Oinarrizko ekuazioaren eraikuntza
Zenbait ariketa, adibide
Kontuz ibili behar da oinarrizko ekuazioa eraiki behar denean egoera-ekuazioetatik abiatuta. Oinarrizko ekuazioaren zer forma, energetikoa edo entropikoa, lortu(behar)ko dugun, egoera-ekuazioek esango digute. Beraz, ondo erreparatu egoera-ekuazioetan agertzen diren aldagai naturalei, hortxe baitago gakoa.
Hurrengo sei ariketetan, lortu oinarrizko ekuazioa. Kontuan izan sistema denak hidrostatikoak direla, baina bina egoera-ekuazio baino ez dela ezagutzen enuntziatuetan.
1. ARIKETA
T = 3\,A\,\frac{s^{2}}{v}
p = A\,\frac{s^{3}}{v^{2}}
U = A \frac{S^{3}}{V}\frac{1}{N} + N \cdot k
\mu = - A\frac{s^{3}}{v} + k
2. ARIKETA
U = P\,V
P = B\,T^{2}
d\left(\frac{\mu}{T}\right) = 0 \Rightarrow \left(\frac{\mu}{T}\right) = k \Rightarrow \left(\frac{\mu}{T}\right) = \left(\frac{\mu}{T}\right)_{0}
S = 2 B^{\frac{1}{2}} V^{\frac{1}{2}} U^{\frac{1}{2}} - \left(\frac{\mu}{T}\right)_{0} N
3. ARIKETA
u = \frac{3}{2}\, p\,v
u^{\frac{1}{2}} = B\,T\,v^{\frac{1}{3}}
\left(\frac{\mu}{T}\right) = - \frac{1}{3}B v^{\frac{1}{3}} u^{\frac{1}{2}}
S = 2 B V^{\frac{1}{3}} U^{\frac{1}{2}} N^{\frac{1}{6}} + Ns_{0}
\begin{array}{rcclccl} d u & = & T & d s & + & p & d v \\ d u & = & \frac{3 A s^{2}}{v} & d s & - & \frac{A s^{3}}{v^{2}} & d v \\\\ \int d u\equiv u & = & \int \frac{3 A s^{2}}{v} & d s & + & f(v) & \end{array}
d u = \int \frac{3 A s^{2}}{v} d s + f(v) \Rightarrow\left(\frac{\partial u}{\partial v}\right)_{s}=-\frac{A s^{3}}{v^{2}}+f^{\prime}(v)\Rightarrow
\cancel{-\frac{A s^{3}}{v^{2}}}+f^{\prime}(v) = \cancel{-\frac{A s^{3}}{v^{2}}} \\ \Rightarrow f^{\prime}(v)=0 \Rightarrow f(v)=\textrm { ktea }
\boxed {u = \frac{A s^{3}}{v} + \textrm { ktea } }
\begin{array}{rcclccl} u \equiv \frac{U}{N} & = & A \frac{\left(S/N\right)^{3}}{\left(V/N\right)}+ N \cdot\textrm{ktea} \Rightarrow \\ U & = & A \frac{S^{3}}{N^{\cancel{3}}} \frac{\cancel{N}}{V} \cancel{N} + N \cdot\textrm{ktea} \Rightarrow \end{array}
\boxed{U=A \frac{S^{3}}{V N}+N \cdot\textrm{ktea}}
\left[dp\right]_{U}={\underbrace{\left(\frac{\partial p}{\partial T}\right)_{U}}{\frac{pV\alpha - C_{p}}{V\left(-T\alpha + p\kappa_{T}\right)}}} \left[{\color{red}{dT}}\right]_{U}\left(\equiv \color{red}{\left(\frac{\partial T}{\partial V}\right)_{U}dV}\right)
4. ARIKETA
U = \frac{1}{2}\, P\,V
T^{2} = A\, \frac{U^{\frac{3}{2}}}{V\,N^{\frac{1}{2}}}
\left(\frac{\mu}{T}\right) = -A^{-\frac{1}{2}} v^{\frac{1}{2}} u^{\frac{1}{4}}
S = 4 A^{-\frac{1}{2}} U^{\frac{1}{4}} V^{\frac{1}{2}} N^{\frac{1}{4}} + S_{0}
5. ARIKETA
Gas ideala
Gas ideala monoatomikoa da; ez da orokortasunik galtzen.
P V = N R T
U = \frac{3}{2} N R T
6. ARIKETA
Van der Waals-en jariakina
\frac{1}{T} = \frac{cR}{\left(u + \frac{a}{v}\right)}
\frac{P}{T} = \frac{R}{v-b} - \frac{acR}{\left( uv^{2} + av\right)}
\boxed{S = NR\ln\left[\left(v-b\right) \left(u+\frac{a}{v}\right)^{c}\right] + N s_{0}}
Oinarrizko ekuazioaren propietateak eta gehiago…
7. ARIKETA
Zenbait sistema termodinamikori dagokion oinarrizko ekuazioa agertuko da ondoren. Dena den, horietariko bost, II, III eta IV postulatuetatik baten edo gehiagoren aurka dago; eta, ondorioz, fisikoki onartezinak dira.
Adieraz itzazu aipatu bostak, eta betegabeko postulatua(k). Adierazpen horietan guztietan, v_{0}, \theta eta R konstante positiboak dira.
Onartu erro positibo erreala soilik, berretura frakzionarioa ageri den guztietan.
S = \left( \dfrac{R^2}{v_0 \theta } \right)^{\frac{1}{3}} \left[ {NVU} \right]^{\frac{1}{3}}
S = \left( \dfrac{R}{\theta^2} \right)^{\frac{1}{3}} \left[ \dfrac{NU}{V}\right]^{\frac{2}{3}}
S = \left( \dfrac{R}{\theta} \right)^{\frac{1}{2}} \left[ NU - \dfrac{R\theta V^2 }{v_0^{2} } \right]^{\frac{1}{2}}
S = \left( \dfrac{R^2 \theta }{v_0^3 } \right)\dfrac{V^3}{NU}
S = \left( \dfrac{R^3 }{v_0\theta^2} \right)^{\frac{1}{5}} \left[ N^2 VU^2 \right]^{\frac{1}{5}}
S = NR\ln \left( \dfrac{UV}{N^{2} R\theta v_0 } \right)
S = \left( \dfrac{R}{\theta } \right)^{\frac{1}{2}} \left[ NU \right]^{\frac{1}{2}} \exp \left(-\dfrac{V^{2}}{2N^{2} v_{0}^{2}} \right)
S = \left( \dfrac{R}{\theta} \right)^{\frac{1}{2}} \left[ NU \right]^{\frac{1}{2}} \exp \left( -\dfrac{UV}{NR\theta v_{0}} \right)
U = \left( \dfrac{v_0\theta }{R} \right)\dfrac{S^{2}}{V}\exp \left( \dfrac{S}{NR} \right)
U = \left( \dfrac{R\theta }{v_0} \right) NV \left( 1 + \dfrac{S}{NR} \right)\exp \left( - \dfrac{S}{NR} \right)
C 1.9.1
8. ARIKETA
Lortu U barne-energia S, V eta N aldagaien funtzioan, aurreko ariketako fisikoki onargarriak diren oinarrizko ekuazioen kasuetarako.
C 1.9.2
9. ARIKETA
A sistemaren oinarrizko ekuazioa honako hau da:
S_{A} = \left( \frac{R^{2}}{v_{0} \theta} \right)^{\frac{1}{3}} \left[ N_{A} V_{A} U_{A} \right]^{\frac{1}{3}}
B sistemarena, berriz:
S_{B} = \left( \frac{R^{2}}{v_{0} \theta} \right)^{\frac{1}{3}} \left[ N_{B} V_{B} U_{B} \right]^{\frac{1}{3}}
Lortu A + B sistema konposatuaren oinarrizko ekuazioa
C 1.9.3
10. ARIKETA
Onartu aurreko ariketan deskribatutako A eta B azpisistemen arteko bereizte-horma finkoa, iragaztezina eta energia trukatzen utziko duena dela. Modu berean, demagun A sistemaren bolumena 9 cm^3 dela, eta mol-kopurua 3. B sistemaren kasuan, bolumena 4 cm^3 da eta mol-kopurua, 2.
Sistema konposatuaren energia osoa 80 J da.
Adieraz ezazu entropia A sistemari dagokion \frac{U_{A}}{(U_{A}+U_{B})} energia-frakzioaren funtzioan.
Zenbatekoa da azpisistema bakoitzari dagokion barne-energia sistemak oreka lortu duen kasurako?
C 1.9.4
11. ARIKETA
Lortu ondoko oinarrizko ekuazioko sistemari dagozkion hiru egoera-ekuazioak:
U = \left( \frac{v_0 \theta}{R^{2}} \right)\frac{S^{3}}{NV}
C 2.2.1
U=\left(\frac{v_{0} \theta}{R^{2}}\right) \frac{S^{3}}{N V} \Rightarrow U=C \cdot \frac{S^{3}}{N V} \quad U=U(S, V, N)
\left\{\begin{aligned} T &\equiv\left(\frac{\partial U}{\partial S}\right)_{V, N} \Rightarrow T=\frac{3 C S^{2}}{N V}\\ -p &\equiv\left(\frac{\partial U}{\partial V}\right)_{S, N} \Rightarrow-p=-\frac{C S^{3}}{N V^{2}} \Rightarrow p=C \frac{S^{3}}{N V^{2}}\\ \mu &\equiv\left(\frac{\partial U}{\partial N}\right)_{S, V} \Rightarrow \mu=-\frac{C S^{3}}{N^{2} V} \end{aligned} \right.\}
12. ARIKETA
Lortu aurreko sistemari dagokion \mu potentzial kimikoa.
Adieraz ezazu emaitza T, V eta N parametroen funtzioan.
C 2.2.2
\mu=\mu(T, V, N)
T=3 C \frac{S^{2}}{N V} \Rightarrow S=\left(\frac{1}{3} \frac{N V T}{C}\right)^{\frac{1}{2}}
\mu=-C \frac{S^{3}}{N^{2} V} \Rightarrow \mu=-C \frac{1}{N^{2} V} \cdot\left(\frac{1}{3} \frac{N V T}{C}\right)^{\frac{3}{2}}
\mu=-\frac{C}{C^{\frac{3}{2}}} \cdot \frac{T^{\frac{3}{2}2} N^{\frac{3}{2}} V^{\frac{3}{2}}}{N^{2} \cdot V}\left(\frac{1}{3}\right)^{\frac{3}{2}}
\mu=-\left(\frac{1}{3}\right)^{\frac{3}{2}} C^{-\frac{1}{2}} \frac{1}{N^{\frac{1}{2}}} V^{\frac{1}{2}} \cdot T^{\frac{3}{2}}
13. ARIKETA
Lortu aurreko sistemaren kasuan, eta tenperatura finkoa denean, presioak bolumenarekiko duen mendekotasuna, hautazkoa den eskalako diagraman.
Irudikatu bi tenperatura desberdini dagozkien lerro isotermoak, eta adieraz ezazu zein den bietatik tenperatura handienari dagokiona.
C 2.2.3
p =p\left(T, V\right)
T =3 C \frac{S^{2}}{N V} \Rightarrow S=\frac{1}{3}\left(\frac{N V T}{C}\right)^{\frac{1}{2}}
p = C \frac{S^{3}}{N V^{2}} \downarrow p=C \frac{1}{N V^{2}}\left(\frac{N V T}{C}\right)^{\frac{3}{2}} \Rightarrow p=\frac{1}{C^{\frac{1}{2}}} \cdot N^{\frac{3}{2}-1} V^{\frac{3}{2}-2} T^{\frac{3}{2}}
\begin{aligned} &p=\frac{1}{C^{\frac{1}{2}}} \frac{N^{\frac{1}{2}}}{W^{\frac{1}{2}}} \cdot T^{\frac{3}{2}} \\ &T=K \\ &N=K \end{aligned}
p=\frac{N^{\frac{1}{2}} \cdot \frac{1}{C}}{v^{\frac{3}{2}}}
p \cdot V^{\frac{3}{2}}=K
14. ARIKETA
Lortu ondoko oinarrizko ekuazioko sistemari dagozkion hiru egoera-ekuazioak.
u = \left( \frac{\theta }{R} \right)s^{{2}} - \left( \frac{R\theta}{v_{{0}}^{{2}}} \right)v^{{2}}
C 2.2.4
15. ARIKETA
Lortu aurreko sistemari dagokion \mu potentzial kimikoa; adieraz ezazu emaitza T eta p parametroen funtzioan.
16. ARIKETA
Lortu ondoko oinarrizko ekuazioko sistemari dagozkion hiru egoera-ekuazioak.
u = \left(\frac{v_{0}\theta}{R}\right)\frac{s^{2}}{v}\exp \left(\frac{s}{R}\right)
C 2.2.6
17. ARIKETA
kuasiestatikoa den zabaltze adiabatikoan zehar (dS=0), tenperaturak bolumenarekiko duen mendekotasuna.
C 2.2.7
18. ARIKETA
Lortu, adierazpen entropikoan, ondoko oinarrizko ekuazioko sistemari dagozkion hiru egoera-ekuazioak.
u = \left( \frac{v_{o}^{\frac{1}{2}}\theta}{R^{\frac{3}{2}}} \right)\frac{s^{\frac{5}{2}}}{v^{\frac{1}{2}}}
C 2.3.1
19. ARIKETA
Lortu aurreko sistemaren kasuan, eta presioa finkoa den kasuan, tenperaturak bolumenarekiko duen mendekotasuna, hautazkoa den eskalako diagraman.
Irudikatu bi presio desberdini dagozkien lerro isobaroak, eta adierazi zein den bietatik presio handienari dagokiona.
C 2.3.2
20. ARIKETA
Lortu, adierazpen entropikoan, ondoko oinarrizko ekuazioko sistemari dagozkion hiru egoera-ekuazioak:
u = \left( \frac{\theta}{R} \right)s^{2} \exp \left( \frac{v^{2} }{v_0^{2}} \right)
C 2.3.3
21. ARIKETA
Bi osagaiko sistemari dagokion oinarrizko ekuazioa honako hau da,
S = NA + NR\ln \left( \frac{U^{\frac{3}{2}} V}{N^{\frac{5}{2}}} \right) - N_{1} R\ln \left( \frac{N_{1}}{N} \right) - N_{2} R\ln \left( \frac{N_{2}}{N} \right)
N \equiv N_{1} + N_{2}
Adierazpen horretan R=1.986 cal/mol K da, eta A, finkatu gabeko konstantea.
Lehenengo osagaiarekiko iragazkorra baina bigarrenarekiko ez den mintz diatermoak bi gunetan zatitu du 10 \mathscr{l}-ko bolumeneko eta itxita dagoen zilindro zurruna.
- Lehenengo gunean ondoko parametroez ezaugarritutako lagina kokatu da: N_{1}^{(1)} = 0.5, N_{2}^{(1)} = 0.75, V^{(1)} = 5 l eta T^{(1)} = 300 K.
- Bigarrenean, berriz, ondokoen bidez: N_{1}^{(2)} = 1, N_{2}^{(2)} = 0.5, V^{(2)} = 5 l eta T^{(2)} = 250 K.
Oreka lortutakoan, zein dira N_{1}^{(1)}, N_{1}^{(2)}, T, p^{(1)} eta p^{(2)} parametroen balioak?
C 2.8.1
\begin{array}{l} S = NA + NR\ln \left[ {\frac{{U^{\frac{3}{2}} V}}{{N^{\frac{5}{2}} }}} \right] - N_1 R\ln \left[ {\frac{{N_1 }}{N}} \right] - N_2 R\ln \left[ {\frac{{N_2 }}{N}} \right] \\ N = N_1 + N_2 \\ \\ \frac{1}{T} = \left( {\frac{{\partial S}}{{\partial U}}} \right)_{V,N_1 ,N_2 } \Rightarrow \frac{1}{T} = NR\frac{3}{2}\frac{1}{U} \\ \\ \frac{p}{T} = \left( {\frac{{\partial U}}{{\partial V}}} \right)_{S,N_1 ,N_2 } \Rightarrow \frac{p}{T} = NR\frac{1}{V} \\ \\ \frac{{\mu _1 }}{T} = - \left( {\frac{{\partial U}}{{\partial N_1 }}} \right)_{S,V,N_2 } \Rightarrow \frac{{\mu _1 }}{T} = - A + \frac{5}{2}R - R\ln \left[ {\frac{{U^{\frac{3}{2}} V}}{{N^{\frac{3}{2}} N_1 }}} \right] \\ \\ \frac{{\mu _2 }}{T} = - \left( {\frac{{\partial U}}{{\partial N_2 }}} \right)_{S,V,N_1 } \Rightarrow \frac{{\mu _2 }}{T} = - A + \frac{5}{2}R - R\ln \left[ {\frac{{U^{\frac{3}{2}} V}}{{N^{\frac{3}{2}} N_2 }}} \right] \\ \\ \\ \end{array}
\begin{array}{l} \left\{ \begin{array}{l} S = NA + NR\ln \left[ {\frac{{U^{\frac{3}{2}} V}}{{N^{\frac{5}{2}} }}} \right] - N_1 R\ln \left[ {\frac{{N_1 }}{N}} \right] - N_2 R\ln \left[ {\frac{{N_2 }}{N}} \right] \\ \\ N = N_1 + N_2 \\ U = U_1 + U_2 \\ \end{array} \right\}^{\left( i \right)} i = 1,2 \Rightarrow \\ \\ \left. \begin{array}{l} S^{\left( i \right)} = N^{\left( i \right)} A + N^{\left( i \right)} R\ln \left[ {\frac{ {{U^{\left( i \right)}}^{\frac{3}{2}} V^{\left( i \right)} }} {{N^{\left( i \right)}}^{\frac{5}{2} }}} \right] - {N_{1}}^{\left( i \right)} R\ln \left[ {\frac{{{N_{1}}^{\left( i \right)} }}{N}} \right] - {N_{2}}^{\left( i \right)} R\ln \left[ {\frac{{{N_{2}}^{\left( i \right)} }}{N}} \right] \\ \\ N^{\left( i \right)} = {N_{1}}^{\left( i \right)} + {N_{2}}^{\left( i \right)} \\ U^{\left( i \right)} = {U_{1}}^{\left( i \right)} + {U_{2}}^{\left( i \right)} \\ \end{array} \right\}i = 1,2 \Rightarrow \\ \\ \end{array}
\begin{array}{l} S^{\left( i \right)} = \left( {N_1^{\left( i \right)} + N_2^{\left( i \right)} } \right)A + N^{\left( i \right)} R\ln \left[ {\frac{{\left( {U_1^{\left( i \right)} + U_2^{\left( i \right)} } \right)^{\frac{3}{2}} V^{\left( i \right)} }}{{\left( {N_1^{\left( i \right)} + N_2^{\left( i \right)} } \right)^{\frac{5}{2}} }}} \right] - N_1^{\left( i \right)} R\ln \left[ {\frac{{N_1^{\left( i \right)} }}{{\left( {N_1^{\left( i \right)} + N_2^{\left( i \right)} } \right)}}} \right] - N_2^{\left( i \right)} R\ln \left[ {\frac{{N_2^{\left( i \right)} }}{{\left( {N_1^{\left( i \right)} + N_2^{\left( i \right)} } \right)}}} \right] \\ i = 1,2 \\ \\ \left( {\frac{1}{T}} \right)^{\left( i \right)} = \left( {\frac{{\partial S^{\left( i \right)} }}{{\partial U^{\left( i \right)} }}} \right)_{V,N_1^{\left( i \right)} ,N_2^{\left( i \right)} } \Rightarrow \left( {\frac{1}{T}} \right)^{\left( i \right)} = N^{\left( i \right)} R\frac{3}{2}\frac{1}{{U^{\left( i \right)} }} \\ \\ \left( {\frac{p}{T}} \right)^{\left( i \right)} = \left( {\frac{{\partial U^{\left( i \right)} }}{{\partial V^{\left( i \right)} }}} \right)_{S,N_1^{\left( i \right)} ,N_2^{\left( i \right)} } \Rightarrow \left( {\frac{p}{T}} \right)^{\left( i \right)} = N^{\left( i \right)} R\frac{1}{{V^{\left( i \right)} }} \\ \\ \left( {\frac{{\mu _1 }}{T}} \right)^{\left( i \right)} = - \left( {\frac{{\partial U^{\left( i \right)} }}{{\partial N_1^{\left( i \right)} }}} \right)_{S,V,N_2^{\left( i \right)} } \Rightarrow \left( {\frac{{\mu _1 }}{T}} \right)^{\left( i \right)} = - A + \frac{5}{2}R - R\ln \left[ {\frac{{{U^{\left( i \right)} }^{\frac{3}{2}} V^{\left( i \right)} }}{{\left( {N_1^{\left( i \right)} + N_2^{\left( i \right)} } \right)^{\frac{3}{2}} N_1^{\left( i \right)} }}} \right] \\ \\ \left( {\frac{{\mu _2 }}{T}} \right)^{\left( i \right)} = - \left( {\frac{{\partial U^{\left( i \right)} }}{{\partial N_2^{\left( i \right)} }}} \right)_{S,V,N_1^{\left( i \right)} } \Rightarrow \left( {\frac{{\mu _2 }}{T}} \right)^{\left( i \right)} = - A + \frac{5}{2}R - R\ln \left[ {\frac{{U^{\frac{3}{2}} V}}{{\left( {N_1^{\left( i \right)} + N_2^{\left( i \right)} } \right)^{\frac{3}{2}} N_2^{\left( i \right)} }}} \right] \\ \\ \end{array}
\begin{array}{l} \left. \begin{array}{l} dS^{{\rm{osoa}}} = 0 \\ \\ dS^{{\rm{osoa}}} = dS^{\left( 1 \right)} + dS^{\left( 2 \right)} \\ \end{array} \right\} \Rightarrow dS^{\left( 1 \right)} + dS^{\left( 2 \right)} = 0 \\ \end{array}
\begin{array}{l} dS^{\left( 1 \right)} = \left( {\frac{1}{T}} \right)^{\left( 1 \right)} dU^{\left( 1 \right)} + \left( {\frac{p}{T}} \right)^{\left( 1 \right)} dV^{\left( 1 \right)} - \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 1 \right)} dN_1^{\left( 1 \right)} - \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 1 \right)} dN_2^{\left( 1 \right)} \\ + \\ dS^{\left( 2 \right)} = \left( {\frac{1}{T}} \right)^{\left( 2 \right)} dU^{\left( 2 \right)} + \left( {\frac{p}{T}} \right)^{\left( 2 \right)} dV^{\left( 2 \right)} - \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 2 \right)} dN_1^{\left( 2 \right)} - \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 2 \right)} dN_2^{\left( 2 \right)} \\ \end{array}
\begin{array}{l} dS^{{\rm{osoa}}} = \left\{ {\left( {\frac{1}{T}} \right)^{\left( 1 \right)} - \left( {\frac{1}{T}} \right)^{\left( 2 \right)} } \right\}dU^{\left( 1 \right)} + \left\{ {\left( {\frac{p}{T}} \right)^{\left( 1 \right)} - \left( {\frac{p}{T}} \right)^{\left( 2 \right)} } \right\}dV^{\left( 1 \right)} - \\ \\ \hspace{1cm} -\left\{ {\left( {\frac{{\mu _1 }}{T}} \right)^{\left( 1 \right)} - \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 2 \right)} } \right\}dN_1^{\left( 1 \right)} - \left\{ {\left( {\frac{{\mu _2 }}{T}} \right)^{\left( 1 \right)} - \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 2 \right)} } \right\}dN_2^{\left( 1 \right)} \\ \\ \end{array}
\begin{array}{l} U^{\left( 1 \right)} + U^{\left( 2 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dU^{\left( 1 \right)} + dU^{\left( 2 \right)} = 0 \Rightarrow dU^{\left( 2 \right)} = - dU^{\left( 1 \right)} \\ \\ V^{\left( 1 \right)} + V^{\left( 2 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dV^{\left( 1 \right)} + dV^{\left( 2 \right)} = 0 \Rightarrow dV^{\left( 2 \right)} = - dV^{\left( 1 \right)} \\ \\ N_1^{\left( 1 \right)} + N_1^{\left( 2 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dN_1^{\left( 1 \right)} + dN_1^{\left( 2 \right)} = 0 \Rightarrow dN_1^{\left( 2 \right)} = - dN_1^{\left( 1 \right)} \\ \\ N_2^{\left( 1 \right)} + N_2^{\left( 2 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dN_2^{\left( 1 \right)} + dN_2^{\left( 2 \right)} = 0 \Rightarrow dN_2^{\left( 1 \right)} = - dN_2^{\left( 2 \right)} \\ \\ \\ \end{array}
\left\{ {\left( {\frac{1}{T}} \right)^{\left( 1 \right)} - \left( {\frac{1}{T}} \right)^{\left( 2 \right)} } \right\} = 0 \Rightarrow \left( {\frac{1}{T}} \right)^{\left( 1 \right)} = \left( {\frac{1}{T}} \right)^{\left( 2 \right)} \Rightarrow N^{\left( 1 \right)} R\frac{3}{2}\frac{1}{{U^{\left( 1 \right)} }} = N^{\left( 2 \right)} R\frac{3}{2}\frac{1}{{U^{\left( 2 \right)} }} \Rightarrow \boxed {\frac{{N^{\left( 1 \right)} }}{{U^{\left( 1 \right)} }} = \frac{{N^{\left( 2 \right)} }}{{U^{\left( 2 \right)} }} }
\left\{ {\left( {\frac{p}{T}} \right)^{\left( 1 \right)} - \left( {\frac{p}{T}} \right)^{\left( 2 \right)} } \right\} = 0 \Rightarrow \left( {\frac{p}{T}} \right)^{\left( 1 \right)} = \left( {\frac{p}{T}} \right)^{\left( 2 \right)} \Rightarrow N^{\left( 1 \right)} R\frac{1}{{V^{\left( 1 \right)} }} = N^{\left( 2 \right)} R\frac{1}{{V^{\left( 2 \right)} }} \Rightarrow \boxed { \frac{{N^{\left( 1 \right)} }}{{V^{\left( 1 \right)} }} = \frac{{N^{\left( 2 \right)} }}{{V^{\left( 2 \right)} }} }
\begin{array}{l} \left\{ {\left( {\frac{{\mu _1 }}{T}} \right)^{\left( 1 \right)} - \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 2 \right)} } \right\} = 0 \Rightarrow \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 1 \right)} = \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 2 \right)} \Rightarrow \\ \\ - A + \frac{5}{2}R - R\ln \left[ {\frac{{{U^{\left( 1 \right)} }^{\frac{3}{2}} V^{\left( 1 \right)} }}{{\left( {N_1^{\left( 1 \right)} + N_2^{\left( 1 \right)} } \right)^{\frac{3}{2}} N_1^{\left( 1 \right)} }}} \right] = - A + \frac{5}{2}R - R\ln \left[ {\frac{{{U^{\left( 2 \right)} }^{\frac{3}{2}} V^{\left( 2 \right)} }}{{\left( {N_1^{\left( 2 \right)} + N_2^{\left( 2 \right)} } \right)^{\frac{3}{2}} N_1^{\left( 2 \right)} }}} \right] \Rightarrow \\ \\ \frac{{{U^{\left( 1 \right)} }^{\frac{3}{2}} V^{\left( 1 \right)} }}{{\left( {N_1^{\left( 1 \right)} + N_2^{\left( 1 \right)} } \right)^{\frac{3}{2}} N_1^{\left( 1 \right)} }} = \frac{{{U^{\left( 2 \right)} }^{\frac{3}{2}} V^{\left( 2 \right)} }}{{\left( {N_1^{\left( 2 \right)} + N_2^{\left( 2 \right)} } \right)^{\frac{3}{2}} N_1^{\left( 2 \right)} }} \Rightarrow \boxed { \frac{{{U^{\left( 1 \right)} }^{\frac{3}{2}} V^{\left( 1 \right)} }}{{N_1^{\left( 1 \right)} }} = \frac{{{U^{\left( 2 \right)} }^{\frac{3}{2}} V^{\left( 2 \right)} }}{{N_1^{\left( 2 \right)} }} } \\ \end{array}
Oharra: koadroko baldintzan \left(N^{(i)}\right)^{\frac{3}{2}} faktorea falta da zorrotz hartuta: baldintza osoa \dfrac{{U^{(1)}}^{\frac{3}{2}} V^{(1)}}{{N^{(1)}}^{\frac{3}{2}} N_1^{(1)}} = \dfrac{{U^{(2)}}^{\frac{3}{2}} V^{(2)}}{{N^{(2)}}^{\frac{3}{2}} N_1^{(2)}} da. Hala ere, orekan T^{(1)} = T^{(2)} denez, U^{(i)} = \frac{3}{2}N^{(i)}RT eta \left(U^{(i)}/N^{(i)}\right)^{\frac{3}{2}} berdina da bi aldeetan; beraz, faktore hori sinplifikatu egiten da eta amaierako emaitzak (V^{(1)} = V^{(2)} izanik, N_1^{(1)} = N_1^{(2)}) ez dira aldatzen.
\begin{array}{l} \left\{ {\left( {\frac{{\mu _2 }}{T}} \right)^{\left( 1 \right)} - \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 2 \right)} } \right\} = 0 \Rightarrow \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 1 \right)} = \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 2 \right)} \Rightarrow \\ \\ - A + \frac{5}{2}R - R\ln \left[ {\frac{{{U^{\left( 1 \right)} }^{\frac{3}{2}} V^{\left( 1 \right)} }}{{\left( {N_1^{\left( 1 \right)} + N_2^{\left( 1 \right)} } \right)^{\frac{3}{2}} N_2^{\left( 1 \right)} }}} \right] = - A + \frac{5}{2}R - R\ln \left[ {\frac{{{U^{\left( 2 \right)} }^{\frac{3}{2}} V^{\left( 2 \right)} }}{{\left( {N_1^{\left( 2 \right)} + N_2^{\left( 2 \right)} } \right)^{\frac{3}{2}} N_2^{\left( 2 \right)} }}} \right] \\ \\ \frac{{{U^{\left( 1 \right)} }^{\frac{3}{2}} V^{\left( 1 \right)} }}{{\left( {N_1^{\left( 1 \right)} + N_2^{\left( 1 \right)} } \right)^{\frac{3}{2}} N_2^{\left( 1 \right)} }} = \frac{{{U^{\left( 2 \right)} }^{\frac{3}{2}} V^{\left( 2 \right)} }}{{\left( {N_1^{\left( 2 \right)} + N_2^{\left( 2 \right)} } \right)^{\frac{3}{2}} N_2^{\left( 2 \right)} }} \Rightarrow \boxed { \frac{{{U^{\left( 1 \right)} }^{\frac{3}{2}} V^{\left( 1 \right)} }}{{N_2^{\left( 1 \right)} }} = \frac{{{U^{\left( 2 \right)} }^{\frac{3}{2}} V^{\left( 2 \right)} }}{{N_2^{\left( 2 \right)} }} } \\ \\ \\ \end{array}
\begin{array}{l} dS^{\left( 1 \right)} = \left( {\frac{1}{T}} \right)^{\left( 1 \right)} dU^{\left( 1 \right)} + \left( {\frac{p}{T}} \right)^{\left( 1 \right)} dV^{\left( 1 \right)} - \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 1 \right)} dN_1^{\left( 1 \right)} - \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 1 \right)} dN_2^{\left( 1 \right)} \\ + \\ dS^{\left( 2 \right)} = \left( {\frac{1}{T}} \right)^{\left( 2 \right)} dU^{\left( 2 \right)} + \left( {\frac{p}{T}} \right)^{\left( 2 \right)} dV^{\left( 2 \right)} - \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 2 \right)} dN_1^{\left( 2 \right)} - \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 2 \right)} dN_2^{\left( 2 \right)} \\ \end{array}
\begin{array}{l} dS^{{\rm{osoa}}} = \left\{ {\left( {\frac{1}{T}} \right)^{\left( 1 \right)} - \left( {\frac{1}{T}} \right)^{\left( 2 \right)} } \right\}dU^{\left( 1 \right)} + \left\{ {\left( {\frac{p}{T}} \right)^{\left( 1 \right)} - \left( {\frac{p}{T}} \right)^{\left( 2 \right)} } \right\}dV^{\left( 1 \right)} - \\ \\ \hspace{1cm} -\left\{ {\left( {\frac{{\mu _1 }}{T}} \right)^{\left( 1 \right)} - \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 2 \right)} } \right\}dN_1^{\left( 1 \right)} - \left\{ {\left( {\frac{{\mu _2 }}{T}} \right)^{\left( 1 \right)} - \left( {\frac{{\mu _2 }}{T}} \right)^{\left( 2 \right)} } \right\}dN_2^{\left( 1 \right)} \\ \\ \end{array}
\begin{array}{l} U^{\left( 1 \right)} + U^{\left( 2 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dU^{\left( 1 \right)} + dU^{\left( 2 \right)} = 0 \Rightarrow dU^{\left( 2 \right)} = - dU^{\left( 1 \right)} \\ \\ \left\{ {U^{\left( 1 \right)} + U^{\left( 2 \right)} } \right\}^{\left( i \right)} = \left\{ {\left( {N_1^{\left( 1 \right)} + \frac{3}{4}} \right)R\frac{3}{2}T^{\left( 1 \right)} + \left( {N_1^{\left( 2 \right)} + \frac{1}{2}} \right)R\frac{3}{2}T^{\left( 2 \right)} } \right\}^{\left( i \right)} \\ \\ \left\{ {U^{\left( 1 \right)} + U^{\left( 2 \right)} } \right\} = \left\{ {\left( {\frac{1}{2} + \frac{3}{4}} \right)R\frac{3}{2}300 + \left( {1 + \frac{1}{2}} \right)R\frac{3}{2}250} \right\} \Rightarrow U^{\left( 1 \right)} + U^{\left( 2 \right)} = R\frac{3}{2}\left( {\frac{5}{4}300 + \frac{3}{2}250} \right) = 1125R \\ \\ U^{\left( 1 \right)} + U^{\left( 2 \right)} = 1125R \\ \\ V^{\left( 1 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dV^{\left( 1 \right)} = 0 \\ V^{\left( 2 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dV^{\left( 2 \right)} = 0 \\ V^{\left( 1 \right)} = V^{\left( 2 \right)} = V \\ \\ N_1^{\left( 1 \right)} + N_1^{\left( 2 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dN_1^{\left( 1 \right)} + dN_1^{\left( 2 \right)} = 0 \Rightarrow dN_1^{\left( 2 \right)} = - dN_1^{\left( 1 \right)} \\ \\ N_2^{\left( 1 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dN_2^{\left( 1 \right)} = 0 \\ N_2^{\left( 2 \right)} = {\rm{Konst}}{\rm{.}} \Rightarrow dN_2^{\left( 2 \right)} = 0 \\ N_2^{\left( 1 \right)} = \frac{3}{4} \\ N_2^{\left( 2 \right)} = \frac{1}{2} \\ \end{array}
\left. \begin{array}{l} \left\{ {\left( {\frac{1}{T}} \right)^{\left( 1 \right)} - \left( {\frac{1}{T}} \right)^{\left( 2 \right)} } \right\} = 0 \Rightarrow \left( {\frac{1}{T}} \right)^{\left( 1 \right)} = \left( {\frac{1}{T}} \right)^{\left( 2 \right)} \Rightarrow \frac{{\left( {N_1^{\left( 1 \right)} + \frac{3}{4}} \right)}}{{U^{\left( 1 \right)} }} = \frac{{\left( {N_1^{\left( 2 \right)} + \frac{1}{2}} \right)}}{{U^{\left( 2 \right)} }} \\ \\ \\ \left\{ {\left( {\frac{{\mu _1 }}{T}} \right)^{\left( 1 \right)} - \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 2 \right)} } \right\} = 0 \Rightarrow \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 1 \right)} = \left( {\frac{{\mu _1 }}{T}} \right)^{\left( 2 \right)} \Rightarrow \frac{{{U^{\left( 1 \right)} }^{\frac{3}{2}} }}{{\left( {N_1^{\left( 1 \right)} + \frac{3}{4}} \right)^{\frac{3}{2}} N_1^{\left( 1 \right)} }} = \frac{{{U^{\left( 2 \right)} }^{\frac{3}{2}} }}{{\left( {N_1^{\left( 2 \right)} + \frac{1}{2}} \right)^{\frac{3}{2}} N_1^{\left( 2 \right)} }} \\ \\ U^{\left( 1 \right)} + U^{\left( 2 \right)} = 1125R \\ \\ N_1^{\left( 1 \right)} + N_1^{\left( 2 \right)} = \frac{3}{2} \\ \end{array} \right\} \Rightarrow
\boxed { \begin{array}{l} N_1^{\left( 1 \right)} = N_1^{\left( 2 \right)} = \frac{3}{4} \\ \\ U^{\left( 1 \right)} = \frac{6}{5}U^{\left( 2 \right)} \Rightarrow U^{\left( 2 \right)} = \frac{5}{{11}}U^{\left( {{\rm{osoa}}} \right)} \Rightarrow U^{\left( 1 \right)} = \frac{6}{{11}}U^{\left( {{\rm{osoa}}} \right)} \\ \\ T = \frac{8}{{33}}\frac{1}{R}U^{\left( {{\rm{osoa}}} \right)} \\ \\ \left( {\frac{p}{T}} \right)^{\left( 1 \right)} = \frac{4}{{55}}U^{\left( {{\rm{osoa}}} \right)} \\ \\ \left( {\frac{p}{T}} \right)^{\left( 2 \right)} = \frac{2}{{33}}U^{\left( {{\rm{osoa}}} \right)} \\ \end{array}}
22. ARIKETA
He gasa duen gordailuak 1000 l-ko bolumenekoa da. Gasa 0.5 atm-n eta 20 ^{\circ}C-ko tenperaturan dago. Bolumen berdineko bigarren gordailuan dagoen He gasa 1 atm-n eta 80 ^{\circ}C-ko tenperaturan dago. Gordailu biak loturik dituen balbula ireki da. Demagun gasa monoatomikoa eta ideala dela, eta gordailuen hormak finkoak eta adiabatikoak.
Lortu sistemari dagozkion amaierako tenperatura eta presioa.
C 3.4.4
23. ARIKETA
Bi sistemari dagozkion egoera-ekuazioak ondokoak dira:
\frac{1}{T^{(1)}} = \frac{3}{2}R\frac{N^{(1)}}{U^{(1)}}
\frac{1}{T^{(2)}} = \frac{5}{2}R\frac{N^{(2)}}{U^{(2)}}
Adierazpen horietan R=1.986 cal/mol K da balioko konstantea. Lehenengo sistemaren mol-kopurua N^{(1)} = 2 da, eta bigarrenarena N^{(2)} = 3. Sistema horiek diatermoa den bereizte-hormak banatu ditu. Sistema konposatuaren energia 6000 cal da.
Oreka lortutakoan, zenbatekoa da sistema bakoitzari dagokion barne-energia?
C 2.6.3
24. ARIKETA
Aurreko ariketako azpisistemak, diatermoa den bereizte-hormaren bidez banatuta daude. Mol kopuruen balioak ondokoak dira: N^{(1)} = 2 eta N^{(2)} = 3. Hasierako tenperaturak T^{(1)} = 250 K eta T^{(2)} = 350 K dira.
Oreka lortutakoan,
- Zein dira U^{(1)} eta U^{(2)} parametroen balioak?
- Zein da oreka-tenperatura?
C 2.6.4
25. ARIKETA
Bi sistemari dagozkion egoera-ekuazioak ondokoak dira:
- \left(\frac{1}{T}\right)^{(1)} = \frac{3}{2}R\frac{N^{(1)}}{U^{(1)}}
- \left(\frac{p}{T}\right)^{(1)} = R\frac{N^{(1)}}{V^{(1)}}
eta
- \left(\frac{1}{T}\right)^{(2)} = \frac{5}{2}R\frac{N^{(2)}}{U^{(2)}}
- \left(\frac{p}{T}\right)^{(2)} = R\frac{N^{(2)}}{V^{(2)}}
Adierazpen horietan R=1.986 cal/mol K da. N^{(1)} = 0.5 eta N^{(2)} = 0.75 dira. Sistema biak isolaturiko zilindroaren barnean daude, higikorra den pistoi isotermoak banandurik. Hasierako tenperaturak T^{(1)} = 200 K eta T^{(2)} = 300 K dira, eta bolumen osoa 20 l.
Oreka lortutakoan:
- zein dira sistema bakoitzari dagozkion energia eta bolumena?
- Zein dira presioa eta tenperatura?
C 2.7.1
Kalkulu termodinamikoa eta zenbait aplikazio
26. ARIKETA
Lortu \left( \frac{\partial T}{\partial v} \right)_{h} koefizientearen adierazpena, (C_{p}, \alpha, \kappa_{T}) sortaren funtzioan.
C 7.4.1
Modu berean ere, lortu, kalkulu termodinamiko hutsa erabiliz, honako koefiziente hauek:
\left(\frac{\partial H}{\partial V}\right)_{T, N}
\left(\frac{\partial v}{\partial s}\right)_{p}
\left(\frac{\partial s}{\partial f}\right)_{v}
\left(\frac{\partial s}{\partial f}\right)_{p}
\left(\frac{\partial g}{\partial f}\right)_{p})
\left(\frac{\partial s}{\partial v}\right)_{h}
C 7.4.6: \left(\frac{\partial h}{\partial v}\right)_{u}
C 7.4.17: \left(\frac{\partial \mu}{\partial v}\right)_{s}
\left( \frac{\partial T}{\partial v} \right)_{h} = -\frac{c_{p}-v\alpha p}{\left(c_{v}\kappa_{T} + v\alpha\right)}
\left(\frac{\partial H}{\partial V}\right)_{T, N}=\frac{1}{\kappa_{T}}(T \alpha-1)
\left(\frac{\partial v}{\partial s}\right)_{p}=T v \frac{\alpha}{c_{p}}
\left(\frac{\partial s}{\partial f}\right)_{v}=-\frac{c_{v}}{T \cdot s}
\left(\frac{\partial s}{\partial f}\right)_{p} = -\frac{T}{c_{p}\left(s + pv\alpha\right)}
\left(\frac{\partial g}{\partial f}\right)_{p}=-\frac{1}{c_{p}}(s+p V \alpha)
\left(\frac{\partial s}{\partial v}\right)_{h}=\frac{c_{p}}{T} \frac{1}{\left(c_{v} \kappa_{T}+v \alpha\right)}
C 7.4.6:
\left(\frac{\partial h}{\partial v}\right)_{u}=\left[p-\frac{c_{p}-v \alpha p}{\kappa_{T} c_{p}-\alpha^{2} Tv}\right]C 7.4.17:
\left(\frac{\partial \mu}{\partial v}\right)_{s}=\frac{1}{\kappa_{T} \mathrm{c}_{v}}\left(T \alpha s-c_{p}\right)
\left.\begin{array}{l} \left(\frac{\partial T}{\partial v}\right)_{h}=-\frac{\left(\frac{\partial h}{\partial v}\right)_{T}}{\left(\frac{\partial h}{\partial T}\right)_{v}} \\ d h=T d s+v d p \\ \left(\frac{\partial h}{\partial v}\right)_{T}=\left(\frac{\partial(T d s+v d p)}{\partial v}\right)_{T}=T\left(\frac{\partial s}{\partial v}\right)_{T}+v\left(\frac{\partial p}{\partial v}\right)_{T}=T\left(\frac{\partial p}{\partial T}\right)_{v}+v \frac{1}{\left(\frac{\partial v}{\partial p}\right)_{T}}=T \frac{\alpha}{\kappa_{T}}-\frac{1}{\kappa_{T}}=\frac{1}{\kappa_{T}}(T \alpha-1) \\ \left(\frac{\partial h}{\partial T}\right)_{v}=\left(\frac{\partial(T d s+v d p)}{\partial T}\right)_{v}=T\left(\frac{\partial s}{\partial T}\right)_{v}+v\left(\frac{\partial p}{\partial T}\right)_{v}=c_{V}+v \frac{\alpha}{\kappa_{T}} \\ c_{p}-c_{V}=\frac{T v \alpha^{2}}{\kappa_{T}} \\ \left(\frac{\partial T}{\partial v}\right)_{h}=-\frac{\frac{1}{\kappa_{T}}(T \alpha-1)}{T c_{V}+v \frac{d}{\kappa_{T}}} \end{array}\right\} \Rightarrow\left(\frac{\partial T}{\partial v}\right)_{h}=-\frac{(T \alpha-1)}{\kappa_{T}{ }^{c_{V}}+v \alpha}
Abiapuntuan, h entalpia funtzioaren diferentzial osoa dago:
d h=T d s+v d p
Dagoeneko, adierazpen horretan badago diferentziala eginda, aldaketa osoa.
Kalkulatu beharreko \left(\frac{\partial h}{\partial v}\right)_{T} koefizientea, beste modu honetan idatz daiteke, ondo interpretatatzeko:
\left(\dfrac{\partial \left[d h=T d s+v d p\right] }{\partial v}\right)_{T}\Rightarrow \left(\frac{\partial h}{\partial v}\right)_{T} = \left(\dfrac{\partial [T d s+v d p]}{\partial v}\right)_{T}
Azken adierazpen horretan, idatzi s eta p, v eta T aldagaien funtzioan. Kontua da idatzitako adierazpenean, erabilitako aldagaiak v eta T direla.
Orduan:
s=s(v, T)\Rightarrow ds=\left(\dfrac{\partial s}{\partial v}\right)_{T} dv + \left(\dfrac{\partial s}{\partial T}\right)_{v} dT
p=s(v, T)\Rightarrow dp=\left(\dfrac{\partial p}{\partial v}\right)_{T} dv + \left(\dfrac{\partial p}{\partial T}\right)_{v} dT
Beraz, berridatz honako adierazpen hau:
\left[d h=T d s+v d p\right]
d h = T \left[\left(\dfrac{\partial s}{\partial v}\right)_{T} dv + \left(\dfrac{\partial s}{\partial T}\right)_{v} dT\right] + \left[\left(\dfrac{\partial p}{\partial v}\right)_{T} dv + \left(\dfrac{\partial p}{\partial T}\right)_{v} dT\right]
Orain, honako honek: \left(\dfrac{\partial(T d s+v d p)}{\partial v}\right)_{\color{red}{T}}, v-rekiko deribatu partziala baina ez duela aukeratzen.
d h = T \left[\left(\dfrac{\partial s}{\partial v}\right)_{T} dv + \left(\dfrac{\partial s}{\partial T}\right)_{v} dT\right]_{\color{red}{T}} + \left[\left(\dfrac{\partial p}{\partial v}\right)_{T} dv + \left(\dfrac{\partial p}{\partial T}\right)_{v} dT\right]_{\color{red}{T}}
Beraz, goiko adierazpenean, dT|_{\color{red}{T}} atalek ez dute ekarpenik egingo.
Hortaz, honako hau izango da azken adierazpena:
\left(\dfrac{\partial \left[d h=T d s+v d p\right] }{\partial v}\right)_{T}\Rightarrow {\color{blue}{\left(\frac{\partial h}{\partial v}\right)_{T}}} = T \left[\left(\dfrac{\partial s}{\partial v}\right)_{T} dv + 0\right]_{\color{red}{T}} + \left[\left(\dfrac{\partial p}{\partial v}\right)_{T} dv + 0\right]_{\color{red}{T}}
27. ARIKETA
Esku artean dugun nitrogeno-gasaren (N_2) 0,80 molek 300 K-eko tenperaturan eta 2 MPa-eko presioan litro bateko gordailu zurruna bete dute. Gordailua 10 cm^3-ko gordailu huts batekin konektatu dugu, eta era askean zabal dadila utzi diogu gasari, bolumen osoa (1010 cm^3) bete arte. Lortu entalpia-aldaketa.
Hasierako egoerari dagozkion datuak honako hauek dira:
1. c_{p}=29,8 J/(mol K), 2. \kappa_{T}=5,02\times10^{-7} Pa^{-1}
3. \alpha=3,49\times10^{-3} K^{-1}
Ikus aurreko ariketako 7. ariketa: \left(\frac{\partial h}{\partial v}\right)_{u}=p-\frac{c_{p}-pv\alpha}{c_{V}\kappa_{T}}. Datuekin (v = 1,25\times10^{-3} m^3/mol, c_V = c_p - Tv\alpha^2/\kappa_T \approx 20,7 J/(mol K)), \left(\frac{\partial h}{\partial v}\right)_{u} \approx -2,8\times10^{4} Pa eta \Delta H = N\left(\frac{\partial h}{\partial v}\right)_{u}\Delta v \approx -0,28 J (gas idealarentzat zero litzateke).
\left.\begin{array}{l} d h=T d s+v d p \\ \left(\frac{\partial h}{\partial v}\right)_{u}=\left(\frac{\partial(T d s+v d p)}{\partial v}\right)_{u} \Rightarrow\left(\frac{\partial h}{\partial v}\right)_{u}=T\left(\frac{\partial s}{\partial v}\right)_{u}+v\left(\frac{\partial p}{\partial v}\right)_{u}=-T \frac{\left(\frac{\partial u}{\partial v}\right)_{s}}{\left(\frac{\partial u}{\partial s}\right)_{v}}-v \frac{\left(\frac{\partial u}{\partial v}\right)_{p}}{\left(\frac{\partial u}{\partial p}\right)_{v}} \\ d u=s d T-p d v \\ \left(\frac{\partial u}{\partial v}\right)_{s}=\left(\frac{\partial(T d s-p d v)}{\partial v}\right)_{s}=T\left(\frac{\partial s}{\partial v}\right)_{s}-p\left(\frac{\partial v}{\partial v}\right)_{s}=-p \\ \left(\frac{\partial u}{\partial s}\right)_{v}=\left(\frac{\partial(T d s-p d v)}{\partial s}\right)_{v}=T\left(\frac{\partial s}{\partial s}\right)_{v}-p\left(\frac{\partial v}{\partial s}\right)_{v}=T \\ \left(\frac{\partial u}{\partial v}\right)_{p}=\left(\frac{\partial(T d s-p d v)}{\partial v}\right)_{p}=T\left(\frac{\partial s}{\partial v}\right)_{p}-p\left(\frac{\partial v}{\partial v}\right)_{p}=T\left(\frac{\partial s}{\partial T}\right)_{p}\left(\frac{\partial T}{\partial v}\right)_{p}-p=c_{p} \frac{1}{v \alpha}-p \\ \left(\frac{\partial u}{\partial p}\right)_{v}=\left(\frac{\partial(T d s-p d v)}{\partial p}\right)_{v}=T\left(\frac{\partial s}{\partial p}\right)_{v}-p\left(\frac{\partial v}{\partial p}\right)_{v}=T\left(\frac{\partial u}{\partial v}\right)_{s}=p \\ \left.-v \frac{\partial u}{\left(\frac{\partial u}{\partial v}\right)_{p}}=-v \frac{c_{p} \frac{1}{v \alpha}-p}{c_{V} \frac{\partial T}{\alpha}}\right)_{v}=c_{V} \frac{\kappa_{T}}{\alpha} \\ \left(\frac{\partial h}{\partial v}\right)_{u}=p-v \frac{c_{p} \frac{1}{c_{V}-p} \frac{\partial_{T}}{\alpha}}{\partial p} \Rightarrow\left(\frac{\partial h}{\partial v}\right)_{u}=p-\frac{v \alpha\left(\frac{c_{p}}{v \alpha}-p\right)}{c_{V} \kappa_{T}} \end{array}\right\} \Rightarrow
28. ARIKETA
Frogatu ondoko berdintza beteko dela eta lortu horri dagokion adierazpena van der Waals-en gasaren kasurako:
\left( \frac{\partial c_{v}}{\partial v} \right)_{T} = T\left( \frac{\partial^{2} p}{\partial T^{2}} \right)_{v}
C 7.4.7
Idatzi c_{V}-ren, bolumen konstanteko bero-ahalamenaren, definizioa, entropiaren funtzioan.
Kontuan izan diferentzial zehatzek; hots, benetan existitzen diren funtzioek, bete beharreko baldintzetako bat, deribatu partzialekin lotutakoa bera.
\begin{gathered} \left(\frac{\partial c_{V}}{\partial v}\right)_{T}=\left(\frac{\partial\left\{T\left(\frac{\partial s}{\partial T}\right)_{v}\right\}}{\partial v}\right)_{T} \Rightarrow\left(\frac{\partial c_{V}}{\partial v}\right)_{T}=\left(\frac{\partial T}{\partial v}\right)_{T}\left(\frac{\partial s}{\partial T}\right)_{v}+T\left(\frac{\partial^{2} s}{\partial v \partial T}\right) \Rightarrow\left(\frac{\partial c_{V}}{\partial v}\right)_{T}=T\left(\frac{\partial^{2} s}{\partial T \partial v}\right) \Rightarrow \\ \left(\frac{\partial c_{V}}{\partial v}\right)_{T}=T\left(\frac{\partial\left(\frac{\partial s}{\partial v}\right)_{T}}{\partial T}\right)_{v} \Rightarrow\left(\frac{\partial c_{V}}{\partial v}\right)_{T}=T\left(\frac{\partial\left(\frac{\partial p}{\partial T}\right)_{v}}{\partial T}\right)_{v} \Rightarrow\left(\frac{\partial c_{V}}{\partial v}\right)_{T}=T\left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{v} \\ \left(p+\frac{a}{v^{2}}\right)(v-b)=R T \Rightarrow p=\frac{R T}{(v-b)}-\frac{a}{v^{2}} \\ \left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{v}=\left(\frac{\partial^{2}\left(\frac{R T}{(v-b)}-\frac{a}{v^{2}}\right)}{\partial T^{2}}\right)_{v} \Rightarrow\left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{v}=\left(\frac{\partial\left(\frac{R T}{(v-b)}\right)}{\partial T}\right)_{v} \Rightarrow\left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{v}=0 \end{gathered}
29. ARIKETA
Frogatu ondoko berdintza beteko dela:
\left( \frac{\partial c_{p} }{\partial p} \right)_{T} = - Tv\left[ {\alpha^{2} + \left( \frac{\partial \alpha}{\partial T} \right)_{p}} \right]
Lortu berdintza horri dagokion adierazpena ondoko egoera-ekuazioa beteko duen sistemaren kasurako:
p\left( v + \frac{A}{T^{2}} \right) = RT
C 7.4.8
Idatzi c_{p}-ren, bolumen konstanteko bero-ahalamenaren, definizioa, entropiaren funtzioan.
Kontuan izan diferentzial zehatzek; hots, benetan existitzen diren funtzioek, bete beharreko baldintzetako bat, deribatu partzialekin lotutakoa bera.
\begin{aligned} &\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=\left(\frac{\partial\left\{T\left(\frac{\partial s}{\partial T}\right)_{p}\right\}}{\partial p}\right)_{T} \Rightarrow\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=\left(\frac{\partial T}{\partial p}\right)_{T}\left(\frac{\partial s}{\partial T}\right)_{p}+T\left(\frac{\partial^{2} s}{\partial p \partial T}\right) \Rightarrow\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=T\left(\frac{\partial^{2} s}{\partial T \partial p}\right) \Rightarrow \\ &\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=T\left(\frac{\partial\left(\frac{\partial s}{\partial p}\right)_{T}}{\partial T}\right)_{p} \Rightarrow\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=T\left(\frac{\partial\left(-\frac{\partial v}{\partial T}\right)_{p}}{\partial T}\right)_{p} \Rightarrow\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=T\left(\frac{\partial(-v \alpha)_{p}}{\partial T}\right)_{p} \Rightarrow \\ &\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=-T\left(\left(\frac{\partial v}{\partial T}\right)_{p} \alpha+v\left(\frac{\partial \alpha}{\partial T}\right)_{p}\right)_{p} \Rightarrow\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=-T\left(v \alpha^{2}+v\left(\frac{\partial \alpha}{\partial T}\right)_{p}\right)_{p} \Rightarrow \\ &\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=-T v\left(\alpha^{2}+\left(\frac{\partial \alpha}{\partial T}\right)_{p}\right)_{p} \end{aligned}
\begin{aligned} &p\left(v+\frac{A}{T^{2}}\right)=R T \Rightarrow v=\frac{R T}{p}-\frac{A}{T^{2}} \\ &\alpha=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_{p} \Rightarrow \alpha=\frac{1}{\left(\frac{R T}{p}-\frac{A}{T^{2}}\right)}\left(\frac{\partial\left(\frac{R T}{p}-\frac{A}{T^{2}}\right)}{\partial T}\right)_{p} \Rightarrow \alpha=\frac{1}{\left(\frac{R T}{p}-\frac{A}{T^{2}}\right)}\left(\frac{R}{p}+\frac{2 A}{T^{3}}\right) \\ &\alpha^{2}=\left(\frac{1}{\left(\frac{R T}{p}-\frac{A}{T^{2}}\right)}\left(\frac{R}{p}+\frac{2 A}{T^{3}}\right)\right)^{2} \\ &\left(\frac{\partial \alpha}{\partial T}\right)_{p}=\left(\frac{\partial\left(\frac{1}{\left(\frac{R T}{p}-\frac{A}{T^{2}}\right)}\left(\frac{R}{p}+\frac{2 A}{T^{3}}\right)\right)}{\partial T}\right) \end{aligned}
30. ARIKETA
Aurreko ariketako mol bati p_{0} hasierako presiotik p_{f} bukaerako presiorainoko espantsio isotermoa eragin diogu.
Lortu prozesuan sistemak trukatu duen bero kantitatea.
C 7.4.9
Q= -\frac{2a}{T^{2}}\Delta p+RT\ln\left(\frac{T^{2}v_{0}+a}{T^{2}v_{f}+a}\right)
\delta Q = T dS, onartuz gertatu den prozesua itzulgarria dela (Prozesuaren adierazpena emanda dago, beraz, onartu daiteke) Kalkulatu behar da S=S(T,p), aldagai independentetzat T eta p onartuz, lehenak konstante dirau eta bigarrenaren aldaketa ezaguna da.
\left[\delta Q\right]_{T}=T\left(-\frac{\partial V}{\partial T}\right)_{p} \left[dp\right]_{T}\Rightarrow Q = -\frac{2a}{T^{2}}\Delta p+RT\ln\left(\frac{p_{0}}{p_{f}}\right)
Lehenengo Printzipioaren adierazpenetik abiatu daiteke ere bai:
\delta Q = {\underbrace{dU}_{=-\frac{2a}{T^{3}}dp}} + {\underbrace{pdV}_{=\frac{RT^{3}}{T^{2}V +a}dV}}
Sistemari, presioa aldatuz, espantsioa eragin diogu; beraz, aldagai independentea p presioa da. Beroaz galdatu dutenez, tenperatura konstanteko preozesuan, ondokoa adierazpena erabil dezakegu: \delta Q=T d S. Orduan, ondokoa da kalkulatu beharreko koefizientea: \left(\frac{\partial S}{\partial p}\right)_{T}
\left(\frac{\partial S}{\partial p}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{p} \Rightarrow\left(\frac{\partial S}{\partial p}\right)_{T}=-V \alpha d S=\left(\frac{\partial S}{\partial p}\right)_{T} d p \Rightarrow d S=(-V \alpha) d p \Rightarrow \int d S=\int(-V \alpha) d p \left(\frac{\partial S}{\partial p}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{p} \Rightarrow\left(\frac{\partial S}{\partial p}\right)_{T}=-V \alpha d S=\left(\frac{\partial S}{\partial p}\right)_{T} d p \Rightarrow d S=(-V \alpha) d p \Rightarrow \int d S=\int(-V \alpha) d p \left.\begin{array}{l}\alpha=\frac{1}{\left(\frac{R T}{p}-\frac{A}{T^{2}}\right)}\left(\frac{R}{p}+\frac{2 A}{T^{3}}\right) \\ v=\frac{R T}{p}-\frac{A}{T^{2}}\end{array}\right\} \Rightarrow-V \alpha=-\left(\frac{R}{p}+\frac{2 A}{T^{3}}\right) \left.\begin{array}{l}\Delta S=\int-\left(\frac{R}{p}+\frac{2 A}{T^{3}}\right) d p \Rightarrow \Delta S=-\left\{R \ln \frac{p_{f}}{p_{i}}+\frac{2 A}{T^{3}} \Delta p\right\} \\ Q=T \Delta S\end{array}\right\} \Rightarrow Q=-T\left\{R \ln \frac{p_{f}}{p_{i}}+\frac{2 A}{T^{3}} \Delta p\right\}
31. ARIKETA
Esku artean dugun sistemaren kasuan ezagunak dira ondoko bi propietate hauek:
C_{V}=AT^3
(v-v_{0})p=B(T)
Adierazpen horretan, A konstantea da eta B(T), tenperaturaren funtzio ezezaguna.
Eztabaidatu zer forma har dezakeen B(T) funtzioak.
Lortu (C_{p}, \alpha, \kappa_{T}) sortakoak, T-ren eta v-ren funtzioan.
C 7.4.22
B''(T)=0 \Rightarrow B(T)=DT+E \Rightarrow p = \frac{\left(DT + E\right)}{v-v_{0}}
c_{p} = A T^{3} + \frac{T^{3}}{\left(DT + E\right)}
Adierazpen horietan D eta E konstanteak dira.
Aurreko ariketaren emaitza erabiliko dugu B(T) funtzioaren forma finkatzeko. Beraz, jakina da ondoko adierazpena beteko dela: \left(\frac{\partial c_{V}}{\partial v}\right)_{T}=T\left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{v} . Orduan, berdintzako atal biak, bakoitza bere aldetik, kalkulatu egingo dugu; eta, ondoren, alderatu.
\begin{aligned} &\left(\frac{\partial c_{V}}{\partial v}\right)_{T}=T\left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{v} \\ &\left(\frac{\partial c_{V}}{\partial v}\right)_{T}=\left(\frac{\partial\left(A T^{3}\right)}{\partial v}\right)_{T} \Rightarrow\left(\frac{\partial c_{V}}{\partial v}\right)_{T}=0 \\ &\left(v-v_{o}\right) p=B(T) \Rightarrow p=\frac{B(T)}{\left(v-v_{o}\right)} \\ &\left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{v}=\left(\frac{\partial^{2}\left(\frac{B(T)}{\left(v-v_{o}\right)}\right)}{\partial T^{2}}\right)_{v}=\frac{1}{\left(v-v_{o}\right)}\left(\frac{\partial^{2}(B(T))}{\partial T^{2}}\right)_{v}=\frac{1}{\left(v-v_{o}\right)} B^{\prime \prime}(T) \\ &T\left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{v}=T \frac{1}{\left(v-v_{o}\right)} B^{\prime \prime}(T) \Rightarrow \\ &0=T \frac{1}{\left(v-v_{o}\right)} B^{\prime \prime}(T) \Rightarrow 0=T B^{\prime \prime}(T) \Rightarrow B^{\prime \prime}(T)=0 \Rightarrow B(T)=B_{1} T+B_{2} \end{aligned}
- \left(C_{p}, \alpha, \kappa_{T}\right) sortakoak.
- \alpha
\left.\begin{array}{l} p=\frac{B_{1} T+B_{2}}{\left(v-v_{o}\right)} \Rightarrow\left(v-v_{o}\right)=\frac{B_{1} T+B_{2}}{p} \\ \alpha=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_{p} \Rightarrow \alpha=\frac{1}{\left(v-v_{o}\right)}\left(\frac{\partial\left(v-v_{o}\right)}{\partial T}\right)_{p} \Rightarrow \\ \left(\frac{\partial\left(v-v_{o}\right)}{\partial T}\right)_{p}=\left(\frac{\partial\left(\frac{B_{1} T+B_{2}}{p}\right)}{\partial T}\right)_{p}=\frac{1}{p} B_{1} \end{array}\right\} \Rightarrow \alpha=\frac{1}{\left(v-v_{o}\right)} \frac{1}{p} B_{1}
- \kappa_{T}
\left.\begin{array}{l} \kappa_{T}=-\frac{1}{v}\left(\frac{\partial v}{\partial p}\right)_{T} \Rightarrow \kappa_{T}=-\frac{1}{\left(v-v_{o}\right)}\left(\frac{\partial\left(v-v_{o}\right)}{\partial p}\right)_{T} \Rightarrow \\ \left(\frac{\partial\left(v-v_{o}\right)}{\partial p}\right)_{T}=\left(\frac{\partial\left(\frac{B_{1} T+B_{2}}{p}\right)}{\partial p}\right)_{T}=-\frac{B_{1} T+B_{2}}{p^{2}}=-\frac{B_{1} T+B_{2}}{\left(\frac{B_{1} T+B_{2}}{\left(v-v_{o}\right)}\right)^{2}}=-\frac{\left(v-v_{o}\right)^{2}}{\left(B_{1} T+B_{2}\right)}=-\frac{\left(v-v_{o}\right)}{p} \\ \kappa_{T}=-\frac{1}{\left(v-v_{o}\right)}\left\{-\frac{\left(v-v_{o}\right)}{p}\right\} \Rightarrow \kappa_{T}=\frac{1}{p} \end{array}\right\} \Rightarrow
- C_{p}
\left.\begin{array}{c} c_{p}-c_{V}=\frac{T v \alpha^{2}}{\kappa_{T}} \Rightarrow c_{p}=c_{V}+\frac{T v \alpha^{2}}{\kappa_{T}} \Rightarrow c_{p}=c_{V}+\frac{T\left(v-v_{o}\right) \alpha^{2}}{\kappa_{T}} \\ \frac{T\left(v-v_{o}\right) \alpha^{2}}{\kappa_{T}}=T\left(v-v_{o}\right) \frac{\left(\frac{1}{\left(v-v_{o}\right)} \frac{1}{p} B_{1}\right)^{2}}{\left(\frac{1}{p}\right)}=T\left(\frac{1}{\left(v-v_{o}\right)} \frac{1}{p} B_{1}^{2}\right)=T B_{1}^{2}\left(B_{1} T+B_{2}\right) \\ c_{p}=c_{V}+T B_{1}^{2}\left(B_{1} T+B_{2}\right) \\ \left(\frac{\partial c_{p}}{\partial p}\right)_{T}=-T v\left\{\alpha^{2}+\left(\frac{\partial \alpha}{\partial T}\right)_{p}\right\} \Rightarrow\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=-T\left(v-v_{o}\right)\left\{\alpha^{2}+\left(\frac{\partial \alpha}{\partial T}\right)_{p}\right\} \\ \left.\alpha=\frac{1}{\left(v-v_{o}\right)} \frac{1}{p} B_{1} \Rightarrow \alpha^{2}=\left(\frac{1}{\left(v-v_{o}\right)} \frac{1}{p} B_{1}\right)^{2}\right) \\ \left(\frac{\partial \alpha}{\partial T}\right)_{p}=\left(\frac{\partial\left(\frac{1}{\left(v-v_{o}\right)} \frac{1}{p} B_{1}\right)}{\partial T}\right)_{p}=0 \\ \left(\frac{\partial c_{p}}{\partial p}\right)_{T}=-T \frac{1}{\left(v-v_{o}\right)}\left(\frac{1}{\left(B_{1} T+B_{2}\right)} B_{1}\right)^{2} \end{array}\right\} \Rightarrow\left(\frac{\partial c_{p}}{\partial p}\right)_{T}=-T\left(v-v_{o}\right)\left(\frac{1}{\left(v-v_{o}\right)} \frac{1}{p} B_{1}\right)^{2} \Rightarrow
32. ARIKETA
Van der Waalsen egoera-ekuazioa esleitu zaion 1 mol-eko sistemari v_{0} hasierako bolumenetik v_{f} bukaerako bolumenerako espantsio isotermoa eragin diogu.
Lortu prozesuan trukaturiko bero kantitatea.
C 7.4.10
\left[\delta Q\right]_{T}=T\left(\frac{\partial p}{\partial T}\right)_{V} \left[dV\right]_{T}
Q = 3RT \ln\left(\frac{v_{f} - b}{v_{i} - b}\right) + \frac{2a}{v}
\begin{aligned} &\left(\frac{\partial S}{\partial V}\right)_{T}=-\left(\frac{\partial p}{\partial T}\right)_{V}=-\frac{\left(\frac{\partial V}{\partial T}\right)_{p}}{\left(\frac{\partial V}{\partial p}\right)_{T}}=\frac{\alpha}{\kappa_{T}} \\ &d S=\left(\frac{\partial S}{\partial V}\right)_{T} d V \Rightarrow d S=\frac{\alpha}{\kappa_{T}} d V \Rightarrow \int d S=\int \frac{\alpha}{\kappa_{T}} d V \end{aligned}
van der Waals-en fluidoaren kasuan, hobea da \left(\frac{\partial p}{\partial T}\right)_{V} koefizientea kalkulatzea: egoera-ekuazioa hirugarren graduko polinomioa, bolumenean, baita. Orduan,
\begin{aligned} &p=\frac{R T}{(v-b)}-\frac{a}{v^{2}} \\ &-\left(\frac{\partial p}{\partial T}\right)_{V}=-\left(\frac{\partial\left(\frac{R T}{(v-b)}-\frac{a}{v^{2}}\right)}{\partial T}\right)_{V} \Rightarrow-\left(\frac{\partial p}{\partial T}\right)_{V}=-\frac{R}{(v-b)} \\ &d S=-\left(\frac{\partial p}{\partial T}\right)_{V} d V \Rightarrow d S=-\frac{R}{(v-b)} d V \Rightarrow \int d S=\int\left(-\frac{R}{(v-b)}\right) d V \\ &\left.\Delta S=-R \ln \frac{\left(v_{f}-b\right)}{\left(v_{i}-b\right)}\right\} \Rightarrow \begin{array}{l} Q=-R T \ln \frac{\left(v_{f}-b\right)}{\left(v_{i}-b\right)} \\ Q=T \Delta S \end{array} \end{aligned}
33. ARIKETA
Esku artean dugun oxigenoaren bi molak 0 ^{\circ}\mathrm{~C}-ko tenperaturan eta 10^{5} Pa-eko presioan daude. Konpresio adiabatikoaren bidez bukaerako tenperatura 300 ^{\circ}C-ko balioraino eraman da.
Lortu bukaerako presioaren balioa ondoko ekuazioa integratuz:
dT = \frac{Tv\alpha}{c_p}dp
Onartu ondoko bero-ahalmeneko gas ideala dela oxigenoa:
c_{p} = 26.20 + 11.49 \times 10^{-3} T - 3.223 \times 10^{-6} T^2
C 7.4.11
Lehenengo eta behin, kalkulatu dezagun ondoko koefizientea: \left(\frac{\partial T}{\partial p}\right), aurreko adierazpenaren berdina dela konprobatzeko.
\begin{aligned} &\left(\frac{\partial T}{\partial p}\right)_{S}=-\frac{\left(\frac{\partial S}{\partial p}\right)_{T}}{\left(\frac{\partial S}{\partial T}\right)_{p}}=-\frac{-\left(\frac{\partial V}{\partial T}\right)_{p}}{\frac{C_{p}}{T}}=\frac{T}{C_{p}} V \alpha \Rightarrow\left(\frac{\partial T}{\partial p}\right)_{S}=\frac{T v \alpha}{c_{p}} \\ &d T=\left(\frac{\partial T}{\partial p}\right)_{S} d p \Rightarrow d T=\frac{T v \alpha}{c_{p}} d p \\ &c_{p}=a+b T+c T^{2} \\ &\Delta p=\frac{1}{v \alpha} \int\left(\frac{a}{T}+b+c T\right) d T \Rightarrow \Delta p=\frac{c_{p} d T}{T}=v \alpha d p \Rightarrow \frac{\left(a+b T+c T^{2}\right) d T}{v}=v \alpha d p \\ &p_{f}=p_{i}+\frac{1}{v \alpha}\left\{a \ln \frac{T_{f}}{T_{i}}+b\left(T_{f}-T_{i}\right)+\frac{1}{2} c\left(T_{f}^{2}-T_{i}^{2}\right)\right\} \end{aligned}
34. ARIKETA
Erraztu honako adierazpen hau:
\left( \frac{\partial u}{\partial p} \right)_h
C 7.4.13
Nahiz eta liburuko (7.4.13 ariketa) enuntziatua honako hau den: Kalkulatu zenbatekoa den barne-energia molarraren aldaketa estrangulazio-prozezu batean, (h konstantekoa bera) zeinean presio-aldaketa dp izan den.
\left(\frac{\partial u}{\partial p}\right)_{h}=-v +p\frac{v\alpha}{c_{p}}\left(c_{v}\frac{\kappa_{T}}{\alpha} + v\right)
35. ARIKETA
Zabaltze askearen ondorioz esku artean dugun gasaren tenperatura-aldaketa dT da.
Lortu presio-aldaketa.
C 7.4.14
Sistemari, bolumena aldatuz, zabaltze askea eragin diogu; beraz, aldagai independentea V bolumena da.
Zabaltze askea denez, U konstanteko prozesua da; eta presioaz galdetu dute, baina tenperatura-aldaketa ezagututa; orduan, honako hau da kalkulatu beharreko koefizientea:
\left(\frac{\partial p}{\partial T}\right)_{u}
\begin{aligned} &d u=T d s-p d v \\ &\left(\frac{\partial p}{\partial T}\right)_{u}=-\frac{\left(\frac{\partial u}{\partial T}\right)_{p}}{\left(\frac{\partial u}{\partial p}\right)_{T}}=-\frac{\left(\frac{\partial(T d s-p d v)}{\partial T}\right)_{p}}{\left(\frac{\partial(T d s-p d v)}{\partial p}\right)_{T}}=-\frac{T\left(\frac{\partial s}{\partial T}\right)_{p}-p\left(\frac{\partial v}{\partial T}\right)_{p}}{T\left(\frac{\partial s}{\partial p}\right)_{T}-p\left(\frac{\partial v}{\partial p}\right)_{T}}=-\frac{c_p-p v \alpha}{-T\left(\frac{\partial v}{\partial T}\right)_{p}-p\left(\frac{\partial v}{\partial p}\right)_{T}}=-\frac{c_p-p v \alpha}{-T v \alpha+p v \kappa_{T}} \\ &d p=\left(\frac{\partial p}{\partial T}\right)_{u} d T \Rightarrow d p=\left\{-\frac{c_p-p v \alpha}{-T v \alpha+p v \kappa_{T}}\right\} d T \end{aligned}
36. ARIKETA
Esku artean van der Waals-en fluidoaren 1 mol dugu: T_{i} tenperaturan, eta V_{i} bolumena betez. Balbula bat irekiz gasa dagoen gordailua hasieran hutsik zegoen beste gordailu batekin konektatu dugu, gasa V_{f} bolumen osora hedatuz. Gordailuen hormak adiabatikoak dira. Lortu T_{f} bukaerako tenperatura.
C 7.4.15
\left(\frac{\partial T}{\partial V}\right)_{U}=-\frac{T\left(\frac{\partial S}{\partial V}\right)_{T} - p}{C_{V}}\Rightarrow \left[dT\right]_{U}=- \frac{1}{C_{V}}\frac{a}{V^{2}}\left[dV\right]_{U}
Sistemari, bolumena aldatuz, zabaltza askea eragin diogu; beraz, aldagai independentea v bolumena da.
Zabaltze askea denez, U konstanteko prozesua da; eta tenperaturaz galdetu dute; orduan, ondokoa da kalkulatu beharreko koefizientea:
\left(\frac{\partial T}{\partial v}\right)_u
\begin{aligned} &d u=T d s-p d v \\ &\left(\frac{\partial T}{\partial v}\right)_{u}=-\frac{\left(\frac{\partial u}{\partial v}\right)_{T}}{\left(\frac{\partial u}{\partial T}\right)_{v}}=-\frac{\left(\frac{\partial(T d s-p d v)}{\partial v}\right)_{T}}{\left(\frac{\partial(T d s-p d v)}{\partial T}\right)_{v}}=-\frac{T\left(\frac{\partial s}{\partial v}\right)_{T}-p\left(\frac{\partial v}{\partial v}\right)_{T}}{T\left(\frac{\partial s}{\partial T}\right)_{v}-p\left(\frac{\partial v}{\partial T}\right)_{v}}=-\frac{T\left(\frac{\partial p}{\partial T}\right)_{v}-p}{T\left(\frac{\partial s}{\partial T}\right)_{v}}=-\frac{1}{c_V}\left(T \frac{\alpha}{\kappa_{T}}-p\right) \\ &d T=\left(\frac{\partial T}{\partial v}\right)_{u} d v \Rightarrow d T=\left\{-\frac{1}{c_V}\left(T \frac{\alpha}{\kappa_{T}}-p\right)\right\} d v \end{aligned}
\begin{aligned} &\left(p+\frac{a}{v^2}\right)(v-b)=R T \Rightarrow p=\frac{R T}{(v-b)}-\frac{a}{v^2} \\ & \end{aligned}
\left.\begin{array}{l} \left(p+\frac{a}{v^2}\right)(v-b)=R T \Rightarrow p=\frac{R T}{(v-b)}-\frac{a}{v^2}\\\\ \left.\begin{array}{l} \alpha=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_p=-\frac{1}{v} \frac{\left(\frac{\partial p}{\partial T}\right)_v}{\left(\frac{\partial p}{\partial v}\right)_T} \\ \kappa_T=-\frac{1}{v}\left(\frac{\partial v}{\partial p}\right)_T=-\frac{1}{v} \frac{1}{\left(\frac{\partial p}{\partial v}\right)_T} \end{array} \right\} \Rightarrow \frac{\alpha}{\kappa_T}=\frac{\left(-\frac{1}{v} \frac{\left(\frac{\partial p}{\partial T}\right)_v}{\left(\frac{\partial p}{\partial v}\right)_T}\right)}{\left(-\frac{1}{v} \frac{1}{\left(\frac{\partial p}{\partial v}\right)_T}\right)} \Rightarrow \frac{\alpha}{\kappa_T}=\left(\frac{\partial p}{\partial T}\right)_v\end{array} \right\} \Rightarrow \frac{\alpha}{\kappa_T}=\frac{R}{(v-b)} \Rightarrow
d T=\left\{-\frac{1}{c_{V}}\left(T \frac{R}{(v-b)}-p\right)\right\} d v \Rightarrow d T=\left\{-\frac{1}{c_{V}} \frac{a}{v^{2}}\right\} d v \Rightarrow \int d T=\int\left\{-\frac{1}{c_{V}} \frac{a}{v^{2}}\right\} d v \Rightarrow
\Delta T=\frac{a}{c_V}\left(\frac{1}{v_f}-\frac{1}{v_i}\right) \Rightarrow \boxed{T_f=T_i+\frac{a}{c_V}\left(\frac{1}{v_f}-\frac{1}{v_i}\right)}
37. ARIKETA
Ebatz ezazu aurreko ariketa berbera; baina, kasu honetan, espantsioa adiabatikoa izateaz gain kuasiestatikoa dela onartuz.
C 7.4.16
\begin{aligned} &\left(\frac{\partial T}{\partial v}\right)_s=-\frac{\left(\frac{\partial s}{\partial v}\right)_T}{\left(\frac{\partial s}{\partial T}\right)_v}==-\frac{\left(\frac{\partial p}{\partial T}\right)_v}{\frac{T}{T}\left(\frac{\partial s}{\partial T}\right)_v}=-\frac{1}{\frac{c_V}{T}} \frac{\alpha}{\kappa_T}=-\frac{T}{c_V} \frac{\alpha}{\kappa_T} \\ &d T=\left(\frac{\partial T}{\partial v}\right)_s d v \Rightarrow d T=\left\{-\frac{T}{c_V} \frac{\alpha}{\kappa_T}\right\} d v \\ &\left(p+\frac{a}{v^2}\right)(v-b)=R T \Rightarrow p=\frac{R T}{(v-b)}-\frac{a}{v^2} \end{aligned}
\left.\begin{array}{l} \left(p+\frac{a}{v^2}\right)(v-b)=R T \Rightarrow p=\frac{R T}{(v-b)}-\frac{a}{v^2}\\\\ \left.\begin{array}{l} \alpha=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_p=-\frac{1}{v} \frac{\left(\frac{\partial p}{\partial T}\right)_v}{\left(\frac{\partial p}{\partial v}\right)_T} \\ \kappa_T=-\frac{1}{v}\left(\frac{\partial v}{\partial p}\right)_T=-\frac{1}{v} \frac{1}{\left(\frac{\partial p}{\partial v}\right)_T} \end{array} \right\} \Rightarrow \frac{\alpha}{\kappa_T}=\frac{\left(-\frac{1}{v} \frac{\left(\frac{\partial p}{\partial T}\right)_v}{\left(\frac{\partial p}{\partial v}\right)_T}\right)}{\left(-\frac{1}{v} \frac{1}{\left(\frac{\partial p}{\partial v}\right)_T}\right)} \Rightarrow \frac{\alpha}{\kappa_T}=\left(\frac{\partial p}{\partial T}\right)_v\end{array} \right\} \Rightarrow \frac{\alpha}{\kappa_T}=\frac{R}{(v-b)} \Rightarrow
\begin{aligned} &d T=\left(\frac{\partial T}{\partial v}\right)_{s} d v \Rightarrow d T=\left\{-\frac{T}{c_{V}} \frac{\alpha}{\kappa_{T}}\right\} d v \Rightarrow d T=-\frac{T R}{c_{V}(v-b)} d v \Rightarrow \frac{d T}{T}=-\frac{R}{c_{V}(v-b)} d v \Rightarrow \\ &\ln \frac{T_f}{T_i}=-\frac{R}{c_V} \ln \frac{\left(v_{f}-b\right)}{\left(v_{i}-b\right)} \Rightarrow \boxed{T_{f}=T_{i}\left[\frac{\left(v_{f}-b\right)}{\left(v_{i}-b\right)}\right]^{-\frac{R}{c_V}}} \end{aligned}
38. ARIKETA
Bolumenaren \% 1-eko beheratze adiabatikoak aldaketa jakina eragin du \mu potentzial kimikoaren gainean. Zein bolumen-portzentajek izango du eragin berbera beheratze isotermoa bada?
C 7.4.17
Ondoko bi koefiziente hauek kalkulatu eta alderatu behar ditugu: \left(\frac{\partial \mu}{\partial v}\right)_{s} eta \left(\frac{\partial \mu}{\partial v}\right)_{T}.
d \mu=-s d T+v d p
\left(\frac{\partial \mu}{\partial v}\right)_T=\left(\frac{\partial(-s d T+v d p)}{\partial v}\right)_T=-s\left(\frac{\partial T}{\partial v}\right)_T+v\left(\frac{\partial p}{\partial v}\right)_T=v\left(\frac{\partial p}{\partial v}\right)_T=v \frac{1}{\left(\frac{\partial v}{\partial p}\right)_T}=-\frac{1}{\kappa_T}
\left(\frac{\partial \mu}{\partial v}\right)_s=s \frac{\left(\frac{\partial p}{\partial T}\right)_v}{\frac{c_V}{T}}-v \frac{\left(\frac{\partial s}{\partial T}\right)_p\left(\frac{\partial T}{\partial v}\right)_p}{\left(\frac{\partial s}{\partial T}\right)_v\left(\frac{\partial T}{\partial p}\right)_v}=s \frac{T}{c_V} \frac{\alpha}{\kappa_T}-v \frac{c_p \frac{1}{\left(\frac{\partial v}{\partial T}\right)_p}}{c_V\left(-\frac{\left(\frac{\partial v}{\partial p}\right)_T}{\left(\frac{\partial v}{\partial T}\right)_p}\right)}=s \frac{T}{c_V} \frac{\alpha}{\kappa_T}-\frac{c_p}{c_V \kappa_T}=-\frac{1}{c_V \kappa_T}\left(T \alpha s-c_p\right)
\left(\frac{\partial \mu}{\partial v}\right)_T=\left(\frac{\partial(-s d T+v d p)}{\partial v}\right)_T=-s\left(\frac{\partial T}{\partial v}\right)_T+v\left(\frac{\partial p}{\partial v}\right)_T=v\left(\frac{\partial p}{\partial v}\right)_T=v \frac{1}{\left(\frac{\partial v}{\partial p}\right)_T}=-\frac{1}{\kappa_T}
\left.\begin{array}{l} \left(\frac{\partial \mu}{\partial v}\right)_s=-\frac{1}{\kappa_T}\left\{\frac{1}{c_V}\left(T \alpha s-c_p\right)\right\} \Rightarrow\left(\frac{\partial \mu}{\partial v}\right)_s=\left(\frac{\partial \mu}{\partial v}\right)_T\left\{\frac{1}{c_V}\left(T \alpha s-c_p\right)\right\}\\\\ \left.\begin{array}{l} d \mu=\left(\frac{\partial \mu}{\partial v}\right)_s d v \\ d \mu=\left(\frac{\partial \mu}{\partial v}\right)_T d v \end{array}\right\} \Rightarrow(d \mu)_s=(d \mu)_T \Rightarrow\left(\frac{\partial \mu}{\partial v}\right)_s=\left(\frac{\partial \mu}{\partial v}\right)_T \frac{d v}{v_o} 100 \Rightarrow \frac{d v}{v_o} 100=\frac{\left(\frac{\partial \mu}{\partial v}\right)_s}{\left(\frac{\partial \mu}{\partial v}\right)_T} \end{array}\right\}
\boxed{\frac{d v}{v_o} 100=\left\{\frac{1}{c_V}\left(T \alpha s-c_p\right)\right\}}
39. ARIKETA
Helio gasa erabiliz beteriko zilindroa pistoi batek itxi du. Zilindroaren aldeko horma adiabatikoa, higiezina eta helioarekiko iragaztezina da. Zilindroaren beheko aldea, berriz, isotermoa, higiezina eta helioarekiko iragazkorra da. Horma horren bidez, zilindroa T tenperaturako bero-iturriarekin eta \mu_{He} potentzial kimikoko materia-iturriarekin ukipenean dago.
- Lortu sistemaren (-\frac{1}{V}\frac{dV}{dp}) konprimigarritasuna.
- Frogatu konprimigarritasunak dibergentzia izango duela.
- Zein da dibergentzia horren jatorri fisikoa?
C 7.4.18
Baldintza esperimentalen arabera, sistemaren tenperatura eta sisteman dagoen helioaren potentzial kimikoaren balioak finkoak dira: bero-iturri batekin eta materia-iturri batekin kontaktuan baitago sistema. Beraz, ondorioz, enuntziatuan aipaturiko -\frac{1}{V}\frac{dV}{dp} konpresibilitatea tenperatura konstantekoa eta \mu_{\textrm{He}} potentzial kimiko konstantekoa da; hots, ondoko hau: -\frac{1}{V}\frac{dV}{dp} \to - \frac{1}{V}\left(\frac{\partial V}{\partial p}\right)_{T,\mu_{\rm{He}} }.
Kasu honetan, edozer kalkulatzeko, potentzial termodinamikorik erabilgarriena ondoko hau da:
U[T,\mu _ {\textrm{He}}]=U[T,\mu _ {\textrm{He}}](T,V,\mu _{\textrm{He}})
hots, aldagai independenteen sorta naturala \{T,V,\mu_{\textrm{He}}\} duena. Orduan, Legendreren bi transformazio burutu behar dugu U barne-energiaren adierazpenetik U[T,\mu_{\textrm{He}}] potentzial termodinamikoa lortzeko: S-ren ordez T tenperatura ordezkatuko duena, lehenik; eta, ondoren, N_{\textrm{He}}-ren ordez \mu_{\rm{He}} ordezkatuko duena.
Jakina, abiapuntua U izan beharrean F bada, Legendreren transformazio bakarra burutu behar da, N_{\textrm{He}} \to \mu_{\rm{He}}; baina, horrez gain, aldagai-aldaketa bat egin behar da:
F=F(T,V,N) \to F=F(T,V,N(T,V,\mu_{\textrm{He}}))
Abiatu gaitezen F potentzial termodinamikotik; ondokoa da adierazpena:
F = - N_{\rm{He}} RT\ln \left[\frac{T^{\frac{3}{2}}V}{N_{\rm{He}}}\right]
Lortu dezagun:
F[\mu_{\textrm{He}}](=U[T,\mu _{\textrm{He}}])
Horretarako, \mu_{\textrm{He}} lortuko dugu.
\mu_{ \mathrm { He } } \equiv \left( \frac { \partial F } { \partial N _ { \mathrm { He } } } \right) _ { T , V } \Rightarrow \mu _ { \mathrm { He } } =\left( \frac { \partial \left( - N _ { \mathrm { He } } R T \ln \left[ \frac { T ^ { 3 } V } { N _ { \mathrm { He } } } \right] \right) } { \partial N _ { \mathrm { He } } } \right) _ { T , V }\Rightarrow \mu_ { \mathrm { He } } = - R T \ln \left[ \frac { T ^ { 3 } V } { N _ { \mathrm { He } } } \right] + R T
\mu _ { \mathrm { He } } = R T \left[ 1 - \ln \left[ \frac { T ^ { 3 } V } { N _ { \mathrm { He } } } \right] \right]
\frac { \mu _ { \mathrm { He } } } { R T } = 1 - \ln \left[ \frac { T ^ { \frac { 3 } { 2 } } V } { N _ { \mathrm { He } } } \right] \Rightarrow \ln \left[ \frac { T ^ { \frac { 3 } { 2 } } V } { N _ { \mathrm { He } } } \right] = 1 - \frac { \mu _ { \mathrm { He } } } { R T } \Rightarrow \frac { T ^ { \frac { 3 } { 2 } } V } { N _ { \mathrm { He } } } = e ^ { \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } \Rightarrow N _ { \mathrm { He } } = \frac { T ^ { \frac { 3 } { 2 } } V } { e ^ { \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } }
F \left[ \mu _ { \mathrm { He } } \right] = F - N _ { \mathrm { He } } \mu _ { \mathrm { He } } \Rightarrow F \left[ \mu _ { \mathrm { He } } \right] = F \left( T , V , \mu _ { \mathrm { He } } \right) - N _ { \mathrm { He } } \left( T , V , \mu _ { \mathrm { He } } \right) \mu _ { \mathrm { He } }
F \left[ \mu _ { \mathrm { He } } \right] = \left\{- \left( \frac { T ^ { \frac { 3 } { 2 } } V } { e ^ { \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } } \right) R T \ln \left[ \frac { T ^ { \frac { 3 } { 2 } } V } { \left( \frac { T ^ { \frac { 3 } { 2 } } V } { e ^ { \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } } \right) } \right] \right\} - \left\{ \frac { T ^ { \frac { 3 } { 2 } } V } { e ^ { \left( 1 - \frac { \mu _ { \mathrm { HE } } } { R T } \right) } } \mu _ { \mathrm { He } } \right\} \Rightarrow
F \left[ \mu _ { \mathrm { He } } \right] = \left\{- \left( \frac { T ^ { \frac { 3 } { 2 } } V } { e ^ { \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } } \right) R T \ln \left[ e ^ { \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } \right] \right\} - \left\{ \frac { T ^ { \frac { 3 } { 2 } } V } { e ^ { \left( 1 - \frac { \mu _ { \mathrm { HE } } } { R T } \right) } } \mu _ { \mathrm { He } } \right\} \Rightarrow
F \left[ \mu _ { \mathrm { He } } \right] = \left\{- \left( \frac { T ^ { \frac { 3 } { 2 } } V } { e ^ { \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } } \right) R T \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) \right\} - \left\{ T ^ { \frac { 3 } { 2 } } V \mu _ { \mathrm { He } } e ^ { - \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } \right\} \Rightarrow
F \left[ \mu _ { \mathrm { He } } \right] = - R T ^ { \frac { 5 } { 2 } } V e ^ { - \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) }
p \equiv - \left( \frac { \partial F \left[ \mu_{ \mathrm { He } } \right] } { \partial V } \right) _ { T , \mu _ { \mathrm { He } } } \Rightarrow p = - \left( \frac { \partial \left( - R T ^ { \frac { 5 } { 2 } } V e ^ { - \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } \right) } { \partial V } \right) _ { T , \mu _ { \mathrm { He } } } \Rightarrow p = R T ^ { \frac { 5 } { 2 } } e ^ { - \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) }
Badago beste modu bat ariketa ebazteko, hainbesteko kalkulu egin gabe:
\left( \frac { \partial V } { \partial p } \right) _ { T , \mu _ { \mathrm { He } } } = - \frac { \left( \frac { \partial \mu _ { \mathrm { He } } } { \partial p } \right) _ { T , V } } { \left( \frac { \partial \mu _ { \mathrm { He } } } { \partial V } \right) _ { T , p } } = \frac { v } { 0 } = \infty
\left( \frac { \partial \mu _ { \mathrm { He } } } { \partial p } \right) _ { T , V } = v eta \left( \frac { \partial \mu _ { \mathrm { He } } } { \partial V } \right) _ { T , p } = 0 definizioak baino ez dira.
He eta Ne gasen arteko nahastura dugunean, sistema konposatuak konpresibilitatea izango du: Ne-ak ezin baitu zilindrotik atera. Oraingo honetan, kalkulatu beharreko potentzial termodinamikoa ondoko hau da: F[\mu_{\textrm{He}}]=F(T,V,\mu_{\textrm{He}},N_{\textrm{Ne}})(=U[T,\mu_{\textrm{He}},N_{\textrm{Ne}}])
Jakina, sistema koposatuari, nahasturari, dagokion F[\mu] potentzial termodinamiko, azpisistema bakoitzari dagokionaren batura da: F=F_{\textrm{He}}+F_{\textrm{Ne}}. Orduan, F-ren aurreko adierazpenetik abiatuz, ondokoa izango dugu:
F = F _ { \mathrm { He } } + F _ { \mathrm { Ne } }
F = - N _ { \mathrm { He } } R T \ln \left[ \frac { T ^ { 3 } V } { N _ { \mathrm { He } } } \right] - N _ { \mathrm { Ne } } R T \ln \left[ \frac { T ^ { 3 } V } { N _ { \mathrm { Ne } } } \right]
F \left[ \mu _ { \mathrm { He } } \right] = - R T ^ { \frac { 5 } { 2 } } V e ^ { - \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } - N _ { \mathrm { Ne } } R T \ln \left[ \frac { T ^ { \frac { 3 } { 2 } } V } { N _ { \mathrm { Ne } } } \right]
p \equiv - \left( \frac { \partial F \left[ \mu _ { \mathrm { He } } ] \right. } { \partial V } \right) _ { T , \mu _ { \mathrm { He } } , N _ { \mathrm { Ne } } } \Rightarrow p = -\left( \frac { \partial \left( - R T ^ { \frac { 5 } { 2 } } V e ^ { - \left( 1 - \frac { \mu _ { \mathrm { HP } } } { R T } \right) } - N _ { \mathrm { Ne } } R T \ln \left[ \frac { T ^ { \frac { 3 } { 2 } } V } { N _ { \mathrm { Ne } } } \right] \right) } { \partial V } \right) _ { T , \mu _ { \mathrm { He } } } \Rightarrow
p = R T ^ { \frac { 5 } { 2 } } e ^ { - \left( 1 - \frac { \mu _ { \mathrm { He } } } { R T } \right) } + \left[ \frac { N _ { \mathrm { Ne } } R T } { V } \right]
p = p \left( T , V , \mu _ { \mathrm { He } } , N _ { \mathrm { Ne } } \right) \Rightarrow \left( \frac { \partial p } { \partial V } \right) _ { T , \mu _ { \mathrm { He } } , N _ { \mathrm { Ne } } } = - \frac { V ^ { 2 } } { N _ { \mathrm { Ne } } R T } \Rightarrow - \frac { 1 } { V } \left( \frac { \partial V } { \partial p } \right) _ { T , \mu _ { \mathrm { He } } , N _ { \mathrm { Ne } } } = \frac { V ^ { 2 } } { N _ { \mathrm { Ne } } R T } \Rightarrow
- \frac { 1 } { V } \left( \frac { \partial V } { \partial p } \right) _ { T , \mu _ { \mathrm { He } } , N _ { \mathrm { Ne } } } = \frac { 1 } { p _ { \mathrm { Ne } } }
40. ARIKETA
Aurreko ariketako zilindroa Ne gasa erabiliz bete dugu: \frac{1}{10} mol sartuz. Zilindroaren beheko horma ez da neonarekiko iragazkorra.
Lortu zilindroko presioaren balioa eta sistemaren konprimigarritasuna.
C 7.4.19
Ondorengo bi ariketak modu berean abatzi behar dira. Lehenengo eta behin, ebaztean erabiliko diren aldagaiak finkatu behar dira; printzipioz, enuntziatuek berek finkatuko dute: bien kasuan, p / v diagraman gertatuko den prozesua dugu; beraz, aldagai independenteak p eta v dira. Gainera, bietan ere bai, prozesua bera nola gertatuko den ere aipatuta dago: p / v diagramako kurba baten adierazpenaren bidez; hots, aldagaien arteko lotura baten bidez (aldakuntzak ere lotuta daude). Beraz, aldagaietatik bat baino ez da independentea, gure esku dago zein. Horrez gain, ariketa bietan, hasierako zein bukaerako oreka-egoerak ezagunak dira: A eta B etiketen bidez adierazita. Azkenik, A eta B egoerak lotuko dituen kurban zehar ezagunak dira \left(c_{p}, \alpha, \kappa_{T}\right) sortako koefiziente esperimentalak; orduan, kalkulu termodinamikoaren bidez lortuko ditugun adierazpenetan ezagunak diren funtzioak ordezkatu ondoren, integrazioz, kalkulatu beharreko parametroen aldakuntzak lortu ahal izango ditugu.
Prozedura laburbilduz:
1. Identifikatu aldagai independenteak 2. Identifikatu hasierako eta bukaerako oreka-egoerak 3. Identifikatu prozesuaren adierazpena: aldagai independenteen arteko lotura Lortu aldagien aldakuntza diferentzialen arteko lotura 4. Identifikatu kalkulatu beharreko koefizientea Kalkulatu koefizientea: lortu adierazpen orokorra 5. Ordezkatu adierazpen orokorrean enuntziatuko datuak Integratu
41. ARIKETA
Esku artean substantzia baten mol bakarra dugu. p/V diagramako A eta B puntuak pv^2 = {\rm{konstantea}} lerroko puntuak dira. Aipatutako lerroan zehar ezagunak dira sistemaren ondoko ezaugarriak: C_p = Cv^2, \alpha = \frac{D}{v} eta \kappa _T = Ev. C, D eta E konstanteak dira.
Lortu T_{B} tenperatura (T_{A}, p_{A}, v_{A}, v_{B}, C, D, E) sortaren funtzioan.
C 7.4.20
Identifikatu aldagai independenteak:
p / v \Rightarrow p, v
Identifikatu hasierako eta bukaerako oreka-egoerak:
\text { A, } B
Identifikatu prozesuaren adierazpena: aldagai independenteen arteko lotura Lortu aldagien aldakuntza diferentzialen arteko lotura
\begin{aligned}&p v^2=\text { konst. } \Rightarrow v^2 d p+2 p d v=0 \Rightarrow \frac{d p}{p}=-2 \frac{d v}{v^2} \\&d p=-2\left(p_A v_A^2\right) \frac{d v}{v^3}\end{aligned}
Identifikatu kalkulatu beharreko koefizientea
Kalkulatu koefizientea: lortu adierazpen orokorra\begin{aligned}&T=T(p, v) \\&d T=\left(\frac{\partial T}{\partial p}\right)_v d p+\left(\frac{\partial T}{\partial v}\right)_p d v \Rightarrow d T=\frac{\kappa_T}{\alpha} d p+\frac{1}{v \alpha} d v\end{aligned}
Ordezkatu adierazpen orokorrean enuntziatuko datuak Integratu
\left.\begin{array}{l}d T=\frac{\kappa_T}{\alpha}\left(-2 p_A v_A^2 \frac{d v}{v^3}\right)+\frac{1}{v \alpha} d v \\\alpha=\frac{D}{v} \\\kappa_T=E v\end{array}\right\} \Rightarrow d T=\frac{E v}{\left(\frac{D}{v}\right)}\left(-2 p_A v_A^2 \frac{d v}{v^3}\right)+\frac{1}{v\left(\frac{D}{v}\right)} d v\Rightarrow
d T=-\frac{2 E}{D} p_A v_A^2 \frac{d v}{v}+\frac{1}{D} d v \Rightarrow d T=\left\{-\frac{2 E}{D} p_A v_A^2 \frac{1}{v}+\frac{1}{D}\right\} d v \Rightarrow \int d T=\int\left\{-\frac{2 E}{D} p_A v_A^2 \frac{1}{v}+\frac{1}{D}\right\} d v \Rightarrow
\Delta T=-\frac{2 E}{D} p_A v_A^2 \ln \frac{v_B}{v_A}+\frac{1}{D}\left(v_B-v_A\right) \Rightarrow \boxed{T_B=T_A-\frac{2 E}{D} p_A v_A^2 \ln \frac{v_B}{v_A}+\frac{1}{D}\left(v_B-v_A\right)}
42. ARIKETA
Esku artean substantzia baten mol bakarra dugu. p/V diagramako A eta B puntuak pv = {\rm{konstantea}} lerroko puntuak dira. Aipatu lerroan zehar ezagunak dira sistemaren ondoko ezaugarriak: C_p = Cv, \alpha = \frac{D}{v^2} eta \kappa _T = Ev. C, D eta E konstanteak dira.
Lortu (u_{A}-u_{B}) kendura (T_{A}, p_{A}, v_{A}, v_{B}, C, D, E) sortaren funtzioan.
C 7.4.21
Identifikatu aldagai independenteak:
p / v \Rightarrow p, v
Identifikatu hasierako eta bukaerako oreka-egoerak:
A, B
Identifikatu prozesuaren adierazpena: aldagai independenteen arteko lotura Lortu aldagien aldakuntza diferentzialen arteko lotura
\begin{aligned}&p v=\text { konst. } \Rightarrow v d p+p d v=0 \Rightarrow \frac{d p}{p}=-\frac{d v}{v} \\&d p=-2\left(p_A v_A\right) \frac{d v}{v^2}\end{aligned}
Identifikatu kalkulatu beharreko koefizientea
Kalkulatu koefizientea: lortu adierazpen orokorrau=u(p, v) \Rightarrow d u=\left(\frac{\partial u}{\partial p}\right)_v d p+\left(\frac{\partial u}{\partial v}\right)_p d v
d u=T d s-p d v
\left.\begin{array}{l}\left(\frac{\partial u}{\partial p}\right)_v=\left(\frac{\partial(T d s-p d v)}{\partial p}\right)_v=T\left(\frac{\partial s}{\partial p}\right)_v-p\left(\frac{\partial v}{\partial p}\right)_v=T\left(\frac{\partial s}{\partial T}\right)_v\left(\frac{\partial T}{\partial p}\right)_v=c_V \frac{\kappa_T}{\alpha} \\\left(\frac{\partial u}{\partial v}\right)_p=\left(\frac{\partial(T d s-p d v)}{\partial v}\right)_p=T\left(\frac{\partial s}{\partial v}\right)_p-p\left(\frac{\partial v}{\partial v}\right)_p=T\left(\frac{\partial s}{\partial T}\right)_p\left(\frac{\partial T}{\partial v}\right)_p-p=c_p \frac{1}{v \alpha}-p\end{array}\right\} \Rightarrow
d u=\left(c_V \frac{\kappa_T}{\alpha}\right) d p+\left(c_p \frac{1}{v \alpha}-p\right) d v
Ordezkatu adierazpen orokorrrean enuntziatuko datuak
Integratu\left.\begin{aligned}& d u=\left(c_V \frac{\kappa_T}{\alpha}\right)\left(-p_A v_A \frac{d v}{v^2}\right)+\left(c_p \frac{1}{v \alpha}-p\right) d v \Rightarrow d u=\left(c_V \frac{\kappa_T}{\alpha}\right)\left(-p_A v_A \frac{d v}{v^2}\right)+\left(c_p \frac{1}{v \alpha}-p\right) d v \\&\\&\left.\begin{aligned}&\left.\begin{array}{l}\alpha=\frac{D}{v^2} \\\kappa_T=E v\end{array}\right\} \Rightarrow \frac{\kappa_T}{\alpha}=\frac{(E v)}{\left(\frac{D}{v^2}\right)}=\frac{E}{D} v^3 \\&c_p=C v \\&c_p-c_V=\frac{T v \alpha^2}{\kappa_T} \Rightarrow c_V=c_p-\frac{T v \alpha^2}{\kappa_T} \\\end{aligned}\right\} \Rightarrow c_V=C v-\frac{T v\left(\frac{D}{v^2}\right)^2}{(E v)} \Rightarrow c_V=C v-T\left(\frac{D}{E}\right)^2 \frac{1}{v^5}&\\&\end{aligned}\right\} \Rightarrow
d u=\left(c_V \frac{\kappa_T}{\alpha}\right)\left(-p_A v_A \frac{d v}{v^2}\right)+\left(c_p \frac{1}{v \alpha}-p\right) d v
d u=\left\{\left(C v-T v \frac{1}{E v}\left(\frac{D}{v^2}\right)^2\right)\left(\frac{E}{D} v^3\right)\right\}\left(-p_A v_A\right) \frac{d v}{v^2}+\left(C v \frac{1}{v\left(\frac{D}{v^2}\right)}-\frac{p_A v_A}{v}\right) d v
d u=\left\{\left(\frac{C E}{D} v^2-T \frac{1}{v^3} D\right)\right\}\left(-p_A v_A\right) d v+\left(\frac{C}{D} v^2-\frac{p_A v_A}{v}\right) d v
d u=\left(\frac{C}{D}\left(1-p_A v_A E\right) v^2+T D p_A v_A \frac{1}{v^3}-\frac{p_A v_A}{v}\right) d v
d T=\frac{\kappa_T}{\alpha}\left(-p_A v_A \frac{d v}{v^2}\right)+\frac{1}{v \alpha} d v
d T=\frac{E}{D} v^3\left(-p_A v_A \frac{d v}{v^2}\right)+\frac{1}{v\left(\frac{D}{v^2}\right)} d v \Rightarrow d T=-p_A v_A \frac{E}{D} v d v+\frac{1}{D} v d v \Rightarrow d T=\left(1-p_A v_A E\right) \frac{1}{D} v d v \Rightarrow
\int d T=\int\left(1-p_A v_A E\right) \frac{1}{D} v d v \Rightarrow T=\left(1-p_A v_A E\right) \frac{1}{D} v^2 \Rightarrow
d u=\left\{\frac{C}{D}\left(1-p_A v_A E\right) v^2+\left(\left(1-p_A v_A E\right) \frac{1}{D} v^2\right) D_A v_A \frac{1}{v^3}-\frac{p_A v_A}{v}\right\} d v
43. ARIKETA
Esku artean dugun sistemaren kasuan, p/v diagramako lerro zuzenean zeharreko espantsioak (p_{0}, v_{0}) eta (p_{f}, v_{f}) egoerak lotu ditu. (c_{p}, \alpha, \kappa_{T}) sortako parametroak v=v_{0} lerro isokoroan zehar eta p=p_{f} lerro isobaroan zehar soilik ezagunak dira; \frac{c_v \kappa _T}{\alpha } = Ap (v = v_0), \frac{c_p}{v\alpha} = Bv (p = p_f).
Lortu prozesuan trukaturiko beroa.
C 7.4.23
Q = \frac{1}{2}A\left(p_{f}^{2}-p_{0}^{2}\right) + \frac{1}{2}B\left(v_{f}^{2}-v_{0}^{2}\right) + \frac{1}{2}\left(p_{0}- p_{f}\right) \left(v_{f} - v_{0}\right)
44. ARIKETA
Aztertuko dugun sistema alboko irudian ageri da; 1 eta 2 zilindroek osatu dute. Irudian ikus daitekeenez, lehenengoa bigarrenaren barnean higi daiteke. Bien artean ez dago marruskadurarik. Marrazturiko hormak adiabatikoak dira eta besteak, aldiz, diatermoak. Hasiera batean sistema osoa oreka-egoeran dago. 1 zilindroa T tenperaturako eta p presioko iturriarekin ukipenean jarri dugu. Efizientzia maximoko punpa dugu, zeinak eskura dugun iturritik beroa ateraz, 1 zilindrotik ateratako bero osoa 2 zilindroari emango dion. Prozesu horrek iturriarekiko oreka lortu arte iraungo du.
Lortu:
- zilindroen bukaerako tenperaturak eta 2 zilindroaren bukaerako bolumena,
- bi sistemetako entropia-aldaketak,
- punpak egindako lana eta
- inguruneari dagokion entropia-aldaketa.
- Eztabaidatu kualitatiboki zenbatekoa izango den sistema osoaren entropia-aldaketa.
45. ARIKETA
Aztertu beharreko sistema honako hau da: V bolumeneko tangan dauden, T\approx0 K tenperaturan, s=\frac{1}{2} spineko partikulak (esaterako, metaletako tenperatura txikietako elektroiz osatutako gas ideala).
Sistemari dagokion oinarrizko ekuazioa honako hau da:
U = \frac{3}{5}\, A\, N^{\frac{5}{3}}\, V^{-\frac{2}{3}}\,\left[ 1 + \frac{5}{3}\,\left(\frac{S}{\pi\,N\,R}\right)^{2} \right]
A da konstante positiboa.
Lortu honako hauek:
- F potentzial termodinamikoa
- p presioa, (S, V) eta (T, V) aldagai-sorten funtzioan
- C_{V} bolumen konstanteko bero-ahalmena
46. ARIKETA
Aztertu beharreko sistemari dagokion oinarrizko ekuazioa honako hau da:
F = - A\, N^{\frac{2}{3}}\,V^{\frac{1}{3}}\,T^{2}
A konstante positiboa da.
Sistemaren bolumena zortzi aldiz handitu dion zabaltze adiabatiko itzulgarriaren ondorioz, zenbatekoa da tenperatura-aldaketa?
47. ARIKETA
Erraztu honako adierazpen hau:
\left(\frac{\partial T}{\partial p}\right)_{H}
Koefiziente horri Joule-ren eta Kelvin-en koefiziente deritzo, \mu_{J-K}.
48. ARIKETA
Aztertu beharreko sistemari dagozkion bi egoerak, p/v planokoak haiek, honako hauek dira: 1. p_{A} = 10^{5} Pa., v_{A} = 2\times 10^{-2} \frac{\textrm{m}^{3}}{\textrm{mol}}, T_{A}=350.9 K 2. p_{B} = 10^{4} Pa., v_{B} = 10^{-1} \frac{\textrm{m}^{3}}{\textrm{mol}}
A egoeratik B egoerara eramndakoan sistema, zenbatekoa da lor daitekeen lan maximoa? Prozesua aurrera eramateko, erabilgarria da 150 Ketan dagoen bero-iturria.
Ezagunak dira baita ere honako datua hauek:
- lerro adiabatiko itzulgarrien adierazpena: pv^{2}=k
- presio konstantean, p=p_{A}: c_{p} = D v^{\frac{2}{3}}
- presio konstantean, p=p_{A}: \alpha=\frac{3}{T}
- D=10^{\frac{3}{8}} \frac{\textrm{J}}{\textrm{m}^{2}K}
