Statistical MolecularThermodynamics University of Minnesota

Homework Week 4

1. In water there are 3 vibrational degrees of freedom with vibrational temperatures of Θvib,1 = 2290 K, Θvib,2 = 5160 K, and Θvib,3 = 5360 K. At a temperature of 3000 K, which vibrational degree of freedom will contribute the least to the molar energy, U¯?

(a) Θvib,1 = 2290 K

(b) Θvib,2 = 5160 K

(c) Θvib,3 = 5360 K (d) They will all contribute equally.

Answer:

For a polyatomic molecule, the vibrational energy is the sum of the contributions from each normal mode,

  −Θ vib/T ¯ X Θvib Θ vibe Uvib(T ) = R + (1) 2 (1 − e−Θ vib/T ) normal modes The ”normal modes” are just the result of a change in the variables we use to describe the internal degrees of freedom (i.e., other than translations and rotations) and there- fore there are as many normal modes as vibrational degrees of freedom. So, for water we have: 9 total - 3 translational - 3 rotation = 3 vibrational. In the problem, we are given three values for Θvibr, one for each normal mode. Thus, for water, equation (1) has three terms:

" ! ! 5360 e−5360/T 5160  e−5160/T U¯ (T ) =R + 5360 + + 5160 vib 2 (1 − e−5360/T ) 2 (1 − e−5160/T ) !# 2290 e−2290/T + + 2290 2 (1 − e−2290/T )

The figure below is a plot including each of the three terms independently, so that we can compare the relative contribution of each mode at each T . 7000

6000

5000 /R

) 4000 T ( 3000 vib ¯ U 2000

1000

0 0 2000 4000 6000 Temperature/ K

¯ Figure 1: The contribution of the three normal modes of vibration of water to Uvib.

2. You have two identical containers at a fixed temperature. In one you have wa- ter vapor (H2O) and in the other you have carbon dioxide (CO2). In the high- temperature, classical limit for vibrations and rotations, meaning that (Θrotation  T ) and (Θvibration  T ), and assuming you can ignore all excited electronic states, which of the following statements is true.

(a) The molar heat capacity of the water is larger than the molar heat capacity of the carbon dioxide. (b) The molar heat capacity of the carbon dioxide is larger than the molar heat capacity of the water. (c) The molar heat capacity of the water and the carbon dioxide are the same.

Answer: 1 1 In the classical limit, each translation adds 2 R, each rotation adds 2 R, and each vibration adds 1R to the molar heat capacity. Water has 3 translation, 3 rotations (non- 12 linear), and 3 vibrations for a total of CV = 2 R. Carbon dioxide has 3 translations, 13 2 rotations (linear), and 4 vibrations for a total of CV = 2 R. 3. The following is the expression for the heat capacity of an ideal diatomic gas, !  2 −Θvib/T ¯ 3 Θvib e CV = R + R + R 2 2 |{z} T (1 − e−Θvib/T ) |{z} B A | {z } | {z } C D

The term D is contributed from the electronic degrees of freedom. (a) TRUE (b) FALSE Answer: This is false. The term D is contributed from the vibrational degrees of freedom. The electronic degrees of freedom are typically very small and neglected in the expression for heat capacity. 4. In one dimension, the energy levels of a particle in a box with length a are given by: h2 n2  = (2) n 8 m a2 and there is only one state per level. As we have seen in lecture video 4.1, the trans- lational partition function depends on T and V . In the one dimensional case, the “volume” is the length of the one-dimensional box, a. So we write

n=∞ X  h2 n2  q (a, T ) = exp −β (3) Transl, 1D 8 m a2 n=1

What is the partition function for a particle in a box of area a × a? q 2π mkB T (a) h2 q 2π mkB T (b) h2 a q 2π mkB T 2 (c) h2 a q 1 2π mkB T 2 (d) 4 h2 a √ 2π mkB T (e) h2 a 2 2π mkB T
4 (f) h2 a (g) none of the above Answer:

And as we’ve seen, for typical conditions, this summation,

n=∞ X  h2 n2  q (a, T ) = exp −β Transl, 1D 8 m a2 n=1 can be very well approximated by an integral:

Z n=∞  h2 n2  q (a, T ) = exp −β dn (4) Transl, 1D 2 n=0 8 m a

β h2 R ∞ 2 p π Let us write α = 8ma2 , and use the fact that 0 exp(−α n ) dn = 4 α . Then we get:

Z n=∞  2 2  Z n=∞ h n  2 exp −β 2 dn = exp −αn dn (5) n=0 8 m a n=0 r 2π mk T = B a (6) h2

Let’s now consider the 2D case. The energy levels of a particle in a 2D box whose sides have length a are given by:

h2 (n2 + n2)  = x y (7) nx,ny 8 m a2 Therefore, the translational partition function is:

 2 2 2  X −β h (nx + ny) q (a, T ) = exp (8) Transl, 2D 8 m a2 all states ∞ ∞  2 2   2 2  X X −β h n −β h ny = exp x exp (9) 8 m a2 8 m a2 nx=1 ny=1 ∞  2 2  ∞  2 2  X −β h n X −β h ny = exp x · exp (10) 8 m a2 8 m a2 nx=1 ny=1

But note that the two factors in the final expression are the same (nx and ny are just dummy variables). Therefore:

∞ !2 X −β h2 n2  q (a, T ) = exp (11) Transl, 2D 8 m a2 n=1 But this is just the square of the summation we’ve computed for the 1-dimensional case. Thus, the 2-dimensional translational partition function is:

2 qTransl, 2D (a, T ) = qTransl, 1D (a, T ) (12) 2π mk T = B · a2 (13) h2

−1 5. The vibrational level spacing in the diatomic molecule N2 is 2330 cm . At room temperature, what fraction of the N2 molecules are vibrationally excited (meaning not in the vibrational ground state)?

(a) < 5% (b) ∼ 10% (c) ∼ 50% (d) > 90% Answer:

2 −1 −1
The average thermal energy at room temperature is kBT ≈ 3 cm · K (300 K) = 200 cm−1. This is an order of magnitude below the spacing of the vibrational levels −1 in N2 of 2330 cm , and therefore much less than 5% would be excited. Note that −1 ν˜ the vibrational temperature for N2 is (using cm for energy units) Θvib = = kb 2330 cm−1 2/3 cm−1·K−1 = 3495 K. This is well above room temperature. 6. Given one mole of lithium atoms in a dark box (i.e. no light shining on them), calculate the fraction of atoms that are in the first excited state at 300 K. The energy of each excited state with respect to the ground state is 14903 cm−1, 14904 cm−1, and 27206 cm−1 for the first, second, and third excited states, respectively. The degeneracy of the ground state is 2, 2 for the first excited state,4 for the second excited state, and 2 for the third excited state. The fraction of atoms in the nth electronic level is given by: g(n) e−β n fn = (14) qelectr. where g(n) is the degeneracy of that electronic level. (a) .062 (b) 9 × 10−32 (essentially zero) (c) 0.5 (d) (6.022 × 1023)/2 (e) .00001 (f) cannot be determined Answer: We take as the zero of electronic energy that of the ground state, and hence the partition −β2 −β3 function is qelectr. = g(1) + g(2)e + g(3)e + ··· where of course n is the energy of the nth level with respect to the ground state. Note that in order to compute the electronic partition function, the sum over the first few levels is enough, since the gaps in energy between levels are very large, and hence the summation converges very quickly. Let’s do the computations: T=300K: Considering the numerator in equation (14) for the first excited state, the degeneracy is 2 and the energy with respect to the ground state is 14903.66 cm−1:  −14903.66 cm−1  g(n) e−β n = 2 exp (15) 0.695cm−1K−1 · 300 K = 1.809 · 10−31 (16) And for the partition function:

−β2 −β3 qelectr. = g(1) + g(2)e + g(3)e + ··· (17)  −14903.66 cm−1   −14904.00 cm−1  = 2 + 2 exp + 4 exp (18) 0.695cm−1K−1 · 300 K 0.695cm−1K−1 · 300 K  −27206.12 cm−1  + 2 exp (19) 0.695cm−1K−1 · 300 K = 2 + 1.809 · 10−31 + 3.613 · 10−31 + 4.287 · 10−57 ≈ 2 (20) Note how vanishingly small the contributions are for even the very first few terms (due to the first excited states).So, we have that: 1.809 · 10−31 f (2) = = 9.04 · 10−32 (21) 300K 2 So for 1 mole of lithium ions in the gas phase at 300K (in the dark), there is ≈ 6.023 · 1023 × 9.04 · 10−32 = 5.4 · 10−8 atoms in the first excited state, that is, most likely not one single atom! 7. In general the fraction of molecules in some level is given by the Boltzmann factor times the degeneracy of that level divided by the partition function. For the case of rotational levels in the classical limit, this leads to the following expression developed in lecture video 4.5:

(2J + 1) Θ f = rot e−J(J+1)Θrot/T J T

Which is greater, the fraction of NO(g) molecules in the 10th rotational level at 300 K, or the fraction of NO(g) molecules in the 10th rotational level at 1000 K ? The rotational temperature of NO(g) is 2.39 K.

(a) the fraction at 300 K (b) the fraction at 1000 K (c) there is no fraction at 1000 K (d) the fractions are the same at both temperatures (e) cannot be determined

Answer:

This question is fairly straightforward, and we can directly substitute in the appropriate numbers, (2 × 10 + 1) × 2.39K f = e−10(10+1)×2.39K/300K J 300K = 0.07 (2 × 10 + 1) × 2.39K f = e−10(10+1)×2.39K/1000K J 1000K = 0.039

8. Which of the following systems meets the required criteria to use the equality,

q(V,T )N Q(N,V,T ) = N!

(a) a gas described by the ideal gas equation of state (b) a gas described by the van der Waals equation of state (c) a gas described by the Redlick-Kwong equation of state (d) a gas described by the Peng-Robinson equation of state (e) all of the above

Answer: No interaction between the particles is one of the criteria for the above to be a valid partition function. No interaction between particles is only true for the ideal gas.

9. What is the most populated rotational level, J, of an ensemble of NO (g) molecules at 300 K? The rotational temperature of NO(g) is 2.39 K. To solve this problem, treat J as a continuous variable, then round your answer.

(a) 6 (b) 7 (c) 0 (d) 2 (e) 17 (f) 1 (g) 50

We want to find which is the most populated rotational state for a molecule at a given temperature. Of course J is a discrete variable, but we will treat it as a continuous one so that we can use calculus. To find the maximum, we need to find the derivative of fJ (see equation on last lecture slide of video 4.5) with respect to J, set it equal to zero and solve for J. Of course, we are likely to get a non-integer value of J: one of the two values of J beside it will be the J that we are looking for. Let’s proceed:

d 2Θ (2J + 1)Θ   Θ f = e−J(J+1)Θ/T + e−J(J+1)Θ/T · −(2J + 1) = 0 (22) dJ J T T T Now, e−J(J+1)Θ/T 6= 0 for any J, so we can divide through by it to get: 2Θ (2J + 1)2Θ2 − = 0 (23) T T 2 r T 1 J = − (24) max 2Θ 2 Let’s now apply this result to the case of the NO molecule in the previous example. At 300 K we get: r 300 K 1 J = − max 2 · 2.39 K 2 Jmax = 7.4

From figure 2, you can see that Jmax for 300 K is 7. At 1000 K: r 1000 K 1 J = − max 2 · 2.39 K 2 Jmax = 14.0

We can also construct a plot of fJ vs. J to see the maximum value of J. This plot is shown below.

Figure 2: Population of vibrational levels in NO(g) at 300 K and 1000 K

10. The following quantities associated with particular thermodynamic variables were de- rived in the lecture videos for week 4:

3 3 RT RT, R, . 2 2 V¯ In the low temperature limit, these formulas represent, respectively, (a) The molar energy, heat capacity, and pressure of a polyatomic ideal gas. (b) The molar energy, molar heat capacity, and pressure of a dense monatomic real gas. (c) The molar heat capacity, molar energy, and pressure of a monatomic ideal gas. (d) The molar heat capacity, molar energy, and pressure of a diatomic ideal gas. (e) The molar energy, molar heat capacity, and pressure of a monatomic ideal gas. (f) The molar heat capacity, molar energy, and pressure of a diatomic ideal gas and monatomic ideal gas. (g) The molar heat capacity, molar energy, and pressure of a polyatomic ideal gas and monatomic ideal gas. (h) The molar energy, molar heat capacity, and pressure of a diatomic ideal gas. (i) The molar energy, molar heat capacity, and pressure of a diatomic ideal gas and monatomic ideal gas. (j) none of the above

Answer: These are the expressions for the molar quantities for a monatomic ideal gas summa- rized in lecture video 4.3. For polyatomic and diatomic gases, there are contributions to the energy and heat capacity from rotational and vibrational degrees of freedom. A dense real gas will not have the same expressions for any of these quantities. It is assumed that the molecules of an ideal gas do not occupy any volume, nor that they have any mutual attraction or repulsion. A real gas molecule has some volume, and repels and attracts its neighbors, and these characteristics are not taken into account in the formula above.