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Introduction to Statistical Physics Solution Manual Kerson Huang ii Chapter 1 1.1 Mass of water =106g, temperature raised by 20 C. ◦ Heat needed Q=2 107cal = 8.37 107J.=23.2 kwh. × × Work needed = mgh= 14 150 29000= 6.09 107 ft-lb =22.9 kwh. × × × 1.2 Work done along various paths are as follows ab: b b dV V PdV =Nk T =Nk T ln b B 1 V B 1 V Za Za a cd: V P (V V )=Nk T 1 b d d− b B 3 − V µ d¶ de: e dV V Nk T =Nk T ln a B 3 V B 3 V Zd d No work is done along bc and ea. The total work done is the sum of the above. Heatabsorbedequalstotalworkdone,sinceinternalenergyisunchanged in a closed cycle. 1.3 (a) 1 ∂V bV Tb 1 α= = 0 − V ∂T TbV 0 (b) bV Tb 1 ∆V = 0 − ∆T Tb 0 Nk T Nk Tb P = B = B 0T1 b − V V 0 Work done = P∆V =bNk ∆T B 1 2 CHAPTER 1. 1.4 Consider an element of the column of gas, of unit cross section, and height betweenzandz+dz. Theweightoftheelementis gdM,wheredM isthemass − of the element: dM = mndz, where m is the molecular mass, and n = P/k T B is the local density, with P the pressure. For equilibrium, the weight must equal the pressure differential: dP = gdM .Thus, dP/P = (mg/k T)dz. At B − − constant T , we have dp/P =dn/n.Therefore n(z)=n(0)e−mgz/kBT 1.5 No change in internal energy, and no work is done. Therefore total heat absorbed ∆Q=∆Q +∆Q =0. That is, heat just pass from one body to the 1 2 other. Suppose the final temperature is T. Then ∆Q =C (T T ), ∆Q =C (T T ). Therefore 1 1 1 2 2 2 − − C T +C T T = 1 1 2 2 C +C 1 2 1.6 Work done by the system is HdM. Thus the work on the system is − R κ κH2 HdM = HdH = T 2T Z Z 1.7 Consider the hysteresis cycle in the sense indicated in Fig.1.6. Solve for the magnetic field: H = H +tanh 1(M/M ) 0 − 0 ± (+forlowerbranch, forupperbranch.). UsingW = HdM,weobtain − − R W = M0 dM[H +tanh 1(M/M )] −M0dM[ H +tanh 1(M/M )] 0 − 0 0 − 0 − − − Z−M0 ZM0 = 4M H 0 0 − 1.8 3 A log log plot of mass vs. A is shown in the following graph. The dashed line is a straightline for reference. M g o l 1 10 100 1000 10000 A . 4 CHAPTER 1. Chapter 2 2.1 Use the dQ equation with P,T as independent variables: dQ=C dT +[(∂U/∂P) +P(∂V/∂P) ]dP P T T For an ideal gas (∂U/∂P) =0, P(∂V/∂P) = V. Thus T T − dQ=C dT VdP. P − The heat capacity is given by C =C V(∂P/∂T) . P path − The path is P =aVb, or equivalently Pb+1 =a(Nk T)b by the equation of B state. Hence V(∂P/∂T) =[ab/(b+1)]V(Nk T)bT 1 =bNk /(b+1). Therefore path B − B b C =C Nk P − b+1 B This correctly reduces to C for b=0. P 2.2 Use a Carnot engine to extracted energy from 1 gram of water between 300 K and 290 K. Max efficiency η =1 (290/300)=1/30. − 1 W =ηC∆T = (4.164 J g 1K 1 1 g 10 K)=1.39 J − − 30 × × Gravitational potential energy=1 g 9.8 kg s 2 110 m=1.08 J − × × 2.3 The highest and lowest available temperatures are, 600 F = 588.7 K and 70 F = 294.3 K. The efficiency of the power plant is W/Q =0.6[1 (294.3/588.7)]=0.3. 1 − In one second: W =106 J. So Q =2.33 106 J =C ∆T. Use C =4.184 J g 1K 1, 2 V V − − × 5 6 CHAPTER 2. Flow rate = 6000 (0.305m)3 × 2.33 106 J ∆T = (4.184 J g 1K 2)6000×(0.305 m)3 106 cm2/m3 =3.27×10−3K − − 2.4 (a) Since water is incompressible, a unit mass input gives a unit mass output. The net heat supplied per unit mass is ∆Q=C(T T) C(T T ), 1 2 − − − whereC isthespecificheatofwater(perunitmass.) Insteadystatev2/2= ∆Q. This gives v = 2∆Q= 2C(T +T 2T) 1 2 − (b) p p The entropy depends on the temperature like lnT. A unit volume of water from each of the input streams has total entropy lnT +lnT This makes two 1 2 unit volumes in the output stream, with entropy 2lnT. Therefore the change in entropy is ln T2/T T 0. Thus T √T T , and 1 2 1 2 ≥ ≥ ¡ ¢ v =√2C T T max 1 2 − ¯p p ¯ ¯ ¯ ¯ ¯ 2.5 (a) PVγ =2P Vγ, PVγ =2 P Vγ (V 1/V )γ =020. (L¯+2 a)/(\L 0 a0) γ =2. 1 2 − £ a¤ 21/γ 1 = − L 21/γ +1 (b) ∆U =∆Q W, ∆Q=0. − C ∆T = W, ∆T = W/C . V V − − T =2T +∆T =2T (W/C ), T =T ∆T =T +(W/C ). 1 0 0 V 2 0 0 V − − RT R[2T (W/C )] P = 1 = 0− V V A(L+a) 1 (c) a W =A dx(P P ) P =2P0Vγ/[A1(L−+2x)]γ, P =P Vγ/[A(L x)]γ. 1 R0 0 2 0 0 − P V a 2a W = 0 0 1 1 γ 1 − L γ − L − ³ − ´µ ¶ 7 2.6 (a) PV =U/3, U =σVT4. P =σT4/3. dS =dQ/T =(dU +PdV)/T. Integrate along paths with T=const, V=const. 4 S = σVT3. 3 (b) S =Constant. ∴T3 V−1. Thus ∼ T R 1 − ∼ 2.7 The heat absorbed by an ideal gas in an isothermal process is ∆Q=NkT ln(V /V ) f i where V and V are respectively the final and initial volume.The temperature f i T in this formula is the ideal-gas temperature. Draw a Carnot cycle on the PV diagram, and label the corners 1234 clock- wise from the upper left. The heat absorbed at the upper temperature T , and the heat rejected at 2 the lower temperature T , are 1 Q =NkT ln(V /V ) 2 2 2 1 Q =NkT ln(V /V ) 1 1 3 4 Because 23 and 12 lie on adiabatic lines, we have V2T2γ−1 =V3T1γ−1 V1T2γ−1 =V4T1γ−1 Dividing one equation by the other yields V /V =V /V . 2 1 3 4 The efficiency of the cycle is therefore Q T η =1 1 =1 1 − Q − T 2 2 2.8 Diesel cycle: Q =C (T T ) 2 P 3 2 − Q =C (T T ) 1 V 4 1 − η =1 (Q /Q )=1 γ 1[(T T )/(T T )] 1 2 − 4 1 3 2 − − − − We have P =P , hence 3 2 T /T =V /V =r 3 2 3 2 c The processes 12 and 34 are adiabatic, with TVγ 1 = constant. V =V . − 4 1 Thus 8 CHAPTER 2. T3V3γ−1 =T4V1γ−1 T2V2γ−1 =T1V1γ−1 Using the three relations derived, we obtain 1 rγ 1 η =1 c − − γrγ 1(r 1) − c− 2.9 Otto cycle: Q =C (T T ) 2 V 3 2 − Q =C (T T ) 1 V 4 1 − η =1 (Q /Q )=1 [(T T )/(T T )] 1 2 4 1 3 2 − − − − The processes 12 and 34 are adiabatic, with TVγ 1 = constant. We have − V =V , V =V Thus 4 1 3 2 T1V1γ−1 =T2V2γ−1. T3V2γ−1 =T4V1γ−1. Taking the ratio of these equations, we have T /T =T /T =rγ 1. 2 1 3 4 − Thus η =1 r1 γ − − 2.10 First note T /T =V /V =2. b a b a Work done Heat absorbed a b P (V V )=P V =NkT C ∆T =C T a b a a a a P P a → − b c 0 C T V a → − c a PdV = NkT ln2 NkT ln2 a a → − − − R W (Net work done) =NkT (1 ln2) a − Q (Heat absorbed) =C T = 5NkT 2 P a 2 a W 2 η = = (1 ln2)=0.12 Q 5 − 2 In comparison, η =1 (T /T )=0.5. Carnot − b a 2.11 First note T =4T . The P,V,T for the points A,B,C,D are as follows: 2 1 P V T A P V =NkT /P T 1 1 1 1 1 B 2P 2V 4T 1 1 1 C 2P V 2T 1 1 1 D P 2V 2T 1 1 1