Solutions to Problem Set 3 in Statistical Physics, Study notes of Thermodynamics

Download Solutions to Problem Set 3 in Statistical Physics and more Study notes Thermodynamics in PDF only on Docsity! Cornell University, Physics Department Fall 2014 PHYS-3341 Statistical Physics Prof. Itai Cohen Solutions to Problem Set 3 David C. Tsang, Woosong Choi 3.1 Boundary Conditions (a) We have the energy of N point particles with periodic Boundary Conditions (BCs) given by E = 2~2π2 mL3 N ∑ i=1 (n2 xi + n2 yi + n2 zi ). Now nxi , nyi, nzi ∈ Z, therefore the number of states with an energy less than E is given by the “volume” of a 3N dimensional sphere. R = √ EmL2 2~2π2 thus we have the number of states with energy less than E, Φ(E) Φ(E) = V3N(R) = π3N/2R3N Γ ( 3N 2 + 1 ) where we’ve used the volume of the 3N -dimensional sphere derived in class. Using Stirling’s approximation we see: Φ(E) ≈ π3N/2 ( EmL2 2~2π2 )3N/2 ( 3N 2 ) −3N/2 e3N/2 √ 2π 3N 2 = √ 3πN ( EmL2e 3~2πN )3N/2 We want the number of states between energy E and E + δE. This should be the volume of the shell between the corresponding radii, ie the hypersurface area times the width of the shell: ΩP (E) = dΦ dR δR = dΦ dE δE = 3N 2 √ 3πN ( EmL2e 3~2πN )3N/2−1 mL2e 3~2πN δE 1 which gives ΩP (E) = 3N 2E √ 3πN ( EmL2e 3~2πN )3N/2 (1) The number of states for the Dirichlet Boundary Conditions can be inferred from comparing the energies of the different systems. We see that ED = 4EP ⇒ RD = 2RP (2) Noting the factor of 2−3N that limits the volume considered to the positive “quadrant” of the 3Nsphere, we see ΩD(E) = 2−3NΩP (E) ( RD RP )3N = ΩP (E) (3) Thus the number of states with the periodic boundary conditions is the same as the number of states with Dirichlet boundary conditions. (b) The Neumann boundary conditions require that the derivative of the wavefunction be zero at the boundaries. This yields a cosine like solution (Dirichlet boundary conditions lead to a sine like solution). Both the Dirichlet and the Neumann solutions have the same wavelength, and thus the same energies in corresponding states. Thus the number of states for the Neumann boundary conditions must be the same as the number for the Dirichlet boundary conditions. 0 L 0 L Dirichlet BCs Neumann BCs Figure 1: The solution to the Dirichlet boundary conditions have the same wave- lengths and energies as the solutions to the Neumann boundary conditions, thus there will be the same number of states with energy less than E in both cases. ✷ 2 β must be equal in the two systems ∂ ln Ω ∂E = ∂ lnΩ′ ∂E ′ ∂ ∂E −E2 2µ2H2N = ∂ ∂E ′ −E ′2 2µ′2H2N ′ Ẽ µ2N = Ẽ ′ µ′2N ′ ⇒ Ẽ = Ẽ ′ µ2N µ′2N ′ (b) Noting that the total energy is conserved we have Ẽ = Etot − Ẽ ′ Ẽ = H(bNµ + b′N ′µ′)− Ẽ µ′2N ′ µ2N ⇒ Ẽ = H(bNµ + b′N ′ u′) 1 + µ′2N ′/µ2N Hence Ẽ = µ2NH(bNµ + b′N ′µ′) µ2N + µ′2N ′ (5) (c) Since the system is siolated and only in thermal contact with each other all the energy transferred must be heat, therefore Q = ∆E = H(bNµ + b′N ′ u′) 1 + µ′2N ′/µ2N − bNµH Q = NN ′H(b′µ′µ2 − bµ′2µ) µ2N + µ′2N ′ (6) (d) We have Ω(E) ∝ δE exp[− E2 2µ2H2N ] Ω′(E ′) ∝ exp[− E ′2 2µ′2H2N ′ ] The probability for system A to be in a state with energy between E and E + dE is PA(E) ∝ Ω(E)Ω′(E(o)− E) = δE exp[− E2 2µ2H2N − (E(o)−E)2 2µ′2H2N ′ ] = δE exp[−µ′N ′E2 + µ2N(E(o)−E)2 2µ2µ′2H2NN ′ ] = δE exp[−E2(µ′2N ′ + µ2N)− 2ẼE(µ′2N ′ + µ2N) 2µ2µ′2H2NN ′ ] exp[− µ2N(E(o))2 2µ2µ′2H2NN ′ ] 5 Since E(o) and Ẽ are constants we can mulitply by exponentials of them and still maintain proportionality, thus dropping the (E(o))2 term and completin the square with Ẽ: PA(E) ∝ δE exp[−(E2 − 2ẼE − E2)(µ2N + µ′2N ′) 2µ2µ′2NN ′H2 ] Normalizing we see: PA(E) = δE√ 2πσ e−(E−Ẽ)2/2σ2 where σ2 ≡ µ2µ′2H2NN ′ µ2N+µ′2N ′ . (e) As we can see in the formulation above, the probability PA(E) is a gaussian, with variance σ2 as defined above. (f) Using the results from above we see ∣ ∣ ∣ ∣ ∆ ∗ E Ẽ ∣ ∣ ∣ ∣ = µµ′H ( NN ′ µ2N+µ′2N ′ )1/2 µ2NH(bNµ+b′N ′µ′ µ2N+µ′2N ′ = µµ′H(NN ′)1/2 µ′2NH [bNµ + b′N ′µ′] (µ2N + µ′2N ′)1/2 ≈ µ′ µb′N1/2 ✷ 6 3.4 Mixture of Ideal Gases Reif §3.5: A system consists of N1 molecules of type 1 and N2 molecules of type 2 confined within a box of volume V . The molecules are supposed to interact very weakly so that they constitute an ideal gas mixture. (a) How doe the total number of states Ω(E) in the range between E and E + δE depend on the volume V of this system? You may treat the problem classically. (b) Use this result to find the equation of state of this system, i.e., to find its mean pressure p̄ as a function of V and T . (a) According to Reif §2.5.14, the total number of states has the following dependence on V: Ω(E) ∝ V N1+N2 (b) We have p̄ = 1 β ∂ ln Ω ∂V = 1 β ∂ ∂V [(N1 +N2) lnV ] = 1 β N1 +N2 V (7) Hence we have the equation of state: p̄ = N1 +N2 V kT ✷ 7