Statistical MolecularThermodynamics University of Minnesota Homework Week 3 1. You are given the following partition function for a fictitious gas \Vikonium". What is the correct equation of state for Vikonium? In the equation A and B are constants, the other variables have their usual meanings. 2 3N=2 A 2m β 2 Q(N; V; T ) = (V − 2NB)N eβN =V N! 3πh2 (a) PV = NkBT A2N 2 N (b) P + V 2 V −2NB = NkBT m2N 2 A (c) P − V 2 (V − 2NB) = N! kBT A2N 2 (d) P + V 2 (V − 2NB) = kBT N 2 (e) P − V (V − 2NB) = NkBβT N 2 N (f) P + V 2 V = NBkBT N 2 1 N (g) P − V 2 V = NkBT N 2 1 (h) 2B P − V 2 V −2N = 3NkBT N 2 (i) P + (V − 2NB) = Nk T V 2 B Answer: Since we want the equation of state, calculate the pressure in terms of volume and @ ln Q temperature using P = kBT @V N;T . Start by separating ln Q into pieces that include V and those that do not, βN 2 ln Q = ln [terms no V ] + N ln (V − 2NB) + V @ ln Q N βN 2 P = kBT = kBT 0 + − 2 @V N;T V − 2NB V Nk T N 2 N 2 P = B − ; P + (V − 2NB) = Nk T V − 2NB V 2 V 2 B 2. In an Einstein crystal, one principal assumption is that each of the N atoms of the crystal vibrate independently about their lattice positions. The crystal is thus pictured in three dimensions as N independent three dimensional harmonic oscillators. Using the partition function for a harmonic oscillator, 1 1 X −β(j+ 1 )hν −βhν=2 X −βjhν qho(T ) = e 2 = e e ; j=0 j=0 determine the partition function for an Einstein crystal. It will be helpful to make use of the geometric series, 1 X 1 xj = 1 − x j=0 to derive this result. 3 e−βhν=2 (a) Q = 1−e−βhν 3N e−βhν=2 (b) Q = 1−e−βhν e−βhν=2 (c) Q = 1−e−βhν 3N 1 (d) Q = 1−e−βhν 3N e−βhν=2 (e) Q = 1+e−βhν 3 eβhν=2 (f) Q = 1+e−βhν Answer: The partition function for a one-dimensional harmonic oscillator is given as 1 1 X −β(j+ 1 )hν −βhν=2 X −βjhν qho(T ) = e 2 = e e j=0 j=0 If we let x = e−βhν we can simplify the partition function to 1 1 −βhν=2 X −βjhν −βhν=2 X j qho(T ) = e e = e x j=0 j=0 Note that this summation is the geometric series, 1 X 1 xj = : 1 − x j=0 We can then write the partition function as 1 X 1 q (T ) = e−βhν=2 xj = e−βhν=2 ho 1 − x j=0 e−βhν=2 q (T ) = ho 1 − e−βhν To find the partition function for a three-dimensional harmonic oscillator, think of the three-dimensional harmonic oscillator as three independent one-dimensional harmonic oscillators. The partition function for a three-dimensional harmonic oscillator can then be written as e−βhν=2 3 q (T ) = [q (T )]3 = 3D ho 1 − e−βhν Each particle in the system is distinguishable. Therefore, the partition function for a system of N particles can be written as N Q = [q3D(T )] Substituting the partition function for the three-dimensional oscillator gives: e−βhν=2 3N Q = 1 − e−βhν q(V;T )N 3. One of the requirements for the statistical result that Q(N; V; T ) = N! is that the system have many more available states than particles. Usually, the number of translational states alone are enough to satisfy this condition. Use the translational partition function to determine which of the following conditions leads to an increased number of accessible translational states: (a) small molar volume (b) low temperature (c) small mass (d) all of the above (e) none of the above Answer: In lecture video 3.6 it was discussed that the condition under which the number of available states exceeds the number of particles is N h2 3=2 1 V 8mkBT Large molar volumes, high temperatures, and large masses favor this condition, so the answer is none of the above. 4. The average energy for a monatomic van der Waals gas is 3 aN 2 hE¯i = N k T − A 2 A B V Use the definition of the constant volume molar heat capacity discussed in lecture video 3.4 to determine a formula for the constant volume molar heat capacity of a monatomic van der Waals gas. Here, NA is Avogadro's number and R is the universal gas constant. ¯ 3 (a) CV = 2 NAkB aN 2 ¯ 3 A (b) CV = 2 NAkBT − V ¯ 3 2 (c) CV = 2 NAkB − aNA aN 2 ¯ 3 A (d) CV = 2R − V ¯ 3 (e) CV = 2 R ¯ 3 2 (f) CV = 4 RT aN 2 T ¯ 3 2 A (g) CV = 4 NAkBT − V (h) a) and e) (i) b) and g) (j) none of the above Answer: The constant-volume molar heat capacity is defined in lecture video 3.4 as ¯ ¯ ¯ @U @hEi CV = = @T N;V @T N;V Insert the average energy that is given 3 aN 2 hE¯i = N k T − A 2 A B V into the expression for the heat capacity: ¯ 2 ¯ @hEi @ 3 aNA CV = = NAkBT − @T N;V @T 2 V Differentiating with respect to temperature gives the heat capacity for a monatomic van der Waals gas as 3 C¯ = N k V 2 A B Substituting R = NAkB, we can also see that 3 C¯ = R V 2 This heat capacity is the same as that for a monatomic ideal gas. 5. In the lecture videos, a general formula for the average energy, 2 @ ln Q hEi = kBT ; @T N;V was derived. Determine the average energy, hEi, for a monatomic ideal gas given the partition function for this gas, 1 2πmk T 3N=2 Q(N; V; T ) = B V N : N! h2 2 N (a) hEi = (3=4)NkBT + V (b) hEi = (3=2)NkBT 2 (c) hEi = (3=4)NkBT N (d) hEi = (3=2)NkBT + V 3N h2 N (e) hEi = (3=4)NkBT + + 2 2πmKB V 3N (f) hEi = 2T Answer: First, rearrange this equation, 1 2πmk 3N=2 Q(N; V; T ) = T 3N=2 B V N : N! h2 Take the natural log of both sides of the partition function and separate the terms to obtain 3N 3N 2πmk ln Q(N; V; T ) = −ln N! + ln T + ln B + NlnV 2 2 h2 Note that only the second term in the above equation depends on T . Therefore, differentiating with respect to T produces the following: @ ln Q 3N = @T N;V 2T Insert this result into the equation for hEi given above: 2 @ ln Q 2 3N hEi = kBT = kBT @T N;V 2T 3 hEi = Nk T 2 B 6. We found that under certain conditions, the partition function for a collection of par- ticles, Q(N; V; T ), could be written in terms of the partition function for individual particles, q(V; T ), and the number of particles, N, q(V; T )N Q(N; V; T ) = : N! This equation is correct for a gas described by the van der Waals equation of state. (a) TRUE (b) FALSE Answer: We actually accept either answer as correct. The formula given is for a partition func- tion involving particles that are indistinguishable, and all of which have identical and independent energy levels, because they do not interact with one another. In a van der Waals gas, the particles are indistinguishable, but the particles do interact with one another (as measured by the a and b parameters), so each particle's energy depends upon its interactions with all other particles, making the given partition function equa- tion inappropriate. However, note that the van der Waals gas partition function, Q, given in problem 1 of this homework can be rearranged so as to be written in the form q(V; T )N Q(V; T ) = : N! While not derived here, this comes from assuming a uniform distribution of particles which generates a "mean-field” potential energy with which each particle interacts "independently", in which case use of the non-interacting ensemble partition function can be viewed as justified. Clearly, the assumptions are not really internally consistent, since you can't both interact and not interact simultaneously! But, that is how the van der Waals equation of state is derived. 7. For a system that has equally spaced non-degenerate energy levels, at a fixed tempera- ture what will happen to the value of the partition function, Q(N; V; T ), if the spacing between the energy levels INCREASES? (a) Q will increase (b) Q will decrease (c) Q will remain the same (d) either (a) or (b) depending on the temperature (e) none of the above Answer: See the plots in lecture video 3.2. If the temperature is fixed, and the spacing between the non-degenerate energy levels increases, then there are fewer states available to the system at that given temperature. Therefore, the partition function will decrease. 8. The following expresses the partition function of the system in terms of the partition functions of the individual particles (such as atoms or molecules), qN Q = : N! For this to be true, it must be the case that there are many more particles than available states.