| Unibertsoaren $\Delta S$ | $0\ \mathrm{J\,K^{-1}}$ | ||
| Lan galdua, $T\,\Delta S_\mathrm{univ}$ | $0\ \mathrm{J}$ |