Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap03

3 The First Law: the machinery Solutions to exercises Discussion questions E3.1(b) The following list includes only those state functions that we have encountered in the first three chapters More will be encountered in later chapters Temperature, pressure, volume, amount, energy, enthalpy, heat capacity, expansion coefficient, isothermal compressibility, and Joule–Thomson coefficient E3.2(b) One can use the general expression for πT given in Justification 3.3 to derive its specific form for a van der Waals gas as given in Exercise 3.14(a), that is, πT = a/Vm2 (The derivation is carried out in Example 5.1.) For an isothermal expansion in a van der Waals gas dUm = (a/Vm )2 Hence Um = −a(1/Vm,2 − 1/Vm,1 ) See this derivation in the solution to Exercise 3.14(a) This formula corresponds to what one would expect for a real gas As the molecules get closer and closer the molar volume gets smaller and smaller and the energy of attraction gets larger and larger E3.3(b) The solution to Problem 3.23 shows that the Joule–Thomson coefficient can be expressed in terms of the parameters representing the attractive and repulsive interactions in a real gas If the attractive forces predominate then expanding the gas will reduce its energy and hence its temperature This reduction in temperature could continue until the temperature of the gas falls below its condensation point This is the principle underlying the liquefaction of gases with the Linde Refrigerator which utilizes the Joule–Thomson effect See Section 3.4 for a more complete discussion Numerical exercises E3.4(b) A function has an exact differential if its mixed partial derivatives are equal That is, f (x, y) has an exact differential if ∂f ∂y ∂ ∂x (a) (b) E3.5(b) E3.6(b) ∂f ∂x ∂f ∂y ∂f ∂s ∂f ∂t dz = (a) (b) = ∂ ∂y ∂f ∂x ∂ ∂y ∂ = 2x y and ∂x ∂ = tes + and ∂t ∂ = 2t + es and ∂s = 3x y and ∂f = 6x y ∂x ∂f = 6x y ∂y ∂f = es ∂s ∂f = es ∂t Therefore, exact Therefore, exact ∂z ∂z dx 2x dy dx + dy = − ∂x ∂y (1 + y)2 (1 + y)3 dz = ∂z ∂z dx + dy = (3x − 2y ) dx − 4xy dy ∂x ∂y ∂ ∂ 2z = (3x − 2y ) = −4y ∂y∂x ∂y ∂ 2z ∂ and = (−4xy) = −4y ∂x∂y ∂x THE FIRST LAW: THE MACHINERY E3.7(b) dz = 45 ∂z ∂z dx + dy = (2xy + y ) dx + (x + 2xy) dy ∂x ∂y ∂ 2z ∂ = (2xy + y ) = 2x + 2y ∂y∂x ∂y and ∂ 2z ∂ = (x + 2xy) = 2x + 2y ∂x∂y ∂x ∂Cp = ∂p T E3.8(b) ∂ ∂p ∂ 2H ∂H = ∂T p ∂p∂T T ∂ ∂T ∂H ∂p T p ∂H = for a perfect gas, its temperature derivative also equals zero; thus ∂p T Because ∂Cp = ∂p T ∂(U +pV ) ∂V (∂H /∂V )p (∂U/∂V )p + p ∂H p p = = = = 1+ ∂U p (∂U/∂V )p (∂U/∂V )p (∂U/∂V )p (∂U/∂V )p E3.9(b) E3.10(b) ∂p dV + ∂V T dp = dp = p p d ln p = We express ∂T ∂V p so d ln p = − E3.11(b) U= ∂p dT ∂T V −1 ∂p ∂p ∂V =− V so =− ∂p T ∂V T ∂V T κT V ∂p in terms of κT and the expansion coefficient α = ∂T V V V ∂p ∂T V ∂p dV + ∂V T p ∂p in terms of the isothermal compressibility κT ∂V T We express κT = − ∂p dT ∂T V ∂V = −1 ∂p T α 1 + = pV κT pκT pκT nRT so (∂V /∂T )p ∂p α =− = ∂T V (∂V /∂p)T κT so ∂U = ∂p T α dT − dV V by direct differentiation H = U + pV = 23 nRT + nRT = 25 nRT , so E3.12(b) ∂H = by direct differentiation ∂p T ∂V nRT α= V = V ∂T p p α= V × V T = T ∂V V nR = = ∂T p p T ∂V ∂T p INSTRUCTOR’S MANUAL 46 E3.13(b) κT = − V κT = − V × − nRT p2 = p The Joule–Thomson coefficient µ is the ratio of temperature change to pressure change under conditions of isenthalpic expansion So µ= E3.14(b) ∂V nRT =− ∂p T p ∂V ∂p T ∂T ≈ ∂p H T −10 K = 0.48 K atm−1 = (1.00 − 22) atm p Um = Um (T , Vm ) dUm = dUm dT + ∂T Vm ∂Um ∂Vm dVm dT = in an isothermal process, so dUm = a ∂Um dVm = dVm ∂Vm T Vm Vm2 −1 −1 Vm2 22.1 L mol dV a a 22.1 L mol m Um = dUm = dV = a = − m 2 Vm 1.00 L mol−1 Vm1 Vm1 Vm 1.00 L mol−1 Vm a a 21.1a =− + = = 0.95475a L−1 mol −1 −1 22.1 L mol 1.00 L mol 22.1 L mol−1 a = 1.337 atm L2 mol−2 Um = (0.95475 mol L−1 ) × (1.337 atm L2 mol−2 ) = (1.2765 atm L mol−1 ) × (1.01325 × 105 Pa atm−1 ) × m3 103 L = 129 Pa m3 mol−1 = 129 J mol−1 w=− so w = − pex dVm RT Vm − b and p = dVm + RT a − for a van der Waals gas Vm − b V m a dVm = −q + Um Vm2 Thus q=+ 22.1 L mol−1 1.00 L mol−1 RT Vm − b dVm = +RT ln(Vm − b) 22.1 L mol−1 1.00 L mol−1 = +(8.314 J K−1 mol−1 ) × (298 K) × ln 22.1 − 3.20 × 10−2 1.00 − 3.20 × 10−2 = +7.7465 kJ mol−1 w = −7747 J mol−1 + 129 J mol−1 = −7618¯ J mol−1 = −7.62 kJ mol−1 THE FIRST LAW: THE MACHINERY E3.15(b) 47 The expansion coefficient is α= = V V (3.7 × 10−4 K −1 + × 1.52 × 10−6 T K−2 ) ∂V = ∂T p V V [3.7 × 10−4 + × 1.52 × 10−6 (T /K)] K −1 V [0.77 + 3.7 × 10−4 (T /K) + 1.52 × 10−6 (T /K)2 ] [3.7 × 10−4 + × 1.52 × 10−6 (310)] K −1 = 1.27 × 10−3 K −1 0.77 + 3.7 × 10−4 (310) + 1.52 × 10−6 (310)2 Isothermal compressibility is V ∂V V so p=− ≈− κT = − V ∂p T V p V κT A density increase 0.08 per cent means V /V = −0.0008 So the additional pressure that must be applied is = E3.16(b) p= E3.17(b) 0.0008 = 3.6 × 102 atm 2.21 × 10−6 atm−1 The isothermal Joule–Thomson coefficient is ∂H = −µCp = −(1.11 K atm−1 ) × (37.11 J K −1 mol−1 ) = −41.2 J atm−1 mol−1 ∂p T If this coefficient is constant in an isothermal Joule–Thomson experiment, then the heat which must be supplied to maintain constant temperature is H in the following relationship H /n = −41.2 J atm−1 mol−1 p so H = −(41.2 J atm−1 mol−1 )n p H = −(41.2 J atm−1 mol−1 ) × (12.0 mol) × (−55 atm) = 27.2 × 103 J E3.18(b) The Joule–Thomson coefficient is µ= ∂T ≈ ∂p H T p so p= T −4.5 K = −3.4 × 102 kPa = µ 13.3 × 10−3 K kPa−1 Solutions to problems Assume that all gases are perfect and that all data refer to 298 K unless stated otherwise Solutions to numerical problems P3.1 atm 1.013 × 105 Pa For the change of volume with pressure, we use κT = (2.21 × 10−6 atm−1 ) × = 2.18 × 10−11 Pa−1 ∂V dp[constant temperature] = −κT V dp κT = − ∂p T V V = −κT V p [If change in V is small compared to V ] dV = ∂V ∂p T p = (1.03 × 103 kg m−3 ) × (9.81 m s−2 ) × (1000 m) = 1.010 × 107 Pa INSTRUCTOR’S MANUAL 48 Consequently, since V = 1000 cm3 = 1.0 × 10−3 m3 , V ≈ (−2.18 × 10−11 Pa−1 ) × (1.0 × 10−3 m3 ) × (1.010 × 107 Pa) = −2.2 × 10−7 m3 , or −0.220 cm3 For the change of volume with temperature, we use ∂V dT [constant pressure] = αV dT α= ∂T p V V = αV T [if change in V is small compared to V ] dV = ∂V ∂T p ≈ (8.61 × 10−5 K −1 ) × (1.0 × 10−3 m3 ) × (−30 K) ≈ −2.6 × 10−6 m3 , or − 2.6 cm3 V ≈ −2.8 cm3 V = 997.2 cm3 Overall, Comment A more exact calculation of the change of volume as a result of simultaneous pressure and temperature changes would be based on the relationship dV = ∂V dp + ∂p T ∂V dT = −κT V dp + αV dT ∂p p This would require information not given in the problem statement P3.5 Use the formula derived in Problem 3.25 Cp,m − CV ,m = λR which gives γ = (3Vr − 1)2 =1− λ 4Vr3 Tr Cp,m CV ,m + λR λR = =1+ CV ,m CV ,m CV ,m In conjunction with CV ,m = 23 R for a monatomic, perfect gas, this gives γ = + 23 λ Vm Vm T 27RbT , Tr = (Table 1.6) with a = = = Vc 3b Tc 8a 4.137 L2 atm mol−2 and b = 5.16 × 10−2 L mol−1 (Table 1.6) Hence, at 100◦ C and 1.00 atm, RT where Vm ≈ = 30.6 L mol−1 p For a van der Waals gas Vr = Vr ≈ 30.6 L mol−1 = 198 (3) × (5.16 × 10−2 L mol−1 ) Tr ≈ (27) × (8.206 × 10−2 L atm K−1 mol−1 ) × (5.16 × 10−2 L mol−1 ) × (373 K) ≈ 1.29 (8) × (4.317 L2 atm mol−2 ) Hence [(3) × (198) − (1)]2 =1− = − 0.0088 = 0.9912, λ (4) × (198)3 × (1.29) γ ≈ (1) + 23 × (1.009) = 1.67 λ = 1.009 THE FIRST LAW: THE MACHINERY 49 Comment At 100◦ C and 1.00 atm xenon is expected to be close to perfect, so it is not surprising that γ differs only slightly from the perfect gas value of 53 P3.7 See the solution to Problem 3.6 It does not matter whether the piston between chambers and is diathermic or adiabatic as long as the piston between chambers and is adiabatic The answers are the same as for Problem 3.6 However, if both pistons are diathermic, the result is different The solution for both pistons being diathermic follows See Fig 3.1 Diathermic piston Diathermic piston Figure 3.1 Initial equilibrium state n = 1.00 mol diatomic gas in each section pi = 1.00 bar Ti = 298 K For each section nRTi (1 mol) × (0.083 145 L bar K−1 mol−1 ) × (298 K) Vi = = pi 1.00 bar = 24.8 L Vtotal = 3Vi = 74.3 L = constant Final equilibrium state The diathermic walls allow the passage of heat Consequently, at equlibrium all chambers will have the same temperature T1 = T2 = T3 = 348 K The chambers will also be at mechanical equlibrium so p1 = p2 = p3 = (n1 + n2 + n3 )RT1 Vtotal (3 mol) × (0.083 145 L bar K−1 mol−1 ) × (348 K) 74.3 L = 1.17 bar = p2 = p3 = The chambers will have equal volume INSTRUCTOR’S MANUAL 50 Vtotal = Vi = 24.8 L = V1 = V2 = V3 V1 = U1 = n1 CV T1 = n1 25 R T1 = (1 mol) × 25 × (8.314 51 J K −1 mol−1 ) × (348 K − 298 K) U1 = 1.04 kJ = U2 = U3 Utotal = U1 = 3.12 kJ = Utotal Solutions to theoretical problems P3.11 dw = ∂w dx + ∂x y,z ∂w dy + ∂y x,z ∂w dz ∂z x,y dw = (y + z) dx + (x + z) dy + (x + y) dz This is the total differential of the function w, and a total differential is necessarily exact, but here we will demonstrate its exactness showing that its integral is independent of path Path a dw = 2x dx + 2y dy + 2z dz = 6x dx (1,1,1) (0,0,0) dw = 6x dx = Path b dw = 2x dx + (y 1/2 + y) dy + (z1/2 + z) dz = (2x + 2x + 2x 1/2 ) dx (1,1,1) (0,0,0) dw = (2x + 2x + 2x 1/2 ) dx = +1+ =3 3 Therefore, dw is exact P3.12 U = U (T , V ) ∂U ∂U dU = dT + dV = CV dT + ∂T V ∂V T For U = constant, dU = 0, and CV dT = − ∂U dV ∂V T or CV = − ∂U dV ∂V T ∂U ∂V T dV ∂U =− dT U ∂V T ∂V ∂T U This relationship is essentially the permuter [Relation 3, Further information 1.7] P3.13 H = H (T , p) ∂H dH = dT + ∂T p ∂H dp = Cp dT + ∂p T For H = constant, dH = 0, and ∂H dp = −Cp dT ∂p T ∂H dT = −Cp = −Cp ∂p T dp H ∂H dp ∂p T ∂T = −Cp µ = −µCp ∂p H This relationship is essentially the permuter [Relation 3, Further information 1.7] THE FIRST LAW: THE MACHINERY P3.16 51 The reasoning here is that an exact differential is always exact If the differential of heat can be shown to be inexact in one instance, then its differential is in general inexact, and heat is not a state function Consider the cycle shown in Fig 3.2 A Isotherm at B Isotherm at Isotherm at Figure 3.2 The following perfect gas relations apply at points labelled 1, 2, and in Fig 3.2 (1) p1 V1 = p2 V2 = nRT , Define T =T −T , (2) p2 V1 = nRT , T =T −T Subtract (2) from (1) −nRT + nRT = −p2 V1 + p1 V1 V1 (p1 − p2 ) RT Subtracting (1) from (3) we obtain giving T = T = V2 (p1 − p2 ) RT Since V1 = V2 , T = T qA = Cp T − CV T = (Cp − CV ) T qB = −CV T + Cp T = (Cp − CV ) T giving qA = qB and q(cycle) = qA − qB = dq = and dq is not exact a RT − p = p(T , V ) = Vm − b Vm2 Therefore P3.18 dp = ∂p dT + ∂T V ∂p dV ∂V T (3) p1 V2 = nRT INSTRUCTOR’S MANUAL 52 In what follows adopt the notation Vm = V ∂p R ; = ∂T V V −b then, dp = ∂p RT 2a =− + ∂V T (V − b)2 V R V −b dT + RT 2a − V3 (V − b)2 dV ∂V is more readily evaluated with the use ∂T p Because the van der Waals equation is a cubic in V , of the permuter ∂p R ∂T V ∂V V −b =− =− ∂p ∂T p + V2a3 − (VRT ∂V T −b)2 RV (V − b) RT V − 2a(V − b)2 = For path a T2 ,V2 T1 ,V1 dp = = T2 T1 V2 R RT2 2a − + dT + V1 − b (V − b) V V1 dV R RT2 RT2 (T2 − T1 ) + − −a V1 − b (V2 − b) (V1 − b) =− RT1 RT2 + −a V1 − b V − b V22 − V22 − V12 V12 For path b T2 ,V2 T1 ,V1 dp = = V2 V1 − T2 RT1 R 2a dV + + dT (V − b) V T1 V2 − b RT1 RT1 − −a V − b V1 − b =− RT1 RT2 + −a V1 − b V − b V22 − V22 V12 − + R (T2 − T1 ) V2 − b V12 Thus, they are the same and dp satisfies the condition of an exact differential, namely, that its integral between limits is independent of path P3.20 p = p(V , T ) Therefore, dp = ∂p dV + ∂V T ∂p dT ∂T V with p = ∂p −nRT 2n2 a −p + = + = ∂V T V − nb (V − nb) V ∂p p n2 a nR = + = ∂T V V − nb T TV2 n2 a nRT − [Table 1.6] V − nb V n2 a V3 × V − 2nb V − nb THE FIRST LAW: THE MACHINERY 53 Therefore, upon substitution dp = = −p dV V − nb n2 a V3 + × (V − 2nb) × (n2 a) × (V − nb)/V − p V − nb = a(Vm − b)/Vm3 − p Vm − b dV + dV V − nb p + n2 a/V T p + a/Vm2 T dVm + + p dT T + n2 a V2 × dT T dT dT Comment This result may be compared to the expression for dp obtained in Problem 3.18 P3.21 p= nRT n2 a − (Table 1.6) V − nb V Hence T = p × (V − nb) + nR na RV ∂T V − nb Vm − b = = = nR ∂p V R × (V − nb) ∂p ∂T V For Euler’s chain relation, we need to show that Hence, in addition to ∂T and ∂p V which can be found from ∂p ∂V T ∂T ∂p V ∂V = −1 ∂T p ∂p [Problem 3.20] we need ∂V T ∂T p = + ∂V p nR = T V − nb na RV − − 2na RV 2na RV ∂V = ∂T p ∂T ∂V p × (V − nb) × (V − nb) Therefore, ∂T ∂p V ∂p ∂V T ∂p ∂T ∂p V ∂V T ∂T ∂V p ∂V = ∂T p = V −nb nR a × (V−nRT + 2n −nb)2 V3 T V −nb 2na − RV × (V − nb) = −T V −nb 2na + RV × (V − nb) T V −nb 2na − RV × (V − nb) = −1 P3.23 µCp = T ∂V −V = ∂T p T ∂T ∂V p − V [Relation 2, Further information 1.7] ∂T 2na T − = (V − nb) [Problem 3.21] ∂V p V − nb RV INSTRUCTOR’S MANUAL 54 Introduction of this expression followed by rearrangement leads to (2na) × (V − nb)2 − nbRT V ×V RT V − 2na(V − nb)2 µCp = Then, introducing ζ = µCp = − nbζ V ζ −1 RT V to simplify the appearance of the expression 2na(V − nb)2 V = − Vbζm ζ −1 V For xenon, Vm = 24.6 L mol−1 , T = 298 K, a = 4.137 L2 atm mol−2 , b = 5.16 × 10−2 L mol−1 , b 5.16 × 10−2 L mol−1 nb = = = 2.09 × 10−3 V Vm 24.6 L mol−1 ζ = (8.206 × 10−2 L atm K−1 mol−1 ) × (298 K) × (24.6 L mol−1 )3 = 73.0 (2) × (4.137 L2 atm mol−2 ) × (24.6 L mol−1 − 5.16 × 10−2 L mol−1 )2 Therefore, µCp = − (73.0) × (2.09 × 10−3 ) × (24.6 L mol−1 ) = 0.290 L mol−1 72.0 Cp = 20.79 J K−1 mol1 [Table 2.6], so à= 0.290 ì 103 m3 mol1 0.290 L mol−1 = −1 −1 20.79 J K mol 20.79 J K−1 mol−1 = 1.393 × 10−5 K m3 J−1 = 1.393 × 10−5 K Pa−1 = (1.393 × 10−5 ) × (1.013 × 105 K atm−1 ) = 1.41 K atm−1 The value of µ changes at T = T1 and when the sign of the numerator − is positive) Hence bζ = at T = T1 Vm that is, T1 = For xenon, 2a Rb or × 1− RT1 bV =1 2na(V − nb)2 Vm nbζ changes sign (ζ − V implying that T1 = 2a(Vm − b)2 RbVm2 b b 27 Tc − = Vm Vm 2a (2) × (4.137 L2 atm mol−2 ) = = 1954 K Rb (8.206 × 10−2 L atm K−1 mol−1 ) × (5.16 × 10−2 L mol−1 ) 5.16 × 10−2 and so T1 = (1954 K) × − 24.6 = 1946 K Question An approximate relationship for µ of a van der Waals gas was obtained in Problem 3.17 Use it to obtain an expression for the inversion temperature, calculate it for xenon, and compare to the result above THE FIRST LAW: THE MACHINERY P3.25 Cp,m − CV ,m = 55 α2 T V [3.21] = αT V κT ∂p [Justification 3.3] ∂T V ∂p nR = [Problem 3.20] ∂T V V − nb ∂V αV = = ∂T ∂T p ∂V p Substituting, T Cp,m − CV ,m = ∂p ∂T V ∂T ∂V p 2na ∂T T − = (V − nb) [Problem 3.21] ∂V p V − nb RV so Substituting, nRT (V −nb) Cp,m − CV ,m = T (V −nb) − 2na RV × (V − nb) = nλR with λ = 1− 2na RT V × (V − nb)2 For molar quantities, Cp,m − CV ,m = λR with 2a(Vm − b)2 =1− λ RT Vm3 Now introduce the reduced variables and use Tc = 8a , Vc = 3b 27Rb After rearrangement, (3Vr − 1)2 =1− λ 4Tr Vr3 For xenon, Vc = 118.1 cm3 mol−1 , Tc = 289.8 K The perfect gas value for Vm may be used as any error introduced by this approximation occurs only in the correction term for λ Hence, Vm ≈ 2.45 L mol−1 , Vc = 118.8 cm3 mol−1 , Tc = 289.8 K, and Vr = 20.6 and Tr = 1.03; therefore (61.8 − 1)2 = 0.90, =1− λ (4) × (1.03) × (20.6)3 and giving λ ≈ 1.1 Cp,m − CV ,m ≈ 1.1R = 9.2 J K−1 mol−1 P3.27 (a) µ=− Cp ∂H = ∂p T Cp RT + aT p R ∂Vm = + 2aT ∂T p p Vm = T ∂Vm − Vm ∂T p [Justification 3.1 and Problem 3.24] INSTRUCTOR’S MANUAL 56 µ= µ= (b) Cp RT RT + 2aT − − aT p p aT Cp ∂p ∂T V ∂p ∂Vm ∂T p ∂T V CV = Cp − αT Vm = Cp − T RT Vm − aT ∂p R RT (−2aT ) = − ∂T V (Vm − aT )2 Vm − aT R 2aRT = + (RT /p) (RT /p)2 2ap p = + T R Therefore R p 2ap CV = Cp − T + 2aT × + p T R But, p = = Cp − RT p 1+ CV = Cp − R + 2apT R × 1+ 2apT R × p T 2apT R Solutions to additional problems P3.29 (a) The Joule–Thomson coefficient is related to the given data by µ = −(1/Cp )(∂H /∂p)T = −(−3.29 × 103 J mol−1 MPa−1 )/(110.0 J K−1 mol−1 ) = 29.9 K MPa−1 (b) The Joule–Thomson coefficient is defined as µ = (∂T /∂p)H ≈ ( T / p)H Assuming that the expansion is a Joule–Thomson constant-enthalpy process, we have T = µ p = (29.9 K MPa−1 ) × [(0.5 − 1.5) × 10−1 MPa] = −2.99 K ... 3.4 ì 102 kPa = 13.3 × 10−3 K kPa−1 Solutions to problems Assume that all gases are perfect and that all data refer to 298 K unless stated otherwise Solutions to numerical problems P3.1 atm 1.013... expected to be close to perfect, so it is not surprising that γ differs only slightly from the perfect gas value of 53 P3.7 See the solution to Problem 3.6 It does not matter whether the piston between... = (n1 + n2 + n3 )RT1 Vtotal (3 mol) × (0.083 145 L bar K−1 mol−1 ) × (348 K) 74.3 L = 1.17 bar = p2 = p3 = The chambers will have equal volume INSTRUCTOR S MANUAL 50 Vtotal = Vi = 24.8 L = V1