The entropy change computed the way it is calculated: each gas expands freely on its own, the two are put in thermal contact, and then they are superposed in one vessel. The gases are material points, drawn with a little size so they can be seen.
Two compartments of $10\ \mathrm{L}$ and $15\ \mathrm{L}$ are separated by a fixed, adiabatic and impermeable wall. In the first we put O₂ at $288\ \mathrm{K}$ and $2\ \mathrm{atm}$, that is $0.846\ \mathrm{mol}$; in the second, N₂ at $300\ \mathrm{K}$ and $1\ \mathrm{atm}$, $0.609\ \mathrm{mol}$. Both diatomic ideal gases, $c_v = 5R/2$. The separating wall is suddenly removed; obtain the entropy change of the whole system. If the gases were the same, how would the result change?
A switch below gives the nitrogen the oxygen's amount ($1.39\ \mathrm{atm}$): with equal amounts the second step ends with equal pressures and the superposition is seen at its plainest. The page of the statement runs the wall itself.
The final state is fixed by the closure: $T_\mathrm{f} = 293.0\ \mathrm{K}$ and $p_\mathrm{f} = 1.40\ \mathrm{atm}$ in $25\ \mathrm{L}$ ($294\ \mathrm{K}$ and $1.63\ \mathrm{atm}$ with equal amounts). The route to its entropy is three steps, each one a change of state whose $\Delta S$ is known: a double free expansion, each gas alone into its $25\ \mathrm{L}$ at its own temperature; thermal contact, each gas at its $25\ \mathrm{L}$, to $T_\mathrm{f}$; and superposition, the two vessels made one, where each gas keeps its own pressure, 0.81 and $0.59\ \mathrm{atm}$, and the two add up to $p_\mathrm{f}$ (Dalton). For two different ideal gases the last step changes nothing, and the mixing entropy was all in the first.
Two vessels of $25\ \mathrm{L}$, one for each gas; the shaded part is vacuum. The points fly in straight lines and bounce off the walls; they pass through each other.
Each step is a change of state; its entropy change is a state-function formula, not a path integral.
Hidden. Show it when the class is ready for the balance.
| O₂ and N₂ | one gas | |
|---|---|---|
| 1 · O₂ free expansion, $10\ \mathrm{L}$ to $25\ \mathrm{L}$ | ||
| 1 · N₂ free expansion, $15\ \mathrm{L}$ to $25\ \mathrm{L}$ | ||
| 2 · O₂ from $288\ \mathrm{K}$ to $293.0\ \mathrm{K}$ | ||
| 2 · N₂ from $300\ \mathrm{K}$ to $293.0\ \mathrm{K}$ | ||
| 3 · Superposition | ||
| $\Delta S$ of the system, which is isolated | ||
| The wall freed instead of removed, for comparison |
Entropia-aldaketa kalkulatzen den moduan kalkulatuta: gas bakoitza bere kabuz hedatzen da askeki, biak kontaktu termikoan jartzen dira, eta gero ontzi batean gainjartzen dira. Gasak puntu materialak dira, tamaina pixka batekin marraztuak, ikus daitezen.
10 eta 15 l-ko gordailuak finkoa, adiabatikoa eta iragaztezina den hormak banandu ditu. Lehenengo gunean O₂ gasa sartu dugu: $288\ \mathrm{K}$-ean eta $2\ \mathrm{atm}$-n, hau da, $0.846\ \mathrm{mol}$. Bigarrenean, N₂ gasa; kasu honetan, $300\ \mathrm{K}$-ean eta $1\ \mathrm{atm}$-n, $0.609\ \mathrm{mol}$. Biak gas ideal diatomikotzat hartuko ditugu ($c_v = 5R/2$). Hasierako baldintzetatik abiatuz, bereizte-horma bat-batean kendu dugu. Lortu sistema osoari dagokion entropia-aldaketa. Gordailuetako gasak berdinak izanik, nola aldatuko litzateke emaitza?
Beheko etengailu batek nitrogenoari oxigenoaren kantitatea ematen dio ($1.39\ \mathrm{atm}$): kantitate berdinekin bigarren urratsa presio berdinekin amaitzen da, eta gainjartzea garbien ikusten da. Enuntziatuaren orriak horma bera abiarazten du.
Azken egoera finkatuta dago, ontzia isolatua baita: $T_\mathrm{f} = 293.0\ \mathrm{K}$ eta $p_\mathrm{f} = 1.40\ \mathrm{atm}$, $25\ \mathrm{L}$-tan ($294\ \mathrm{K}$ eta $1.63\ \mathrm{atm}$ kantitate berdinekin). Haren entropiarako bidea hiru urrats da, bakoitza egoera-aldaketa bat, $\Delta S$ ezaguna duena: hedapen aske bikoitza, gas bakoitza bakarrik bere $25\ \mathrm{L}$-etara, bere tenperaturan; kontaktu termikoa, gas bakoitza bere $25\ \mathrm{L}$-etan, $T_\mathrm{f}$-ra; eta gainjartzea, bi ontziak bat eginda, non gas bakoitzak bere presioa gordetzen duen, 0.81 eta $0.59\ \mathrm{atm}$, eta biek $p_\mathrm{f}$ ematen duten batuta (Dalton). Bi gas ideal desberdinentzat azken urratsak ez du ezer aldatzen, eta nahaste-entropia osoa lehenengoan zegoen.
$25\ \mathrm{L}$-ko bi ontzi, bat gas bakoitzeko; itzaldutako zatia hutsa da. Puntuak lerro zuzenean doaz eta hormetan errebotatzen dute; elkar zeharkatzen dute.
Urrats bakoitza egoera-aldaketa bat da; haren entropia-aldaketa egoera-funtzio baten formula da, ez bide-integral bat.
Ezkutatuta. Erakutsi ezazu klasea balantzerako prest dagoenean.
| O₂ eta N₂ | gas bat | |
|---|---|---|
| 1 · O₂-aren hedapen askea, $10\ \mathrm{L}$-etik $25\ \mathrm{L}$-era | ||
| 1 · N₂-aren hedapen askea, $15\ \mathrm{L}$-etik $25\ \mathrm{L}$-era | ||
| 2 · O₂, $288\ \mathrm{K}$-etik $293.0\ \mathrm{K}$-era | ||
| 2 · N₂, $300\ \mathrm{K}$-etik $293.0\ \mathrm{K}$-era | ||
| 3 · Gainjartzea | ||
| Sistemaren $\Delta S$, isolatua baita | ||
| Horma kendu beharrean askatuta, konparatzeko |