2 Thermodynamics : Worked Examples

Transcription

1 2 hermodynamics : Worked Examples (21 ν moles of an ideal diatomic gas are driven along the cycle depicted in Fig 1 Section A is an adiabatic free expansion; section C is an isotherm at temperature A = = C ; CD is an isobar, and DA is an isochore he volume at is given by V = (1 xv A +xv C, where0 x 1 (a Find an expression for the total work W cycle in terms of ν, A, V A,V C, andx (b SupposeV A = 10L,V C = 50L, A = 500K, andν = 5 What is the volume V such that W cycle = 0? (a We have W A = W DA = 0, and W C = W CD = C D C pdv = νr A C dv V = νr Aln ( VC V pdv = p C (V D V C = νr A (1 V A V C hus, W CYC = νr A [ ] VC ln( 1+ V A V V C Figure 1: Cycle for problem 1, consisting of adiabatic free expansion (A, isotherm (C, (b Setting V = (1 xv A +xv C, and defining r V A /V C, we have isobar (CD, and isochore (DA W CYC = νr A ( ln ( x+(1 xr +1 r, and setting W CYC = 0 we obtainx = x, with x = er 1 r 1 r For V A = 10L andv C = 50L, we haver = 1 5 andx = 031, corresponding tov = 22L 1

2 (22 A strange material obeys the equation of statee(s,v,n = as 7 /V 4 N 2, whereais a dimensionful constant (a What are the SI dimensions of a? (b Find the equation of state relatingp,, andn = N/V (c Find the coefficient of thermal expansionα p = 1 V Express your answers in terms of p and ( V p and the isothermal compressibilityκ = ( 1 V V p (d ν moles of this material execute a Carnot cycle between reservoirs at temperatures 1 and 2 Find the heat Q and work W for each leg of the cycle, and find the cycle efficiencyη (a Clearly [a] = K 7 m 12 /J 2 where K are Kelvins, m are meters, and J are Joules (b We have ( E = + S ( E p = V V,N S,N = 7aS6 N 2 V 4 = 4aS7 N 2 V 5 A We must eliminate S Dividing the second of these equations by the first, we find S = 7pV/4, and substituting this into either equation, we obtain the equation of state, with c = 6 7 7/6 a 1/6 p = c ( 1/3 N 7/6, V D Figure 2: he Carnot cycle C (c aking the logarithm and then the differential of the above equation of state, we have dp p + dv 3V 7d 6 dn 3N = 0 hus, α p = 1 V ( V = 7 p,n 2, κ = 1 V ( V = 3 p,n p (d From the results of part (b, we have that ds = 0 means d(n 2 V 4 = 0, so with N constant the equation for adiabats is d(v 4 = 0 hus, for the Carnot cycle of Fig 2, we have 2 V 4 A = 1 V 4 D, 2 V 4 = 1 V 4 C We shall use this relation in due time Another relation we shall use is obtained by dividing out the S 7 factor common in the expressions for E and forp, then substituting forpusing the equation of state: E = 1 4 pv = 1 4 cn1/3 V 2/3 7/6 2

3 A: Consider the A leg of the Carnot cycle We use the equation of state along the isotherm to find V W A = dv p = 3 2 cn1/3 7/6( 2/3 2 V V 2/3 A V A SinceE depends on volume, unlike the case of the ideal gas, there is a change in energy along this leg: ( E A = E E A = 1 4 cn1/3 7/6 ( 2/3 2 V V 2/3 A Finally, the heat absorbed by the engine material during this leg is Q A = ( E A +W A = 7 4 cn1/3 7/6 ( 2/3 2 V V 2/3 A C: Next, consider the C leg ClearlyQ C = 0 since C is an adiabat hus, W C = ( E C = E E C = 1 4 cn1/3( 7/6 2 V 2/3 7/6 1 V 2/3 C ut the fact that C is an adiabat guaranteesv 2/3 C = ( 2 / 1 1/6 V 2/3, hence W C = 1 4 cn1/3 V 2/3 1/6 2 ( 2 1 CD: For thecd leg, we can apply the results froma, mutatis mutandis hus, W CD = 3 2 cn1/3 7/6 ( 2/3 2 V D V 2/3 C We now use the adiabat conditions V 2/3 C = ( 2 / 1 1/6 V 2/3 and V 2/3 D = ( 2 / 1 1/6 V 2/3 A to write W CD as We therefore have Note that bothw CD and Q CD are negative W CD = 3 2 cn1/3 1 1/6 2 ( 2/3 V A V 2/3 Q CD = 7 4 cn1/3 1 1/6 ( 2/3 2 V A V 2/3 DA: We apply the results from the C leg, mutatis mutandis, and invoke the adiabat conditions We find Q DA = 0 and W DA = 1 4 cn1/3 V 2/3 A 1/6 2 ( 2 1 For the cycle, we therefore have and thus W cyc = W A +W C +W CD +W DA = 7 4 cn1/3 1/6 2 ( 2 1 ( V 2/3 V 2/3 A η = W cyc Q A = his is the same result as for an ideal gas, as must be the case as per the Second Law of hermodynamics 3

4 (23 For each of the following situations, explain clearly and fully why it is or is not thermodynamically possible (a Energy functione(s,v,n = asvn withaconstant (b Equation of statev = an p with a constant (c A system where ( V < 0 over some range of andp p,n (d he phase diagram for a single component system depicted in Fig 3 (left panel (You only need know that a superfluid is a distinct thermodynamic phase (e he phase diagram for a single component system in Fig 3 (right panel (You only need know that CC, HCP, and FCC solids are distinct phases (f E(S,V,N = an 2 V 1 exp(s/nb with a andbconstant (g 15 Joules of heat energy are required to raise the temperature of a system by = 1 C at constant volume 10 Joules of heat energy are required to raise the temperature of the same system by = 1 F at constant pressure (h A heat engine operating between reservoirs at temperatures 1 = 400K and 2 = 600K During each cycle, the engine does work W = 300J and the entropy of the upper reservoir decreases by 200J/K Figure 3: Phase diagrams for parts (d and (e of problem 3 (a No! E(λS,λV,λN = λ 3 E(S,V,N is homogeneous of degree 3 not extensive (b No! he isothermal compressibility κ = 1 V ( V p = 1/p is negative, which violates κ > κ S > 0 (c Yes! Many systems, such as water, contract upon a temperature increase over some range of temperature (d No! his one is tricky From the Clapeyron equation, we have ( dp d = s coex v Nernst s law says that the entropy of both the solid and superfluid phases must vanish at = 0 herefore all coexistence curves which intersect the pressure axis at = 0 must do so with zero slope 4

5 (e No! he Gibbs phase rule d = 2+σ ϕ gives the dimension of thermodynamic space over which ϕ distinct phases amongσ species can coexist Forσ = 1 we haveϕ 3, sinced 0 So four phase coexistence with a single component is impossible (f Yes! E is properly extensive and convex One can derive E = pv = Nb, which is the ideal gas law with k replaced by b (d Yes! he heat capacity at constant volume is C V = ( dq d = 15J/K he heat capacity at constant pressure is V C p = ( dq d p = 10J/ 5 9 K = 18J/K Stability requiresc p > C V, which is satisfied (h Yes! he only possible obstacle here is whether the engine s efficiency is greater than that of the corresponding Carnot cycle, for which η C = 1 1 = We have η = W Q and S 2 2 = Q 2 hus, η = W/ [ 2 2 ( S 2 ] = 300J/ [ (600K(200J/K ] = 1 4 < η C 5

6 (24 Using the chain rule from multivariable calculus (see 217 of the lecture notes, solve the following: (a Find ( N/ S,p in terms of,n,s, andc p,n (b Experimentalists can measure C V,N but for many problems it is theoretically easier to work in the grand canonical ensemble, whose natural variables are(,v,µ Show that C V,N = wherez = exp(µ/k is the fugacity ( E V,z ( E z,v ( N V,z /( N z,v, (a We have ( N = (N,S,p S,p (,S,p = (N,S,p (N,,p (N,,p (,S,p = NC p,n S (b Using the chain rule, C V,N = (E,V,N (,V,N = (E,V,N (,V,z (,V,z (,V,N [ ( E ( N = = ( E V,z V,z z ( E z,v,v ( E z ( N V,z,V ( N /( N z V,z,V ] ( z N,V 6

7 (25 he entropy of a thermodynamic system S(E,V,N is given by wherer is a dimensionful constant S(E,V,N = re α V β N γ, (a Extensivity ofs imposes a condition on(α,β,γ Find this constraint (b Even with the extensivity condition satisfied, the system may violate one or more stability criteria Find the general conditions on(α, β, γ which are thermodynamically permissible (a Clearly we must have α + β + γ = 1 in order for S to be extensive (b he Hessian is Q = 2 S = α(α 1S/E2 αβs/ev αγs/en αβs/ev β(β 1S/V 2 βγs/vn X i X j αγs/en βγs/vn γ(γ 1S/N 2 As shown in the notes, ( for any 2 2submatrix of Q, obtained by eliminating a single row and its corresponding a b column, and written, we must havea < 0,c < 0, andac > b b c 2 For example, if we take the upper left 2 2 submatrix, obtained by eliminating the third row and third column ofq, we havea = α(α 1S/E 2,b = αβs/ev, andc = β(β 1S/V 2 he condition a < 0 requiresα (0,1 Similarly,b < 0 requiresβ (0,1 Finally, ac > b 2 requires α + β < 1 Since α + β + γ = 1, this last condition requires γ > 0 Obviously we must have γ < 1 as well, else either α or β would have to be negative An examination of either of the other two submatrices yields the same conclusions hus, α (0,1, β (0,1, γ (0,1 7

8 (26 Consider the equation of state, p = R2 2 a+vr, wherev = N A V/N is the molar volume andais a constant (a Find an expression for the molar energy ε(,v Assume that in the limit v, where the ideal gas law pv = R holds, that the gas is ideal with ε(v, = 1 2 fr (b Find the molar specific heat c V,N (a We fix N throughout the analysis As shown in 2112 of the lecture notes, ( E = V,N ( p p V,N Defining the molar energy ε = E/ν = N A E/N and the molar volume v = V/ν = N A V/N, we can write the above equation as ( ( [ ( lnp ] ε p = p = p 1 v v ln v Now from the equation of state, we have lnp = 2ln ln(a+vr+2lnr, hence ( lnp = 2 vr ln v a+vr Plugging this into our formula for ( ε v, we have ( ε v = ap a+vr = ar2 2 (a+vr 2 Now we integrate with respect to v at fixed, using the method of partial fractions After some grinding, we arrive at ar ε(,v = ω( (a+vr In the limit v, the second term on the RHS tends to zero his is the ideal gas limit, hence we must have ω( = 1 2fR, wheref = 3 for a monatomic gas, f = 5 for diatomic, etc hus, ε(,v = 1 ar 2fR a+vr = 1 2 fr a v + a 2 v(a+vr (b o find the molar specific heat, we compute c V,N = ( ε = 1 2 fr a 2 R v (a+vr 2 8

9 (27 A diatomic gas obeys the equation of state wherea, b, andcare constants p = R v b a v 2 + cr v 3, (a Find the adiabatic equation of state relating temperature and molar volume v (b What is the internal energy per mole, ε(,v? (c What is the Helmholtz free energy per mole, f(,v? (a Letεbe the molar internal energy andv the molar volume We have already shown (see Lecture Notes, 2112 hus, for our system, ( ε = v ( p p v ( ε = a v v 2 ε(,v = 5 2 R a v, where the first term is the result for the rarefied limit v, where the gas presumably becomes ideal Now if s = S/ν is the molar entropy (ν = N/N A is the number of moles, then ds = dε+pdv = 5 dv dv 2Rd +R +cr v b v 3 Dividing by and then integrating, we have [ s(,v = Rln 5/2 (v be c/2v2] +const hus, the equation of the adiabat is 5/2 (v be c/2v2 = const (b We have already obtained the result ε(,v = 5 2 R a v (c Fromf = ε s, wheref = F/ν is the Helmholtz free energy per mole, we have f(,v = 5 2 R a ( ( br v b v 5 2 R ln R ln + cr a b 2v 2 s 0 Here we have inserted constants with the proper dimensions in order to render our expression for f with the appropriate dimensions hus, the constant s 0 has dimensions of J/mol K, the same as the gas constant R Since c/b 2 is dimensionless, there is more than one way to do this Any resulting differences will show up in a different expression fors 0 9

10 (28 A van der Waals gas undergoes an adiabatic free expansion from initial volume V i to final volume V f he equation of state is given in 2113 of the lecture notes he number of particlesn is held constant (a If the initial temperature is i, what is the final temperature f? (b Find an expression for the change in entropy S of the gas (a his part is done for you in 2105 of the notes One finds = f i = 2a ( 1 1 fr v f v i (b Consider a two-legged thermodynamic path, consisting first of a straight leg from( i,v i to( i,v f, and second of a straight leg from( i,v f to( f,v f We then have S 1 S 2 {}}{{}}{ V f ( f ( S S S = dv + d V V i,n V i f,n i Along the first leg we use and we then find Along the second leg, we have f S 2 = d i ( S = V,N ( p = R V,N v b ( vf b S 1 = Rln v i b ( f S C Vf,N = d V f,n i f = 1 2 fr i d = 1 2 frln ( f i hus, ( [ vf b S = Rln + 12 v i b frln 1+ 2a ( 1 1 ] fr i v f v i 10

11 (29 Recall that a van der Waals gas obeys the equation of state ( p+ a v 2 (v b = R, wherev is the molar volume We showed that the energy per mole of such a gas is given by ε(,v = 1 2 fr a v, where is temperature andf is the number of degrees of freedom per particle (a For an ideal gas, the adiabatic equation of state is v f/2 = const Find the adiabatic equation of state (at fixed particle number for the van der Waals gas A (b One mole of a van der Waals gas is used as the working substance in a Carnot engine (see Fig 1 Find the molar volume at v C in terms of v, 1, 2, and constants (c Find the heatq A absorbed by the gas from the upper reservoir (d Find the work done per cycle,w cyc Hint: you only need to knowq A and the cycle efficiencyη (a We have D Figure 4: he Carnot cycle C 0 = ds = dε+pdv = 1 2 frd + ( p+ a v 2 dv = 1 R dv 2fRd + v b = 1 2 fr dln[ (v b f/2], wheres = N A S/N is the molar entropy hus, the adiabatic equation of state for the van der Waals gas is Setting b = 0, we recover the ideal gas result (b SinceC is an adiabat, we have ds = 0 (v b f/2 = const (v b f/2 2 = (v C b f/2 1 v C = b+(v b ( f/2 2 1 (c We have, from the First Law, Q A = E E A +W A ( a = ν a v +ν dv p v A v v A ( a = ν a v [ R2 +ν dv v A v v b a ] v 2, v A 11

12 hence with ν = 1 ( v b Q A = νr 2 ln v A b (d Since the cycle is reversible, we must have η = W ( cyc v b W Q cyc = νr( 2 1 ln A v A b 12

13 (210 he triple point of a single component thermodynamic system is an isolated point( t,p t in the(,p plane where there is three phase coexistence between solid, liquid, and vapor Consider three phase coexistence between a pure solid, a pure vapor, and a solution where the solute fraction is x Find the shift ( t, p t as a function of x, t, and the quantities s S,L,V andv S,L,V, ie the molar entropies and volumes of the three respective phases At the triple point, we haveµ S ( t,p t = µ L ( t,p t = µ V ( t,p t, which gives two equations for the two unknowns t and p t We write t = t 0 + andp t = p0 t + p, and we solve µ 0 L (0 t +,p0 t + p xk (0 t + = µ0 V (0 t +,p0 t + p µ 0 V( 0 t +,p 0 t + p = µ 0 S( 0 t +,p 0 t + p, where the 0 superscript indicates the value for a pure phase We now expand in the notionally small quantities and p, and we use ( ( µ S = = s, p,n N p, N A ( ( µ V = = v, p,n N p, N A wheresandv are the molar entropy and molar volume, respectively his yields the linear system, ( ( ( sv s L v L v V xr 0 = t s S s V v V v S p 0 his yields = p = (v V v S xr 0 t s V (v L v S +s L (v S v V +s S (v V v L (s V s S xr 0 t s V (v L v S +s L (v S v V +s S (v V v L Note that we do not retain terms of orderx, because we have assumedxis small, ie a weak solution 13

14 (211 A grocer starts his day with 4 boxes of pears, 5 boxes of oranges, and 6 boxes of apples Each box contains 24 fruit and is initially completely filled (a At some time, the grocer notes that exactly half the pears, a third of the oranges, and a quarter of the apples have been sold Assuming that customers take fruit from random positions in each of the boxes, find the dimensionless entropy lnw of the fruit distribution (b A clumsy customer then topples the table on which the fruit boxes rest, and all the fruit fall to the ground he customer cleans up the mess, putting all the fruit back into the boxes, but into random locations What is the entropy of the final state? (a he grocer starts with 96 pears, 120 oranges, and 144 apples y the time the grocer checks, 48 pears, 40 oranges, and 36 apples have been removed he number of ways of doing this is hus, lnw = 2163 W = ( ( ( = (b here are a total of = 360 slots for the fruit, which contain the remaining 48 pears, 120 oranges, and 108 apples he rest of the slots, which amount to = 84 in total, are empty herefore, W = and the dimensionless entropy is lnw = ! 94! 48! 120! 108! = , 14

15 (212 In a chemical reaction among σ species, ζ 1 A 1 +ζ 2 A 2 + +ζ σ A σ = 0, where A a is a chemical formula and ζ a is a stoichiometric coefficient When ζ a > 0, the corresponding A a is a product; whenζ a < 0,A a is a reactant (See 2131 of the Lecture Notes he condition for equilibrium is σ ζ a µ a = 0, whereµ a is the chemical potential of thea th species he equilibrium constant for the reaction is defined as / σ wherex a = n a b=1 n b is the fraction of species a κ(,p = σ x ζ a a, (a Working in the grand canonical ensemble, show that κ(,p = σ ( ζa k ξ a ( Note that the above expression does not involve any of the chemical potentials µ a (b Compute the equilibrium constant κ(,p for the dissociative reaction N 2 2N at = 5000K, assuming the following: the characteristic temperature of rotation and that of vibration of the N 2 molecule are Θ rot = 284K and Θ vib = 3350K he dissociation energy, including zero point contributions, is = 1693kcalmol 1 he electronic ground state of N 2 has no degeneracy, but that of the N atom is 4 due to electronic spin pλ 3 a (a In the GCE, we have Ω (,V,{µ a } = k V σ λ 3 a eµ a /k ξ a, where λ a = (2π 2 /m a k 1/2 the thermal wavelength for species a and ξ a ( is the internal coordinate partition function for species a We then have n a = 1 ( Ω = z V µ a λ 3 a ξ a, a,v,µ b a wherez a = e µ a /k OK, so we now define x a = n a σ b=1 n b = z a λ 3 a ξ a p/k = k ξ a z a pλ 3, a since b n b = Ω/Vk = p/k (RememberΩ = pv herefore κ(,p = σ x ζ a a σ ( ζa k ξ a pλ 3 a σ z ζ a a 15

16 However, since σ ζ a µ a = 0 σ z ζ a a = σ ( e ζ a µ a /k 1 = exp k σ ζ a µ a = 1, (b he internal partition function for N is justξ N = (2S+1(2I +1, wheres = 3 2 is the total electronic spin from Hund s rules, and I = 1 is the nuclear spin It turns out that we will never need to know the value of I For N 2 the internal partition function is ξ N2 = (2I +1 2 e / 2Θ rot 1 e Θ vib / his formula requires some explanation We appeal to Eqs 4292 in the Lecture Notes Since Θ rot, we have ζ g ζ u due uθ rot / = 2Θ rot, where the factor of 1 2 comes from summing only over half the allowedlvalues, ie either all even or all odd, and where u = L(L + 1 so du = (2L + 1dL We then have ξ rot = (2I /2Θ rot because g g + g u = (2I he vibrational partition function was derived to be ξ vib = 1 2 csch(θ vib /2, however since we are including the zero point vibrational energy 1 2 ω vib = 1 2 k Θ vib in the dissociation energy, we get the above expression for ξ N 2 According to our result from part (a, we have κ(,p = 32k Θ rot e / (1 e Θ vib / λ3 N 2 pλ 6 N = 16 2 kθ rot pλ 3 N Now we need to evaluate some quantities he gas constant is hence at = 5000 K, we have e / (1 e Θ vib / R = N A k = 8314J/mol K = kcal/mol K, k = (1693kcal/mol(4184J/kcal = 170 (8314 J/mol K(5000 K Furthermore,Θ vib / = 0670 he thermal wavelength of N at this temperature is found to be We also have ( 2π ( gcm 2 /s 2 1/2 λ N = (14g/ ( = 660Å erg/k 5000K k Θ rot pλ 3 N = ( erg/k (284K ( g/cm s 2 ( cm 3 p0 p = 135p 0, p wherep 0 = Pa is atmospheric pressure Putting it all together, we obtain κ( = 5000K,p = p0 p 16

17 (213 he phase diagram for a binary eutectic system is depicted in Fig 5 he liquid phase is completely miscible, but the solid phase separates into A-rich α and -rich β phases over a broad range of temperatures and compositions here is a single chemical composition which solidifies at a temperature lower than any other for this system - the eutectic composition You are invited to model such a system using the Gibbs free energy densities [ ] g L (,p,x = (1 xµ A L(,p+xµ L(,p+k xlnx+(1 xln(1 x +λ L x(1 x ] g S (,p,x = (1 xµ A S (,p+xµ S [xlnx+(1 xln(1 x (,p+k +λ S x(1 x, whereλ L < 0 and λ S > 0 For simplicity, you may assume µ A L(,p µ L(,p µ L (,p µ A S(,p µ S(,p µ S (,p, with µ S (,p = µ L (,p+rk ( 0, wherer > 0 (a y sketching the free energies, show that the phase diagram is as shown in Fig 5 (b Solve numerically for the eutectic temperature assuming λ L = 1,λ S = +1, and k 0 = 1, andr = 08 Figure 5: Eutectic phase diagram (from Wikipedia L denotes the liquid phase, and α and β are two solid phases (a A set of curves illustrating the phenomenon is shown in Fig 6 We have taken the valus in part (b of the problem and varied the quantityk (in dimensionless units For our system, both the liquid and solid free energies are symmetric in x about the point x = 1 2 At high temperatures, g L < g S for all x, as shown in the upper left panel of Fig 6 As the temperature is lowered, g S starts to dip below g L at the endpoints x = 0,1 For our model and parameters, this happens for k = k 0 = 1 ecause λ L > λ S, the curvature of g L (x is greater than that of g S (x, which means that initially there will be two intersections whereg L (x = g S (x, at x = x < 1 2 andx = 1 x > 1 2 o guarantee thermodynamics stability, one must invoke the Maxwell construction which connects the solid curve at some point x 1 < x to the liquid curve at point x 2 > x, withx 2 < 1 2 A similar construction follows on the second half of the curve, between g L (1 x 2 and g S (1 x 1 hese two phase regions represent mixtures of the liquid at intermediate concentration and a low or high concentration solid phase Furthering Figure 6: Gibbs free energies for liquid (blue lower the temperature, the solid curve develops a negative curvature at x = 1 with Maxwell constructions shown and solid (red phases at different temperatures, 2 for k < 1 2 λ S Eventually, the temperature gets so low that g S (x lies below g L (x for all x [0,1] he system is then in the solid phase, but one must nevertheless invoke a Maxwell construction, as shown in the lower left panel in Fig 6, between a low-concentration solid at x = x 3 < 1 2 and a high-concentration solid atx = 1 x 3 > 1 2 At such temperatures, the solid is in a homogeneous phase for x < x 3 orx > 1 x 3, and in a mixed phase for x 3 < x < 1 x 3 (b A crude numerical experiment is performed by successively lowering k until the minima of the g L (x andg S (x curves cross, and then iterating to find the temperature where the minima coincide In this manner, I find a eutectic temperaturek e = 03948, as shown in Fig 7 17 Figure 7: Gibbs free energies for the liquid (blue and solid (red phases at the eutectic temperature