5 The Second Law: the machinery Solutions to exercises Discussion questions E5.1(b) See the solution to Exercise 3.14(a) and Example 5.1, where it is demonstrated that πT = a/Vm2 for a van der Waals gas Therefore, there is no dependence on b for a van der Waals gas The internal pressure results from attractive interactions alone For van der Waals gases and liquids with strong attractive forces (large a) at small volumes, the internal pressure can be very large E5.2(b) The relation (∂G/∂T )p = −S shows that the Gibbs function of a system decreases with T at constant p in proportion to the magnitude of its entropy This makes good sense when one considers the definition of G, which is G = U + pV − T S Hence, G is expected to decrease with T in proportion to S when p is constant Furthermore, an increase in temperature causes entropy to increase according to S= f i dqrev /T The corresponding increase in molecular disorder causes a decline in th Gibbs energy (Entropy is always positive.) E5.3(b) The fugacity coefficient, φ, can be expressed in terms of an integral involving the compression factor, specifically an integral of Z − (see eqn 5.20) Therefore, we expect that the variation with pressure of the fugacity coefficient should be similar, in a very qualitative sense, to the variation with pressure of the compression factor itself Comparison of figures 1.27 and 5.8 of the text shows this to be roughly the case, though the detailed shapes of the curves are necessarily different because φ is an integral function of Z − over a range of pressures So we expect no simple proportionality between φ and Z But we find φ < in pressure regions where attractive forces are expected to predominate and φ > when repulsive forces predominate, which in behavior is similar to that of Z See Section 5.5(b) for a more complete discussion Numerical exercises E5.4(b) E5.5(b) α= × V ∂V ∂T p κT = − V × ∂V ∂p T ∂S ∂V =− = −αV ∂p T ∂T p pf pf Vi at constant temperature, G = nRT ln = pi pi Vf = nRT ln Vi Vf = (2.5 × 10−3 mol) × (8.314 J K −1 mol−1 ) × (298 K) × ln = −2.035 = −2.0 J 72 100 INSTRUCTOR’S MANUAL 76 ∂G = −S ∂T p ∂Gf ∂Gi = −Sf and = −Si ∂T p ∂T p ∂Gf ∂Gi S = Sf − Si = − + ∂T p ∂T p ∂(Gf − Gi ) ∂ G =− =− ∂T ∂T p p ∂ {−73.1 + 42.8 T /K} J =− ∂T = −42.8 J K−1 E5.6(b) E5.7(b) See the solution to Exercise 5.7(a) Without knowledge of the compressibility of methanol we can only assume that V = V1 (1 − κT p) ≈ V1 Then G=V p m m 25 g ρ= = 31.61 cm3 so V = = V ρ 0.791 g cm−3 m3 106 cm3 G = (31.61 cm3 ) × × (99.9 × 106 Pa) = +3.2 kJ E5.8(b) (a) pi Vf = nR ln Vi pf Taking inverse logarithms S = nR ln [Boyle’s Law] pf = pi e− S/nR = (150 kPa) exp − −(−15.0 J K−1 ) (3.00 mol) × (8.314 J K −1 mol−1 ) = 274 kPa (b) G = nRT ln pf pi = −T S [ H = 0, constant temperature, perfect gas] = −(230 K) × (−15.0 J K−1 ) = +3450 J = 3.45 kJ E5.9(b) µ = µf − µi = RT ln pf pi = (8.314 J K−1 mol−1 ) × (323 K) ì ln = 2.71 kJ mol1 E5.10(b) à0 = µ−− + RT ln µ = µ−− + RT ln p p −− f p −− 252.0 92.0 THE SECOND LAW: THE MACHINERY 77 f p µ − µ0 = RT ln f ≡φ p µ − µ0 = RT ln φ = (8.314 J K−1 mol−1 ) × (290 K) × ln(0.68) = −929.8 J mol−1 = −930 J mol−1 or −0.93 kJ mol−1 (160.0 cm3 mol−1 ) × 1016mcm3 B B = =− RT (8.314 J K−1 mol−1 ) × (100 K) E5.11(b) = −1.924 × 10−7 Pa−1 φ = eB p+··· −7 −1 ≈ e− 1.924×10 Pa × 62×10 Pa ≈ e−11.93 = × 10−6 E5.12(b) or of the order of 10−6 G = nVm p = V p = (1.0 L) × m3 103 L × (200 × 103 Pa) = 200 Pa m3 = 200 J E5.13(b) Gm = RT ln pf pi = (8.314 J K−1 mol−1 ) × (500 K) × ln 100.0 kPa 50.0 kPa = +2.88 kJ mol−1 E5.14(b) E5.15(b) ∂G RT + B + C p + D p2 [5.10] = ∂p T p which is the virial equation of state ∂S ∂p = ∂V T ∂T V V = For a Dieterici gas p= RT e−a/RT Vm Vm − b R + RVam T e−a/RVm T ∂p = Vm − b ∂T Vm dS = ∂S dVm = ∂Vm T ∂p dVm ∂T Vm INSTRUCTOR’S MANUAL 78 S= Vm,f Vm,i dS = R + a RVm T e−a/RVm T Vm,f − b Vm,i − b Vm,f Vm,i S for a Dieterici gas may be greater or lesser than S for a perfect gas depending on T and the magnitudes of a and b At very high T , S is greater At very low T , S is less For a perfect gas S = R ln Solutions to problems Solutions to numerical problem P5.2 For the reaction N2 (g) + 3H2 (g) → 2NH3 (g) (a) rG −− rG −− = f G−− (NH3 , g) (500 K) = τ r G−− (Tc ) + (1 − τ ) r H −− (Tc ) = Problem 5.1, τ = T Tc 500 K × (2) × (−16.45 kJ mol−1 ) 298.15 K 500 K + 1− × (2) × (−46.11 kJ mol−1 ) 298.15 K = −55.17 + 62.43 kJ mol−1 = +7 kJ mol−1 (b) rG −− (1000 K) = 1000 K × (2) × (−16.45 kJ mol−1 ) 298.15 K 1000 K + 1− × (2) × (−46.11 kJ mol−1 ) 298.15 K = (−110.35 + 217.09) kJ mol−1 = +107 kJ mol−1 Solutions to theoretical problems P5.5 We start from the fundamental relation dU = T dS − p dV [2] But, since U = U (S, V ), we may also write dU = ∂U dS + ∂S V ∂U dV ∂V S Comparing the two expressions, we see that ∂U =T ∂S V and ∂U = −p ∂V S These relations are true in general and hence hold for the perfect gas We can demonstrate this more explicitly for the perfect gas as follows For the perfect gas at constant volume dU = CV dT THE SECOND LAW: THE MACHINERY 79 and dS = Then CV dT dqrev = T T CV dT ∂U = CV dT ∂S V T dU = dS V =T For a reversible adiabatic (constant-entropy) change in a perfect gas dU = dw = −p dV ∂U = −p ∂V S ∂p ∂T =− [Maxwell relation] ∂S V ∂V S Therefore, P5.8 = ∂S ∂T = P5.10 V ∂S ∂U − = ∂p ∂T V V ∂U ∂T ∂V ∂T ∂V ∂p ∂H = ∂p T ∂V ∂S T ∂U ∂T T ∂S ∂T V [Maxwell relation] = V ∂U ∂S p T [chain relation] = ∂S ∂V V [inversion twice] = [inversion] ∂p − ∂V ∂S ∂U T V ∂V ∂T ∂U ∂T p [chain relation] V αT κT CV V ∂U =T ∂S V ∂S + ∂p T ∂H [Relation 1, Further information 1.7] ∂p S  dH = T dS + V dp [Problem 5.6]  ∂H ∂H dS + dp (H = H (p, S)) compare dH = ∂S p ∂p S Thus, ∂H ∂S p ∂H = T, ∂S p Substitution yields, ∂H = V [dH exact] ∂p S ∂H =T ∂p T ∂S + V = −T ∂p T ∂V + V [Maxwell relation] ∂T p (a) For pV = nRT nR ∂V , = ∂T p p hence ∂H −nRT +V = = ∂p T p (b) For p = an2 nRT − [Table 1.6] V − nb V T = p(V − nb) na(V − nb) + nR RV INSTRUCTOR’S MANUAL 80 ∂T p 2na(V − nb) na = − + ∂V p nR RV RV −T + V [inversion] = p +V 2na(V −nb) na nR + RV − RV p which yields after algebraic manipulation Therefore, ∂H −T = ∂T ∂p T ∂V nb − ∂H = ∂p T 1− When b Vm 2na RT λ2 2na RT V λ2 , λ=1− nb V 1, λ ≈ and 2na 2na p 2pa 2na × ≈ × = 2 = RT V RT nRT RT V R T Therefore, nb − 2na ∂H RT ≈ ∂p T − R2pa 2T For argon, a = 1.337 L2 atm mol−2 , b = 3.20 × 10−2 L mol−1 , 2na (2) × (1.0 mol) × (1.337 L2 atm mol−2 ) = 0.11 L = RT (8.206 × 10−2 L atm K−1 mol−1 ) × (298 K) 2pa (2) × (10.0 atm) × (1.337 L2 atm mol−2 ) = = 0.045 R2 T (8.206 × 10−2 L atm K−1 mol−1 ) × (298 K) Hence, ∂H {(3.20 × 10−2 ) − (0.11)} L = −0.0817 L = −8.3 J atm−1 ≈ ∂p T − 0.045 H ≈ P5.12 ∂H ∂p T p ≈ (−8.3 J atm−1 ) × (1 atm) = −8 J ∂p − p [5.8] ∂T V RT BRT p= + [The virial expansion, Table 1.6, truncated after the term in B] Vm Vm2 πT = T ∂p R = + ∂T V Vm RT Hence, πT = Vm BR RT ∂B p RT + = + T Vm2 Vm ∂T V Vm ∂B RT B ≈ ∂T V Vm2 T ∂B ∂T V Since πT represents a (usually) small deviation from perfect gas behaviour, we may approximate Vm Vm ≈ RT p πT ≈ p2 × R B T THE SECOND LAW: THE MACHINERY From the data 81 B = ((−15.6) − (−28.0)) cm3 mol−1 = +12.4 cm3 mol−1 Hence, (1.0 atm)2 × (12.4 × 10−3 L mol−1 ) = 3.0 × 10−3 atm (8.206 × 10−2 L atm K−1 mol−1 ) × (50 K) (a) πT = (b) πT ∝ p ; so at p = 10.0 atm, πT = 0.30 atm Comment In (a) πT is 0.3 per cent of p; in (b) it is per cent Hence at these pressures the approximation for Vm is justified At 100 atm it would not be P5.13 Question How would you obtain a reliable estimate of πT for argon at 100 atm? ∂H ∂U and Cp = CV = ∂T p ∂T V (a) ∂ 2U ∂ 2U ∂CV = = = ∂V ∂T ∂T ∂V ∂V T ∂CV ∂ 2U ∂ 2U = = = ∂p∂T ∂T ∂p ∂p T ∂V ∂U ∂ = ∂T ∂V T ∂p ∂Cp Since Cp = CV + R, = ∂x T ∂ ∂T ∂ ∂T T V ∂U =0 ∂V T V ∂U ∂p T V =0 [πT = 0] (πT = 0) ∂CV for x = p or V ∂x T d2 U dCV dCV CV and Cp may depend on temperature Since = is nonzero if U depends on , dT dT dT T through a nonlinear relation See Chapter 20 for further discussion of this point However, for a perfect monatomic gas, U is a linear function of T ; hence CV is independent of T A similar argument applies to Cp (b) This equation of state is the same as that of Problem 5.12 ∂ 2U ∂CV = = ∂V T ∂T ∂V = = = P5.15 ∂πT [Part (a)] ∂T V ∂ RT ∂T Vm2 2RT Vm2 RT Vm2 ∂B [Problem 5.12] ∂T V V ∂B RT + ∂T V Vm ∂ (BT ) ∂T ∂ 2B ∂T V V ∂p − p [5.8] ∂T V nRT × e−an/RT V [Table 1.6] p= V − nb nRT nRT na nap ∂p × e−an/RT V + × × e−an/RT V = p + = T ∂T V V − nb RT V V − nb RT V πT = T INSTRUCTOR’S MANUAL 82 Hence, πT = P5.17 nap RT V πT → as p → 0, V → ∞, a → 0, and T → ∞ The fact that πT > (because a > 0) ∂U is consistent with a representing attractive contributions, since it implies that > and the ∂V T internal energy rises as the gas expands (so decreasing the average attractive interactions) ∂G dG = dp = V dp ∂p T G(pf ) − G(pi ) = pf pi V dp In order to complete the integration, V as a function of p is required ∂V = −κT V (given), ∂p T d ln V = −κ dp so Hence, the volume varies with pressure as V V0 d ln V = −κT p pi dp or V = V0 e−κT (p−pi ) (V = V0 when p = pi ) pf Hence, pi dG = V dp = V0 G(pf ) = G(pi ) + (V0 ) × pf pi e−κT (p−pi ) dp − e−κT (pf −pi ) κT = G(pi ) + (V0 ) × − e−κT p κT 1, − e−κT p ≈ − − κT p + 21 κT2 p = κT p − 21 κT2 p If κT p Hence, G = G + V0 p − κT p For the compression of copper, the change in molar Gibbs function is G m = V m p − κT p = = 63.54 g mol−1 8.93 × 106 g m−3 M p ρ × − κT p × (500) × (1.013 × 105 Pa) × − κT p = (360.4 J) × − 21 κT p If we take κT = (incompressible), κT Gm = +360 J For its actual value p = 21 × (0.8 × 10−6 atm−1 ) × (500 atm) = × 10−4 − 21 κT p = 0.9998 Hence, Gm differs from the simpler version by only parts in 104 (0.02 per cent) THE SECOND LAW: THE MACHINERY P5.19 V κS = − × 83 ∂V =− ∂p ∂p S V ∂V S The only constant-entropy changes of state for a perfect gas are reversible adiabatic changes, for which pV γ = const ∂p = ∂V S ∂ const = −γ × ∂V V γ S +1 −1 Therefore, κS = = −γp γp V V Then, const V γ +1 = −γp V Hence, pγ κS = +1 P5.21 S = S(T , p) ∂S dS = dT + ∂T p ∂S dp ∂p T ∂S ∂S dT + T dp ∂T p ∂p T T dS = T Use ∂S = ∂T p ∂S ∂H p ∂H = × Cp ∂T p T ∂H = T , Problem 5.6 ∂S p ∂S ∂V =− [Maxwell relation] ∂p T ∂T p ∂V dp = Cp dT − αT V dp ∂T p For reversible, isothermal compression, T dS = dqrev , dT = 0; hence Hence, T dS = Cp dT − T dqrev = −αT V dp qrev = pf pi −αT V dp = −αT V p [α and V assumed constant] For mercury qrev = (−1.82 × 10−4 K −1 ) × (273 K) × (1.00 × 10−4 m−3 ) × (1.0 × 108 Pa) = −0.50 kJ P5.25 When we neglect b in the van der Waals equation we have p= RT a − Vm Vm and hence Z =1− a RT Vm Then substituting into eqn 5.20 we get ln φ = p o p −a Z−1 dp = dp RT Vm p o p INSTRUCTOR’S MANUAL 84 In order to perform this integration we must eliminate the variable Vm by solving for it in terms of p Rewriting the expression for p in the form of a quadratic we have RT a Vm2 − Vm + = p p The solution is 1 Vm = (RT )2 − 4ap RT /p ± p applying the approximation (RT )2 Vm = 4ap we obtain RT RT ± p p Choosing the + sign we get RT Vm = which is the perfect volume p Then p a ap ln φ = − dp = − RT (RT )2 For ammonia a = 4.169 atm L2 mol−2 4.169 atm L2 mol−2 × 10.00 atm ln φ = − (0.08206 L atm K−1 mol−1 × 298.15 K)2 = −0.06965 f φ = 0.9237 = p f = φp = 0.9237 × 10.00 atm = 9.237 atm P5.27 The equation of state qT pVm =1+ is solved for Vm = RT Vm pVm 2q −1 Z−1 qT R = RT = = p p pVm + + 4pq R ln φ = p Z−1 p Defining, a ≡ + + ln φ = a a−1 a dp[24] = RT 2p 1+ 1+ 1/2 dp 2q p R + + 4pq R 1/2 4pq 1/2 R(a − 1) da, gives , dp = R 2q da [a = 2, when p = 0] 4pq 1/2 4pq 1/2 1 = a − − ln a = + 1+ − − ln + R R Hence, φ = 1/2 2e{(1+4pq/R) −1} + + 4pq R 1/2 4pq 1/2 so R THE SECOND LAW: THE MACHINERY 85 This function is plotted in Fig 5.1(a) when 4pq R 1, and using the approximations 1.2 –1.0 –0.1 –0.01 1.0 0.01 0.1 1.0 0.8 Figure 5.1(a) ex ≈ + x, (1 + x)1/2 ≈ + x, and (1 + x)−1 ≈ − x [x 1] pq φ ≈1+ R When φ is plotted against x = 4pq/R on a linear rather than exponential scale, the apparent curvature seen in Fig 5.1(a) is diminished and the curve seems almost linear See Fig 5.1(b) 2 –2 Figure 5.1(b) Solution to applications P5.28 wadd,max = rG −− rG [4.38] (37◦ C) = τ r G−− (Tc ) + (1 − τ ) r H −− (Tc ) = 310 K 298.15 K Problem 5.1, τ = × (−6333 kJ mol−1 ) + − 310 K 298.15 K T Tc × (−5797 kJ mol−1 ) = −6354 kJ mol−1 The difference is r G−− (37◦ C) − r G−− (Tc ) = {−6354 − (−6333)} kJ mol−1 = −21 kJ mol−1 Therefore, an additional 21 kJ mol−1 of non-expansion work may be done at the higher temperature INSTRUCTOR’S MANUAL 86 Comment As shown by Problem 5.1, increasing the temperature does not necessarily increase the maximum non-expansion work The relative magnitude of r G−− and r H −− is the determining factor P5.31 The Gibbs–Helmholtz equation is ∂ ∂T G T =− H T2 so for a small temperature change rG −− = T so d r G−− T −− r G190 = rH T2 −− r =− T H −− dT T2 −− T190 r G220 T220 + −− r G2 and T2 −− r G190 and −− rH T190 1− = = −− r G1 T1 −− r G220 T220 − −− rH T2 T + rH −− T190 − T220 T190 T220 For the monohydrate −− r G190 = (46.2 kJ mol−1 ) × −− r G190 = 57.2 kJ mol−1 190 K 220 K + (127 kJ mol−1 ) × − 190 K , 220 K 190 K 220 K + (188 kJ mol−1 ) × − 190 K , 220 K 190 K 220 K + (237 kJ mol−1 ) × − 190 K , 220 K For the dihydrate −− r G190 = (69.4 kJ mol−1 ) × −− r G190 = 85.6 kJ mol−1 For the monohydrate P5.32 −− r G190 = (93.2 kJ mol−1 ) × −− r G190 = 112.8 kJ mol−1 The change in the Helmholtz energy equals the maximum work associated with stretching the polymer Then dwmax = dA = −f dl For stretching at constant T ∂A ∂U ∂S =− +T ∂l T ∂l T ∂l T assuming that (∂U/∂l)T = (valid for rubbers) f =− f = T ∂S =T ∂l T =T − 3kB l N a2 ∂ ∂l T =− − 3kB T N a2 3kB l +C 2N a l This tensile force has the Hooke’s law form f = −kH l with kH = 3kB T /N a ... Then substituting into eqn 5.20 we get ln φ = p o p −a Z−1 dp = dp RT Vm p o p INSTRUCTOR S MANUAL 84 In order to perform this integration we must eliminate the variable Vm by solving for it in... very high T , S is greater At very low T , S is less For a perfect gas S = R ln Solutions to problems Solutions to numerical problem P5.2 For the reaction N2 (g) + 3H2 (g) → 2NH3 (g) (a) rG −−... additional 21 kJ mol−1 of non-expansion work may be done at the higher temperature INSTRUCTOR S MANUAL 86 Comment As shown by Problem 5.1, increasing the temperature does not necessarily increase the