Physics 333, Thermal and Statistical Physics: Homework #2 Solutions Manual

Size: px

Start display at page:

Download "Physics 333, Thermal and Statistical Physics: Homework #2 Solutions Manual"

Patricia Lee
6 years ago
Views:

1 Physics 333, Thermal and Statistical Physics: Homework #2 Solutions Manual 1. n 5 = 0 n 5 = 1 n 5 = 2 n 5 = 3 n 5 = 4 n 5 = 5 d n 5,0,0,0, ,1,0,0, ,2,0,0, ,1,1,0, ,2,1,0, ,1,1,1, ,1,1,1, Column Sum Total = 126 Column Probability Total = 1 1 ln P n 5 The relative probability of a certain state for the Boltzmann distribution is given by P (E n ) e βen, where β 1/kT. As we are plotting ln P, we d expect the points to line in a straight line with a slope equal to βe 0. Using the slope of the best-fit line, we see βe Substituting β 1/kT, we can find T = 1.28E 0 /k. 1

2 2. Problem 6.9 of Schroeder (a) Take the ground-state energy (E 1 ) to be zero, so that the excited-state energies are E 2 = 10.2 ev, E 3 = 12.1 ev, and so on up to E = 13.6 ev. Also note that kt at T = 5800 K is 0.50 ev. Including only the first three levels, the partition function would be Z = e E 1/kT + 4e E 2/kT + 9e E 3/kT = e 0 + 4e e 24.2 = 1 + ( ) + ( ) = (b) Since there are n 2 states with energy E n, the full partition function is Z = n 2 e En/kT (This sum includes only bound states, not ionized states; let s say that it s not an atom if its ionized.) All of the E n s are less than E 1 = 13.6 ev, so we can safely conclude that Z > n 2 e E /kt = e E /kt n 2 = Even though the Boltzmann factor e E /kt might be tiny (for example, at T = 5800 K it is only ), the sum diverges because there are infinitely many states with Boltzmann factors at least this large. (c) If we keep the P dv term in Equation 6.3, then the Boltzmann factor acquires an additional P V term in the exponent: New Boltzmann Factor = e (E+P V )/kt, where V is the volume of the system and P is the pressure of the reservoir. For a hydrogen atom in its ground state, V (1Å) 3, so at atmospheric pressure, P V (10 5 Pa)(10 30 m 3 ) = J 10 6 ev. For the n = 10 state, however, the radius is 100 times larger so the volume is a million times larger, that is, P V 1 ev at atmospheric pressure. (I dont know off-hand what the pressure is at the sun s surface, where the density of atoms is probably lower than in our atmosphere but the temperature is much higher. At any reasonable pressure, though, there will be some not-too-large n at which P V 1 ev.) At 5800 K, a P V term of 1 ev reduces the Boltzmann factor by a factor of e 2 = 0.14, so this term is not negligible. And as n grows, the P V terms cause the Boltzmann factors to decrease exponentially. Therefore, the correct partition function will be dominated by the first few energy levels. At the sun s temperature, this means that Z will be very close to 1, as shown in part (a). 2

3 3. (a) The exact partition function is given by Z = j (2j + 1)e j(j+1)ɛβ, but can be approximated as Z = 1/βɛ when βɛ << 1. Using this approximation at room temperature, Z = 110. (b) The average energy can be calculated using Ē = 1 Z Z β = (βɛ) 1 β βɛ = 1 β = kt. At room temperature, this evaluates to Ē = ev. This is in agreement with the equipartition theorem, as there are two degrees of freedom present in a rotating diatomic molecule as shown in the following figure from Chapter 1.3 of Schroeder. Each of these degrees of freedom contributes 1 kt for a total of kt. 2 (c) The rotational quantum number j corresponding to this most likely energy can be found using E j = j(j + 1)ɛ with j Z +. Since j must be an integer, we see j =

4 4. Problem 6.39 of Schroeder (a) For a nitrogen molecule at 1000 K, the most probable speed is 2kT 2RT v max = m = M = 2(8.315 J K 1 )(1000 K) = 771 m s kg (Here M is the mass of a mole of nitrogen molecules.) The escape speed, m s 1, exceeds v max by a factor of /771 = Following the example in the text on pages , we can simply plug in this number for x min in the integral 4 x 2 e x2 dx π x min to obtain the probability of a molecule moving faster than escape speed. I used Mathematica to do the integral numerically and got the answer (Alternatively, you could use the asymptotic expansion derived in Problem B.5.) The age of the earth is of order seconds, so even if a molecule has a trillion (10 12 ) chances to escape per second over the lifetime of the earth, its chance of escaping by now would be less than one in (b) For a hydrogen molecule, M is only kg, so v max is larger by a factor of 14 = 3.7, that is, v max = 2880 m s 1. Escape speed exceeds this number by a factor of /2880 = 3.81 = x min. Evaluating the integral numerically once again, I find that the probability of a hydrogen molecule moving faster than escape speed is This is small but not at all negligible. Over the lifetime of the earth we would expect almost every hydrogen molecule to reach the uppermost atmosphere, achieve this speed, and hence escape. For helium, the mass is twice as great as for hydrogen so v max = 2030 m s 1, x min = 5.39, and the probability evaluates to Again, this is large enough that every atom should have had plenty of chances to escape by now. (c) Consider a nitrogen molecule in the moons (former) atmosphere. Assuming a temperature of 1000 K as in earths upper atmosphere, the most probable speed is again 771 m s 1, but the escape speed of 2400 m s 1 exceeds this by a factor of only 3.1, and, therefore, our integral for the probability evaluates to Nitrogen on the moon should, therefore, escape even faster than hydrogen on earth. Presumably this happened long long ago, leaving the moon with no atmosphere today. 4

5 5. As shown in Section 6.2 of Schroeder, the rotational energy of a diatomic molecule at room temperature is kt, corresponding to two degrees of freedom. Therefore the total thermal energy of 0.5 moles of O 2 is U = 3 2 NkT + NkT = 5 2 NkT = 5 2 nrt = R(300 K) = J Enthalpy is then given by H = U + P V = U + nrt = ( ) 10 3 J = J To compute the remaining quantities we need the internal partition function, which in this case is purely rotational. However, the the electronic ground state for O 2 is threefold-degenerate (as stated in Section 6.7 of Schroeder), which contributes a factor of 3 to the partition function: Z int = 3Z rot = 3 kt 2ɛ = ( ev/k)(300 K) = ( ev) We also need the quantum volume, ( ) 3 ( h J s v Q = = 2πmkT 2π(16)( kg)( J/K)(300 K) = m 3 and the average volume per particle, V N = kt P = ( J/K)(300 K) N/m 2 = m 3. From these numbers, we can compute the logarithm ( ) ( ) V Z int ( m 3 )(216) ln = ln = Nv Q m 3 The Helmholtz free energy is therefore [ ( ) ] V Z int F = nrt ln + 1 = ( J)( ) = J. Nv Q The easiest way to get the entropy is from the definition F = U T S: S = U F T = ( J) ( J) 300K = 19.0 J/K. And the easiest way to get the chemical potential is from G = F + nrt = Nµ: µ = F + nrt N = ( J) + ( J) (0.5)( ) ) 3 = J = ev 5

Part II: Statistical Physics

Part II: Statistical Physics Chapter 6: Boltzmann Statistics SDSMT, Physics Fall Semester: Oct. - Dec., 2014 1 Introduction: Very brief 2 Boltzmann Factor Isolated System and System of Interest Boltzmann Factor The Partition Function