CHEM-UA 652: Thermodynamics and Kinetics Professor M.E. Tuckerman Midterm Examination April 2, 203 NAME and ID NUMBER: There should be 8 pages to this exam, counting this cover sheet. Please check this exam NOW! There are 5 pages of formulae/data, and a periodic table (with molar masses) at the back of this exam. Books, notes, etc. are not permitted, however calculators are. Partial credit will be given but only if your answers can be deciphered. Therefore, make sure that your writing is neat and that your logic is clear, organized, and easy to follow. Unreadable answers will not be given the benefit of the doubt. Choose any 3 of the four questions, and supply your answers in the examination booklets provided. Solve all 4 problems for extra credit, but be sure to indicate clearly which one you wish to have considered as your extra credit problem. Good luck. GRADING. (33 points) 2. (33 points) 3. (33 points) 4. (33 points) TOTAL: 99 points Extra Credit (33 possible points):
2. (33 points) The van der Waals equation of state is and the energy of a van der Waals gas or liquid is P = nrt V nb an2 V 2 E = 3 an2 nrt 2 V Up to irrelevant additive constants, what is its Helmholtz free energy function A(n, V, T), assuming that the particles are indistinguishable? 2. (33 points) Near the liquid-gas critical point, it is observed experimentally that the equation of state behaves as P P c (ρ ρ c ) δ where δ is anotherexample of a critical exponent. Here ρ c = n/v c is the critical density, V c is the critical volume, and P c is the critical pressure. This behavior is observed when T = T c, where T c is the critical temperatuure. In this problem, you will use the van der Waals equation to predict a value for the exponent δ. Recall that in terms of density, the van der Waals equation is P = ρrt ρb aρ2 Given this, the value of δ can be obtained by performing a Taylor expansion of this equation about ρ = ρ c = (/3b) when T = T c = 8a/27Rb. This expansion takes the form P P c = k P k! ρ k (ρ ρ c ) k ρ=ρc,t=t c k= and the first nonzero term in the expansion gives the value of δ. Using this procedure, what value for δ does the van der Waals equation yield? 3. (33 points) Consider two thermodynamic states of a system, denoted A and B, characterized by energy levels E (A) j (N,V) and E (B) j (N,V). One way to drive the system from state A to B is to do so continuously by introducing a parameter λ [0,] and a set of energy levels where f(λ) and g(λ) are two arbitrary functions. E j (λ,n,v) = f(λ)e (A) j (N,V)+g(λ)E (B) j (N,V) a. (4 points) As λ is continuously varied from 0 to, the system continuously transforms from state A, with energy levels E (A) j (N,V) at λ = 0, to B, with energy levels E (B) j (N,V) at λ =. Given this requirement, what conditions must the functions f(λ) and g(λ) satisfy at the endpoints, λ = 0 and λ =? b. (6 points) Write down one possible choice for f(λ) and g(λ) that obey the restrictions you obtained in part a. c. (5 points) Define a λ-dependent partition function Q(N,V,T,λ) = j e βej(λ,n,v)
3 and a λ-dependent Helmholtz free energy A(N,V,T,λ) = k B T lnq(n,v,t,λ) Show that the free energy difference beteween states A and B can be computed from E A AB = dλ 0 λ λ where E λ λ Q(N,V,T,λ) ( Ej (N,V,λ) λ j ) e βej(n,v,λ) Hint: Try starting from the trivial relation from the fundamental theorem of calculus: A AB = 0 A λ dλ d. (8 points) Now consider an ideal gas of N particles at temperature T. Let n,...,n N denote the N quantum numbers. Suppose the system undergoesan expansionfromavolume V A to volumev B at fixedt. Using the functions f(λ) and g(λ) you devised in part b, write down the corresponding expression for the λ-dependent energy levels E n n N (N,V,λ) you would use to carry out the procedure in part c. 4. (33 points) A monatomic ideal gas is taken around the following thermodynamic cycle: i. An isothermal expansion from volume V A to volume V B at temperature T h. ii. An isochoric cooling from temperature T h to temperture T l. iii. An isothermal compression from volume V B to volume V A at temperature T l. iv. an isochoric warming from temperature T l to temperature T h, which returns the system to its initial state. a. ( points) Sketch the cycle in both the P-V and T-S planes, and give the equations that determine curves for each segment of the cycle in both the P-V and T-S planes. b. (5 points) Calculate the energy and entropy change in each step of the cycle, and show that the net change in energy and the net change in entropy around the cycle are both zero. c. (7 points) The partition function for a monatomic ideal gas is Q(N,V,T) = N! ( ) N V λ 3 where λ = βh 2 /2πm. Calculate the Helmholtz free energy change for each step of the cycle, show that the net change in the Helmholtz free energy is zero around the cycle, and find the net work output of the cycle.
4 POSSIBLY USEFUL INFORMATION N 0 = 6.022 0 23 mol R = 8.344J mol K = 0.082057L atm mol K k B =.38066 0 23 J mol K =.36262 0 25 L atm mol K h = 6.62609 0 34 J s, h = h 2π =.05457 0 34 J s Some conversion factors Å = 0 8 cm = 0 0 m L = 0 3 m 3 = 0 3 cm 3 J = kg m 2 s 2 atm =.0325 0 5 Pa = 760 mm Hg at 0 o C = 760 torr Pa = kg m s 2 L atm = 0.325J Formulas PV = nrt, γ = c P c V, c P = c V +R, E = 3 2 nrt = 3 2 Nk BT Ω(N,V,E) = N! ( ) 3N/2 me 3Nπ h 2 V N, n = N, N 0 S(N,V,E) = k B lnω(n,v,e), S AB = B A dq rev T R = N 0 k B = k B ln ( ΩB Ω A ) E = Q+W
5 A = E TS = µn PV, G = A+PV = µn = H TS At a critical point : P V = 0, 2 P V 2 = 0 Q(N,V,T) = j e βej(n,v) E = β lnq(n,v,t), ( ) lnq(n,v,t) P = k BT, V A(N,V,T) = k B T lnq(n,v,t), P = ( ) A V N,T S = ( ) A T N,V = k B lnq+ E T A = W rev (isothermal) (N,P,T) = 0 e βpv Q(N,V,T)dV H = β ln (N,P,T), ( ) ln (N,P,T) V = k BT, P G(N,P,T) = k B T ln (N,P,T), V = S = ( ) G = k B ln + H T N,P T ( ) G P N,T Ĥψ = Eψ, E mc 2 E n = E n = h2 π 2 2mL 2n2, E n = h2 π 2 2mV 2/3 n 2 ( n+ ) ( hω, E n = n x +n y +n z + 3 ) hω 2 2 E JM = h2 J(J +) 2I F i = m i a i = m i d 2 r i dt 2
6 Mathematics Vectors: a θ b a = (a x,a y,a z ) a = a 2 x +a2 y +a2 z a b = a x b x +a y b y +a z b z a+b = (a x +b x,a y +b y,a z +b z ) a+b = a 2 + b 2 +2a b [ ] a b θ = cos a b Identities lnx+lny = ln(xy), e x+y = e x e y, e x = e x lnx lny = ln ( ) x y Sums and integrals: r n = n=0 a m b n = n,m f(x) = r, 0 < r < ( )( ) a m b n n k=0 m dx x = lnx+c, k! f(k) (x 0 )(x x 0 ) k, (Taylor series expansion about x = x 0 ) x n dx = xn+ n+ +C Fundamental Theorem of Calculus : b a f(x)dx = F(b) F(a) df dx = f(x)
7 Euler s Theorem: If f(λx,...,λx k,x k+,...,x M ) = λ n f(x,...,x M ) k f Then nf(x,...,x M ) = x i x i i= Legendre transform: Given f(x) and y = f (x) = g(x), b(y) = f(x(y)) x(y)f (x(y))
FIG. : Periodic table. 8