Introduction to Thermodynamics and Statistical Physics phần 6 pot

Chapter 2. Ideal Gas V p 1 p 2 p a b d c V p 1 p 2 p a b d c Fig. 2.11. V p V 1 V 2 a b c d V p V 1 V 2 a b c d Fig. 2.12. two isentropic processes (constant entropy) b→candd→a, as shown in Fig. 2.12. Assume that the heat capacities C V and C p are temperature independent. Calculate the e fficiency η of this engine. Eyal Buks Thermodynam ics and Statistical Physics 78 2.9. Solutions Set 2 29. Consider two vessels A and B each containing ideal classical gas of par- ticles having no internal degrees of freedom . The pressure and num ber of particles in both ves sels are p and N respectively, and the tempera- ture is τ A in vessel A and τ B in vessel B. The two vessels are brought into thermal contact. No heat is exchanged with the environment during this process. Moreover, the pressure is kept constant at the value p in both vessels during this process. Find the change in the total en tropy ∆σ = σ final − σ initial . 2.9 Solutions Set 2 1. The entropy is give n by σ = N ( log " µ Mτ 2π} 2 ¶ 3/2 V N # + 5 2 ) , (2.162) or using pV = Nτ σ = N ( log " µ M 2π} 2 ¶ 3/2 τ 5/2 p # + 5 2 ) , (2.163) thus C p = τ µ ∂σ ∂τ ¶ p = 5 2 N. (2.164) 2. Since ∂ 2 F ∂V ∂τ = ∂ 2 F ∂τ∂V , (2.165) where F is Helmholtz free energy, one has µ ∂ ∂V µ ∂F ∂τ ¶ V ¶ τ = µ ∂ ∂τ µ ∂F ∂V ¶ τ ¶ V . (2.166) By definition µ ∂F ∂V ¶ τ = −p. Moreover, using F = U −τσ one finds µ ∂F ∂τ ¶ V = µ ∂U ∂τ ¶ V − τ µ ∂σ ∂τ ¶ V − σ = µ ∂U ∂τ ¶ V − µ ∂U ∂σ ¶ V µ ∂σ ∂τ ¶ V − σ = −σ, (2.167) Eyal Buks Thermodynam ics and Statistical Physics 79 Chapter 2. Ideal Gas thus µ ∂σ ∂V ¶ τ = µ ∂p ∂τ ¶ V . (2.168) 3. We have found in class the following relations n Q = µ Mτ 2π~ 2 ¶ 3/2 , (2.169) η = − µ τ , (2.170) log Z gc = e −η Z int Vn Q , (2.171) U = µ 3τ 2 − ∂ log Z int ∂β ¶ log Z gc , (2.172) N =logZ gc . (2.173) a) Using Eqs. (2.171) and (2.173) one finds log n n Q Z int = µ τ , (2.174) thus µ = τ µ log n n Q − log Z int ¶ . (2.175) b) Using Eqs. (2.172) and (2.173) U = N µ 3τ 2 − ∂ log Z int ∂β ¶ . (2.176) c) Using the relations F = U −τσ , (2.177) σ =logZ gc + βU + ηN , (2.178) one finds F = U −τσ (2.179) = Nτ (−η −1) (2.180) = Nτ ³ µ τ − 1 ´ (2.181) = Nτ µ log n n Q − log Z int − 1 ¶ . (2.182) (2.183) Eyal Buks Thermodynam ics and Statistical Physics 80 2.9. Solutions Set 2 d) Using the relation σ = − µ ∂F ∂τ ¶ V , (2.184) one finds σ = − µ ∂F ∂τ ¶ V = N   −   ∂ ³ τ log n n Q ´ ∂τ   V + ∂ (τ log Z int ) ∂τ +1   = N   −τ   ∂ ³ log n n Q ´ ∂τ   V − log n n Q + ∂ (τ log Z int ) ∂τ +1   = N µ 3 2 − log n n Q + ∂ (τ log Z int ) ∂τ +1 ¶ = N µ 5 2 +log n Q n + ∂ (τ log Z int ) ∂τ ¶ . (2.185) e) By definition c V = τ µ ∂σ ∂τ ¶ V = N µ 3 2 + τ ∂ 2 (τ log Z int ) ∂τ 2 ¶ . (2.186) f) The following holds c p = τ µ ∂σ ∂τ ¶ p (2.187) = τ µ ∂σ ∂τ ¶ V + τ µ ∂σ ∂V ¶ τ µ ∂V ∂τ ¶ p = c V + τ µ ∂σ ∂V ¶ τ µ ∂V ∂τ ¶ p . Using Vp= Nτ and Eq. (2.185) one finds c p = c V + τ N V N p = c V + N. (2.188) Eyal Buks Thermodynam ics and Statistical Physics 81 Chapter 2. Ideal Gas 4. Using Eqs. (1.87), (1.70) and (1.71), and noting that ∂ ∂τ = − 1 τ 2 ∂ ∂β , (2.189) one finds that c = τ ∂σ ∂τ = ∂U ∂τ = − 1 τ 2 ∂U ∂β = D (∆U) 2 E τ 2 . (2.190) 5. The internal partition function is given by Z int = 1 2sinh ~ω 2τ ' τ ~ω , (2.191) thus using Eqs. (2.186) and (2.188) c V = N à 3 2 + τ ∂ 2 ¡ τ log τ ~ω ¢ ∂τ 2 ! = 5N 2 , (2.192) c p = 7N 2 . (2.193) 6. Energy conservation requires that the temperature of the mixture will remain τ . The entropy of an ideal gas of density n, which contains N particles, is given by σ (N,n)=N µ log n Q n + 5 2 ¶ thus the change in entropy is given by ∆σ = σ mix − σ A − σ B = σ µ N A , N A V A + V B ¶ + σ µ N B , N B V A + V B ¶ − σ µ N A , N A V A ¶ − σ µ N B , N B V B ¶ = N A log V A + V B V A + N B log V A + V B V B 7. The probabilit y t o find a m olecule in the volume v is given by p = v/V . Thus, p n is given by p n = N! n!(N − n)! p n (1 − p) N−n . (2.194) Using the solution of problem 4 of set 1 p n = λ n n! e −λ , where λ = Nv/V. Eyal Buks Thermodynam ics and Statistical Physics 82 2.9. Solutions Set 2 8. The partition function of a single atom is given b y Z 1 =exp(µ 0 Hβ)+exp(−µ 0 Hβ)=2cosh(µ 0 Hβ) , where β =1/τ , thus the partition function of the entire system is Z =(2cosh(µ 0 Hβ)) N , a) The free energy is given by F = −τ lo g Z = −Nτ log(2cosh(µ 0 Hβ)) . (2.195) The magnetization is given by M = − µ ∂F ∂H ¶ τ = Nµ 0 tanh (µ 0 Hβ) . (2.196) b) The energy U is given by U = − ∂ log Z ∂β = −Nµ 0 H tanh (µ 0 Hβ) , (2.197) thus C = τ µ ∂σ ∂τ ¶ H = µ ∂U ∂τ ¶ H = −Nµ 0 H à ∂ tanh µ 0 H τ ∂τ ! H = N à µ 0 H τ 1 cosh µ 0 H τ ! 2 . (2.198) c) The entropy σ,whichisgivenby σ = β (U −F) = N · log µ 2cosh µ 0 H τ ¶ − µ 0 H τ tanh µ 0 H τ ¸ , (2.199) and which remains constant, is a function of the ratio H/τ,therefore τ 2 = τ 1 H 2 H 1 . (2.200) 9. The partition function of a single atom is given b y Eyal Buks Thermodynam ics and Statistical Physics 83 Chapter 2. Ideal Gas Z = 1 X m=1 exp ( m ) =1+2exp()cosh(à 0 H) , (2.201) where =1/ . T he free energy is given by F = N log Z, (2.202) thus the m agnetization is given by M = à F H ả = 2Nà 0 exp ()sinh(à 0 H) 1+2exp()cosh(à 0 H) . (2.203) and the magnetic su sceptibility is given by = Nà 2 0 Ă 1+ 1 2 exp ()  , (2.204) 10. The partition function of a s ingle atom is given by Z = J X m=J exp (màH) , where =1/ . By multiplying by a factor sinh (àH/2) one nds sinh à àH 2 ả Z = 1 2 ã exp à àH 2 ả exp à àH 2 ảá J X m=J exp (màH) = 1 2 ã exp ãà J + 1 2 ả àH á exp ã à J + 1 2 ả àH áá , (2.205) thus Z = sinh ÊĂ J + 1 2  àH Ô sinh àH 2 . (2.206) a) The free energy is given by F = N log Z = N log sinh ÊĂ J + 1 2  àH Ô sinh àH 2 . (2.207) Eyal Buks Thermodynam ics and Statistical Physics 84 2.9. Solutions Set 2 b) The magnetization is given by M = − µ ∂F ∂H ¶ τ = Nµ 2 ½ (2J +1)coth · (2J +1) µH 2τ ¸ − coth µ µH 2τ ¶¾ . (2.208) 11. The intern al chemical potential µ g is giv en by Eq. (2.131). In thermal equilibrium the to tal chemical potential µ tot = µ g + mgz , (2.209) (m is the mass of the eac h dia tom ic molecule N 2 , g is the gravity accel- eration constant, and z is the height) is z inde pendent. Thus, the density n as a function of height above see lev e l z can be expressed as n (z)=n (0) exp µ − mgz k B T ¶ . (2.210) The condition n (z)=0.5 × n (0) yields z = k B T ln 2 mg = 1.3806568 × 10 −23 JK −1 × 300 K × ln 2 14 × 1.6605402 × 10 −27 kg × 9.8ms −2 =12.6km. (2.211) 12. The Helmholtz free energy of an ideal gas of N particles is given by F = −τN log " µ Mτ 2π} 2 ¶ 3/2 V # + τN log N − τN , (2.212) th us the chemical potential is µ = µ ∂F ∂N ¶ τ,V = −τ log à µ Mτ 2π} 2 ¶ 3/2 V ! + τ log N, (2.213) and the pressure is p = − µ ∂F ∂V ¶ τ,V = Nτ V . (2.214) Using these results the fugacity λ =exp(βµ) can be expressed in terms of p λ = e βµ = µ Mτ 2π} 2 ¶ −3/2 N V = µ M 2π} 2 ¶ −3/2 τ −5/2 p. (2.215) At equilibrium the fugacity of the gas and that of the system of absorb- ing sites is the same. The grand canonical partition function of a single absorption site is given by Eyal Buks Thermodynam ics and Statistical Physics 85 Chapter 2. Ideal Gas Z =1+e à + e (2à) , (2.216) or in terms of the fugacity =exp(à) Z =1+ + 2 e . (2.217) Thus hN a i = N 0 log Z = N 0 +2 2 e 1+ + 2 e , (2.218) where is given by Eq. (2.215). 13. The internal partition function is given by Z int = g 1 + g 2 exp . (2.219) Using Eq. (2.135) c V = 3 2 N + N ã 2 2 ( log Z int ) á V = N ( 3 2 + 2 g 1 g 2 exp Ă Â Ê g 1 + g 2 exp Ă ÂÔ 2 ) . (2.220) Using Eq. (2.136) c p = N ( 5 2 + 2 g 1 g 2 exp Ă Â Ê g 1 + g 2 exp Ă ÂÔ 2 ) . (2.221) 14. Using Maxwells relation à V ả ,N = à p ả V,N , (2.222) and the equation of state one nds that à V ả ,N = N V b . (2.223) a) Using the denitions c V = à ả V,N , (2.224) c p = à ả p,N , (2.225) Eyal Buks Thermodynam ics and Statistical Physics 86 2.9. Solutions Set 2 and the general identit y µ ∂z ∂x ¶ α = µ ∂z ∂x ¶ y + µ ∂z ∂y ¶ x µ ∂y ∂x ¶ α , (2.226) one finds c p − c V = τ µ ∂σ ∂V ¶ τ,N µ ∂V ∂τ ¶ p,N , (2.227) or with the help of Eq. (2.223) and the equation of state c p − c V = N Nτ p (V − b) = N. (2.228) b) Using the identity µ ∂U ∂V ¶ τ,N = µ ∂U ∂V ¶ σ,N + µ ∂U ∂σ ¶ V,N µ ∂σ ∂V ¶ τ,N , (2.229) together with Eq. (2.223) yields µ ∂U ∂V ¶ τ,N = −p + Nτ V − b =0. (2.230) Thus, the energy U is independent on the volume V (it can b e ex- pressed as a function of τ and N only), and therefore for processes for whic h d N = 0 the c hange in energy dU can be expressed as dU = c V dτ. (2.231) For an isentropic pr ocess no heat is exchanged, and therefore dW = −dU,thussincec V is independent on temperature one has W = −∆U = −c V (τ 2 − τ 1 ) . (2.232) 15. Using the definitions c V = τ µ ∂σ ∂τ ¶ V,N , (2.233) c p = τ µ ∂σ ∂τ ¶ p,N , (2.234) and the general iden tity µ ∂z ∂x ¶ α = µ ∂z ∂x ¶ y + µ ∂z ∂y ¶ x µ ∂y ∂x ¶ α , (2.235) Eyal Buks Thermodynam ics and Statistical Physics 87 [...]... from the hot bath per cycle h and delivered −Q0 heat to the cold one per cycle, where W = Q0 + Q0 l l h and ηc = W τl = 1− Q0 τh h (2. 262 ) According to Clausius principle Ql + Q0 ≤ 0 , l (2. 263 ) thus γ= Q0 Q0 − W τh τl Ql ≤− l = h = −1= W W W τh − τl τh − τl (2. 264 ) 21 Using Eq (2. 264 ) τl A (τ h − τ l ) = , P τh − τl (2. 265 ) µ ¶ P τ 2 − 2τ l τ h + + τ2 = 0 , l h 2A (2. 266 ) thus or P ± τl = τh + 2A... σ,N ∂σ V,N ∂V τ ,N ∂V τ ,N Eyal Buks Thermodynamics and Statistical Physics 88 2.9 Solutions Set 2 (2.244) and Maxwell’s relation µ ¶ µ ¶ ∂p ∂σ = , ∂V τ ,N ∂τ V,N (2.245) one finds µ µ ¶ ¶ ∂U ∂p =τ −p ∂V τ ,N ∂τ V,N (2.2 46) In the present case µ ¶ a ∂U Nτ −p= 2 , = ∂V τ ,N V −b V (2.247) thus ∆U = Z V2 V1 µ ∂U ∂V ¶ dV = a Z V2 V1 τ ,N V2 − V1 dV =a , V2 V2 V1 (2.248) and Q = ∆U + W = N τ 0 log V2 − b ... 2A s µ ¶2 P − τ2 τh + h 2A The solution for which τ l ≤ τ h is sµ ¶2 P P − − τ2 τl = τh + τh + h 2A 2A 22 Using Eq (2.244) µ µ ¶ ¶ ∂U ∂p =τ −p, ∂V τ ,N ∂τ V,N Eyal Buks Thermodynamics and Statistical Physics (2. 267 ) (2. 268 ) (2. 269 ) 91 Chapter 2 Ideal Gas thus 3Aτ 3 2Aτ 3 Bτ n = −p= , V V V (2.270) therefore B = 2A , n=3 (2.271) (2.272) 23 In general the following holds µ ¶ ∂ 2 (τ log Zint ) 3 +τ... the following holds µ n ¶ 1 ∂ Zgc hE n i = , − Zgc ∂β n η (2.249) (2.250) and µ ∂ log Zgc U =− ∂β ¶ (2.251) η Thus the variance is given by D E ­ ® (∆E)2 = E 2 − hEi2 µ 2 ¶ µ ¶2 1 1 ∂ Zgc ∂Zgc = − 2 Zgc ∂β η ∂β 2 η Zgc µ 2 ¶ ∂ log Zgc = ∂β 2 η (2.252) Furthermore, the following holds: Eyal Buks Thermodynamics and Statistical Physics 89 Chapter 2 Ideal Gas D E ­ ® (∆E)3 = E 3 − 3E 2 U + 3EU 2 − U 3... given by ∆σ = σfinal − σ1 − σ 2 µ ¶ µ ¶ µ ¶ (p1 + p2 ) τ nQ 5 τ nQ 5 τ nQ 5 = 2N log + + + − N log − N log 2p1 p2 2 p1 2 p2 2 = N log (p1 + p2 )2 4p1 p2 (2.288) Eyal Buks Thermodynamics and Statistical Physics 93 Chapter 2 Ideal Gas 26 Using σ = log Zc + βU = log Zgc + βU + ηN one finds log Zc = log Zgc + ηN (2.289) The following holds for classical ideal gas having no internal degrees of freedom log... (2. 260 ) The function f (x) = x − 1 − log x in the range 0 < x < ∞ satisfy f (x) ≥ 0, where f (x) > 0 unless x = 1 Eyal Buks Thermodynamics and Statistical Physics 90 2.9 Solutions Set 2 19 The efficiency is given by η = 1+ p2 (V1 − V2 ) Qca Cp (τ a − τ c ) Ql = 1−γ , (2. 261 ) = 1+ = 1+ Qh Qab Cv (τ b − τ a ) V2 (p1 − p2 ) where γ = Cp /CV 20 Energy conservation requires that W = Ql + Qh Consider a Carnot... + 3 exp − , τ thus a) CV is given by  ¡ ∆ ¢2 (2.275) −∆ τ  3 e  3 CV = N  + ³ τ ´2  , ∆ 2 1 + 3e− τ b) and Cp is given by   ¡ ¢2 ∆ 3 ∆ e− τ  5 Cp = N  + ³ τ ´2  , ∆ 2 1 + 3e− τ (2.2 76) (2.277) 24 Consider an infinitesimal change in the temperatures of both bodies dτ 1 and dτ 2 The total change in entropy associated with the reversible process employed by the heat engine vanishes, thus µ... heat engine vanishes, thus µ ¶ dQ1 dQ2 dτ 1 dτ 2 + =C + (2.278) 0 = dσ = dσ 1 + dσ 2 = τ1 τ2 τ1 τ2 a) Thus, by integration the equation dτ 2 dτ 1 =− , τ1 τ2 (2.279) one finds Eyal Buks Thermodynamics and Statistical Physics 92 2.9 Solutions Set 2 Z τf τ1 dτ 1 =− τ1 Z τf τ2 dτ 2 , τ2 (2.280) or log τf τ2 = log , τ1 τf (2.281) thus τf = √ τ 1τ 2 (2.282) b) Employing energy conservation law yields √ √... (2.2 56) b) Using Eq (2.253) D E ∂ 3N 3N 8U 3 (∆E)3 = − = 3 = ∂β 2β 2 9N 2 β 18 The entropy change of the body is Z τb τb dτ = C log ∆σ1 = C , τ τa τa (2.257) (2.258) and that of the bath is ∆σ2 = ∆Q C (τ a − τ b ) = , τb τb thus ∆σ = C µ τa τa − 1 − log τb τb ¶ (2.259) (2. 260 ) The function f (x) = x − 1 − log x in the range 0 < x < ∞ satisfy f (x) ≥ 0, where f (x) > 0 unless x = 1 Eyal Buks Thermodynamics. .. process a → b, while Ql is associated with process c → d In both isentropic processes (b → c and d → a) no heat is exchanged Thus η =1+ Eyal Buks Cp (τ d − τ c ) Ql =1+ Qh Cp (τ b − τ a ) Thermodynamics and Statistical Physics (2.294) 94 . (2.225) Eyal Buks Thermodynam ics and Statistical Physics 86 2.9. Solutions Set 2 and the general identit y µ ∂z ∂x ¶ α = µ ∂z ∂x ¶ y + µ ∂z ∂y ¶ x µ ∂y ∂x ¶ α , (2.2 26) one finds c p − c V = τ µ ∂σ ∂V ¶ τ,N µ ∂V ∂τ ¶ p,N ,. per cycle and delivered −Q 0 l heat to the cold one per cycle, where W = Q 0 l + Q 0 h and η c = W Q 0 h =1− τ l τ h . (2. 262 ) According to Clausius principle Q l + Q 0 l ≤ 0 , (2. 263 ) thus γ. τ l . (2. 264 ) 21. Using Eq. (2. 264 ) A (τ h − τ l ) P = τ l τ h − τ l , (2. 265 ) thus τ 2 l − 2τ l µ τ h + P 2A ¶ + τ 2 h =0, (2. 266 ) or τ l = τ h + P 2A ± s µ τ h + P 2A ¶ 2 − τ 2 h . (2. 267 ) The