Homework Week 2

Statistical MolecularThermodynamics University of Minnesota

Homework Week 2

1. The hard-sphere potential is a simple model for real molecular interactions.

This simple model can actually do a very good job of predicting real molecular behavior under what conditions? (a) At very low temperatures compared to the energy of attraction between molecules. (b) At very high temperatures compared to the energy of attraction between molecules. (c) At the critical temperature. (d) (a) and (c) Answer: This simple model is only valid at high temperatures. At high temperatures, the repulsive forces are dominant compared to the attractive forces (to paint a mental picture: they are buzzing around faster, constantly coming in suﬃciently close contact to repel one another strongly. Recall in the hard sphere model there is no attractive potential, thus, at lower temperatures, the model begins to fail as the repulsive molecular forces between molecules contribute to the same degree or less than the attractive forces. See the plot of B2V (T ) vs. Temperature in the lecture slides.

2. At a temperature above the Boyle temperature, the second virial coeﬃcient (B2V ) is, (a) positive (b) negative (c) zero Answer: Above the Boyle temperature, the compressibility factor is greater than 1. We know ¯ P (V ) B2V (T ) ¯ Z = RT = 1 + V¯ ; thus, since V is positive, B2V (T ) must be positive. 3. The plot below shows the compressibility factor of methane at four diﬀerent temperatures. Compressibility of methane

C D B A

Which line is at the lowest temperature?

(a) A (b) B (c) C (d) D

Answer: Recall that at low temperatures, attractive forces between gas molecules dominate, ¯ ¯ and thus VReal is less than VIdeal. For an ideal gas Z = 1, so the compressibility factor of the gas at the lowest temperature should show the maximum deviation below the ideal threshold, and thus A is the correct answer.

4. You have a container that has a volume of V = 2.49 dm3 and in it you have 1 mole of methane (CH4) gas at a temperature of 300 K. If the compressibility factor of methane is 0.9 at this temperature, what is the pressure in this container?

(a) near, but above 1 bar (b) near, but below 1 bar (c) near, but above 10 bar (d) near, but below 10 bar (e) near, but above 100 bar (f) near, but below 100 bar

We can compute, −2 3 −1 −1 ZRT (0.9)(8.31×10 bar·dm ·mol ·K )(300 K) 24.9 P = V¯ = 2.49 dm3·mol−1 = 0.9 · 2.49 bar = 9 bar This pressure is a bit lower than the 10 bar that would be expected for an ideal gas, for which Z = 1.

5. The plot below shows the potential energy of interaction between two molecules of a hypothetical compound given as a function of the separation distance (this is a Lennard-Jones potential). Note that the energy axis is in Joules and is on the 10−21 scale. !!% &'%# %

# ,'('- r + u !% -1.4 x 10-21 J !"# "## $"# %### r '(')*

Which of the following temperatures is closest to the Boyle temperature for this compound?

(a) 30 K (b) 100 K (c) 300 K (d) 500 K (e) 700 K (f) 1400 K

3ε At the Boyle temperature, TBoyle ≈ where ε is the depth of the Lennard-Jones kB −21 attractive well. From the plot, ε = 1.4 × 10 J. Plug this in and you ﬁnd TBoyle ≈ 3(1.4×10−21 J) 1.381×10−23 J·K−1 ≈ 300 K.

6. You have ethane (C2H6) gas at T = 300 K and P = 10 bar. You measure the molar ¯ 3 −1 volume and ﬁnd that it is Vexperiment = 2.2 dm ·mol . Which of the following is closest to the value of the second virial coeﬃcient (B2V ) for ethane at 300 K? (a) −0.8 dm3 · mol−1 (b) −0.3 dm3 · mol−1 (c) −0.1 dm3 · mol−1 (d) 0 dm3 · mol−1 (e) 0.1 dm3 · mol−1 (f) 0.2 dm3 · mol−1 (g) 0.8 dm3 · mol−1

−2 3 −1 −1 ¯ ¯ ¯ RT (8.314×10 dm ·bar·mol ·K )(300 K) Use the relationship, B2V = V −Videal. Calculate Videal = P = 10 bar = 24.94 3 −1 3 −1 10 dm · mol = 2.494 dm · mol . Plug this in with the given experimental value, 3 −1 3 −1 3 −1 B2V = 2.2 dm · mol − 2.494 dm · mol ≈ −0.3 dm · mol . 7. One can derive the following relationship between the a and b coeﬃcients in the van der Waals equation of state and the second virial coeﬃcient, a B = b − . 2V RT Given this, what is the correct expression for the Boyle temperature in terms of a and b?

(a) TBoyle = 0

(b) TBoyle = Rb

Answer Recall that at the Boyle temperature, the intermolecular attractive and repulsive forces B2V (T ) balance. This means that the compressibility is equal to 1. So, from Z = 1 + V¯ , we know that B2V (T ) must be zero at the Boyle temperature. But we know from the derived result, a B2V (TB) = b − , RTB a so if B2V (TB) = 0, the result is TB = Rb

8. At a temperature of 1200 K, which of the following is closest in energy to kBT . It −1 −1 might be helpful to have kB ≈ 0.7 cm · K . (a) The ionization energy of a hydrogen atom

(b) The bond dissociation energy of O2

(d) The energetic gap between adjacent vibrational states of F2 gas in a 1 L volume Answer: We can compare the energies in just about any units, but since we are given kB in cm−1 · K−1, lets use wavenumbers. The thermal energy at 1200 K ≈ 1200 × 0.7 cm−1 · K−1 ≈ 840 cm−1 We know from a previous problem that ionization energy of a hydrogen atom is ≈ 13.6 eV (a good number to remember). So if we convert this to wavenumbers,

1.6 × 10−19 J 1 1 13.6 eV = 109, 470 cm−1 eV 2.99 × 1010 cm · s−1 6.626 × 10−34J · s This is much greater than 840 cm−1, so it is not (a). Typical bond dissociation energies are on the order of hundreds of kilojoules per mol. Let’s use an approximation of 100 kJ/mol. To see what this equals in wavenumbers:

100 kJ 1 mol 103 J 1 1 = 8, 381 cm−1 mol 6.022 × 1023 kJ 2.99 × 1010 cm · s−1 6.626 × 10−34 J · s . Still much larger than 840 cm−1, so it is not (b).

What about the energetic gap between adjacent translational states of N2 gas? We can calculate this from the formula: h2 n2 n2 n2 = x + y + z trans 8m a2 b2 c2 where a, b, and c are the dimensions of a macroscopic box. If we consider a 1 litre box and the nx = ny = nz = 1 quantum level.

2 2 2 2 2 −34 kg m 2 h n n n (6.626 × 10 2 · s) 3 = x + y + z = s = 3.54×10−40J trans 8m a2 b2 c2 8 · 0.28 kg · mol−1/(6.022 × 1023 mol−1) 0.12 m2 !

kg m2 Here, we substituted 1 J = 1 s2 . After all of the units cancel, we are left with the kg m2 answer in s2 , which is in joules. If we convert this answer to wave numbers: 1 s 1 3.54 × 10−40 J = 1.79 × 10−17 cm−1 2.99 × 1010 cm 6.626 × 10−34 J · s . The next level up would involve one of the three quantum numbers being equal to 2, making the numerator 6 instead of 3 in computing the energy, i.e, twice as high, which is then a gap of 1.79 × 10−17 cm−1. As we consider increasing quantum numbers, the various levels will actually become closer together, as there are three free integers that are changing. This is much smaller than 840 cm−1! So the answer is deﬁnitely not (c). This leaves (d) by default as the correct answer. You might have recognized this right away as an energy very close to vibrational spacings in molecules, and indeed the vibrational frequency (which is the spacing between harmonic vibrational levels) for F2 is 917 cm−1. Nevertheless, it was still useful to do the above calculations to illustrate some important concepts.

9. On the plot below your are given measurements and a best fit of the compressibility factor of a gas as a function of the molar volume at a temperature of 200 K, Given the slope of the fit, indicated on the plot, at a temperature of 200 K what is the pressure of this gas at a fixed molar volume of V¯ = 0.2 dm3/mol? Assume that all virial coefficients beyond the second (B2V ) are negligible. '

!"((&

!"((% )*+,-./.!!"' Z !"(($

!"((#

!"(( ! !"!# !"!$ !"!% !"!& !"' 1/V¯ mol/dm3 ! "

(a) P = 0.1 bar (b) P = 0.2 bar (c) P = 0.5 bar (d) P = 1 bar (e) P = 2 bar (f) P = 5 bar (g) P = 21 bar (h) P = 42 bar (i) P = 100 bar

Answer: B2V (T ) ¯ Recall that Z = 1 + V¯ . Since the graph is plotted in units of molar volume 1/V then B2V (T ) must be equal to the slope of the line (think about the equation of a straight line, y = mx + b, where m is the slope. This is an analogous situation, where ¯ m → B2V (T ) and b → 1, and x → 1/V ). We also know that ,

P V¯ B (T ) = Z = 1 + 2V , RT V¯ so P V¯ 0.1dm3/mol = 1 − = 0.5 RT 0.2dm3/mol Now it is an algebra problem, just solve for P and substitute in the proper values:

0.08314 dm3 · bar · mol−1 · K−1 200 K P = 0.5 = 41.6 bar 0.2 dm3/mol

10. The attraction between two molecules (for example two methane or CO2 molecules) at their van der Waals minimum is about the same as the energy of a covalent chemical bond. (a) TRUE (b) FALSE

Answer: Consider the attractive part of the intermolecular potential energy between two neon atoms: /kB = 35.6 K and σ = 235 pm (this is from the table in the lecture slides). The attractive part of the potential is

σ 6 −4 , r which, if we evaluate at r = σ (it makes the calculation easier), leaves us with 4. To obtain units of Joules, we must multiple the number given in the table by Boltzman’s constant. Then, to get this value as a molar quantity, we multiply by Avogadro’s number. We can compare this number to a typical value of a covalent bond (on the order of 200 kJ/mol):

4 × 35.6 K × 1.38 × 10−23J/K × 6.022 × 1023mol−1 × 10−3kJ/J = 1.18 kJ/mol

This is two orders of magnitude less than a typical covalent bond strength.