Chem 4501 Introduction to hermodynamics, 3 Credits Kinetics, and Statistical Mechanics Module Number 2 Active Learning Answers and Optional Problems/Solutions 1. McQuarrie and Simon, 2-6. Paraphrase: How many molecules are there in 1 cm 3 of gas at 298 K and 10 12 torr pressure? What is the molar volume corresponding to this pressure? At so low a pressure, we may certainly assume ideal-gas behavior, i.e., PV / R = n where n is the number of moles of gas. P in atm is (10 12 torr) x (1 / 760 atm torr 1 ) if we want to use the gas constant in units of cm 3 atm mol 1 K 1 (82.057). Solving for n gives 5.38 x 10 20 moles. Multiplying by Avogadro s number (6.02 x 10 23 molecules mol 1 ) gives about 32,400 molecules. Molar volume is volume per mole. In this case, we already know the number of moles occupying 1 cm 3, so, dividing 1 cm 3 by 5.38 x 10 20 moles gives 1.86 x 10 19 cm 3 mol 1, or 1.86 x 10 16 L mol 1. hat is a lot of liters 2. McQuarrie and Simon, 2-7. Paraphrase: We are given 300 K pressure/density values (bar / g L 1 ) of (0.1 / 0.1771), (0.5 / 0.8909), (1.0 / 1.796), (1.01325 / 1.820), (2.0 / 3.652) and asked to find the molecular weight of the corresponding gas. In addition, determine B 2V from these data (think about how the molar volume is related to the density in the virial equation of state). Density ρ is total mass M (g) per volume V (L), and total mass per volume is number of moles n per volume times molecular mass m (g mol 1 ). From the ideal gas law, then, we have P = nr / V = ρr / m, so m = ρr / P. hat is, the ratio of the
2 density and the pressure should be a constant for the ideal gas, multiplication of which by R would give the molecular mass. he ratio in this case varies from 1.771 to 1.826 (so the gas is not behaving ideally). A rough estimate could be had from the lowest pressure value (since here the gas should be most ideal), 1.771 g bar 1 L 1. Multiplying by R (8.31 x 10 2 L bar mol 1 K 1 ) and 300 K gives 44.15 g mol 1. A more accurate approach is to consider not the ideal gas equation, but instead the virial equation of state truncated at the first non-ideal term, i.e., PV R =1+ 1 V B 2V which, given our derivation above, may be written as multiplying both sides by R/ m gives P ρ = R m + R m B 2 2V ρ so that a plot of P / ρ against ρ should have a slope of RB2V / m 2 and an intercept of R / m. As we don t know B2V, the slope is not so helpful, but the indicated plot has a best fit intercept of 0.5656 bar L g 1 (with R 2 = 0.9999, so a good linear fit). Inverting and multiplying by R gives a molecular mass of 44.08 g mol 1. Note that the deviation from ideality is not large, given the comparison of this number to the value of 44.15 computed above. One nice feature of this approach, though, is that it lets us also compute B2V, since we see that the slope in the last equation above is the intercept divided by m times B2V.
3 So, dividing the slope (4.8987 x 10 3 bar L 2 g 2 mol -1 ) by the intercept and multiplying by R provides B2V = 0.382 L mol 1. If we convert this to units of cm 3 mol 1 and compare to Figure 2.12 we see that we have a very sticky gas. Actually, given the molecular weight, the gas seems to be propane. A quick glance at Corner, J. he Second Virial Coefficient of a Gas of Non- Spherical Molecules Proc. Roy. Soc. London Ser. A 1948, 192, 275-292 indicates that this is correct, as the tabulated 300 K second virial coefficient is very close to that determined here. 3. McQuarrie and Simon, 2-9 We are told that a mixture of H2 and N2 has a density of 0.216 g L 1 at 300K and 500 torr and asked to determine the mole fraction composition. From the ideal gas equation, we can determine the total number of moles of gas in a given volume (the molar density, if you like) and, given the mass density and the known molecular weights of the 2 gases (roughly 2 and 28, respectively), we can then determine composition. So, first we have from the ideal gas law n / V = P / R. Using R = 62.36367 L torr K 1, we have n / V = 500 / (62.36367 x 300) = 2.672 x 10 2 L 1. hus, in one liter, we would have 0.216 g and that mass would be equal to 2.672 x 10 2 moles of molecules. If x is the number of moles of H2, then the number of moles of N2 is 2.672 x 10 2 x. Multiplying by the respective molecular weights (rounded for simplicity), we have 2x + 28(2.672 x 10 2 x) = 0.216 Solving for x gives 0.0205 moles H2, leaving 0.00622 moles N2. he mole fractions are thus about 77% and 23%, respectively. 4. McQuarrie and Simon, 2-15.
4 Given temperature and pressure values of 200 K and 1000 bar, respectively, we are asked to solve for the molar volume using the ideal gas equation of state ( = R / P), the van der Waals equation of state (text eq. 2.10), and the Redlich-Kwong equation of state (text eq. 2.9). he latter two are cubic equations, so they must be solved by some numeric procedure (I used Solver with Microsoft Excel). he resulting predictions are 0.01663, 0.04998, and 0.03866 L mol 1, respectively, which represents errors of 58.5%, 24.7%, and 3.6%, respectively. 5. McQuarrie and Simon, 2-36. Paraphrase: Prove B 2V () = RB 2P (). We may take the pressure-dependent virial expansions for the compressibility factor Z if we multiply both sides by R /, we have Note that this isn t a complete solution for P, because P continues to appear as a power expansion on the right-hand side. Of course, if we replaced the first occurrence of P (after B2P()) with this expression, we would have which would simplify as
5 If we followed the same approach repeatedly, we would derive terms in higher inverse orders of V, thus we may say P = R V + R2 2 V 2 B 2P " ( ) + O$ 1 # V 3 % ' If we replace the P in the compressibility factor of the volumedependent virial expansion with this expression, we have ( * ) R V + R2 2 V 2 B 2P R " ( ) + O$ 1 # V 3 % + '- V, =1+ B 2V ( ) V + B 3V ( ) +! V 2 Simplifying the numerator on the left hand side provides 1+ R V B 2P " ( ) + O$ 1 # V 2 ( ) % ' =1+ B 2V V + B 3V ( ) +! V 2 Since this expression holds for all values of V, it must be true that all coefficients of identical powers of V are equal. he coefficients of V 1 are RB 2P () on the left-hand side and B 2V () on the right-hand side, which completes the proof. 6. McQuarrie and Simon, 2-54. Paraphrase: compute the c 6 coefficient for N 2 from its molecular properties (in able 2.8) and its Lennard-Jones parameters (able 2.7). he c 6 coefficient from molecular properties is eq. 2.37. Happily, N 2 has a zero dipole moment, so the first two terms in eq. 2.37 are zero, and from the final term we have c 6 = 3 4 I % α ( ' * 4πε 0 ) 2 = 0.75 ( 2.496 10 18 J) ( 1.77 10 30 m 3 ) 2 = 5.86 10 78 J m 6
6 From the Lennard-Jones expression, the coefficient of the r 6 term is 4εσ 6. he values in able 2.7 lead to 1.35 x 10 77 J m 6, or about double the value from molecular parameters. 7. McQuarrie and Simon, 2-58 and 2-59. Paraphrase: Do some dirt simple differential calculus to derive the coefficient of thermal expansion and isothermal compressibility for an ideal gas. he ideal gas equation of state is V = R / P. hus # V % ( $ ' P = R P and # V % ( $ P ' = R P 2 So, given the definitions of α and κ, we have α = 1 V $ V ' ) % ( P = P R R P = 1 and κ = 1 V $ V ' ) % P ( = P $ R R ' ) = 1 % P 2 ( P 8. McQuarrie and Simon derive on p. 85 that the van der Waals equation of state has an intermolecular potential u(r) associated with it. In particular, it is a hard-sphere potential joined to an attractive r 6 potential, i.e., u( r) =, r < σ ( ' c 6 ) ( r 6, r σ where c 6 is a characteristic coefficient for a given gas and σ is the hardsphere diameter. In this proof, they truncated the power series expansion for the exponential after the second term. Considering the answer to McQuarrie and Simon, 2-54, solved above, for c 6 for N 2 and the σ value for N 2 from able 2.7, was such a truncation justified. If truncation is not acceptable until after the third term in the power series expansion, what are the consequences?
7 Consider the exponential expansion from p. 85 for e x where x is c6/kbr 6. e c 6 / r 6 =1+ c 6 r + 1 " c 6 % $ ' 6 2 # r 6 2 +! he question becomes, what are the relative magnitudes of c 6 r 6 vs 1" c 6 % $ ' 2# r 6 2 Clearly, as r gets larger we may rely on the first term dominating the second, so we can check the worst-case scenario of r = σ. Using a rough value for c6 computed above, and the σ value from able 2.7, we have 1 10 77 J m 6 ( 1.3806 10 23 J K 1 ) 300 K ( ) 6 = 0.941 ( ) 370 10 12 m comparing this to one-half times the square of this value, which is equal to 0.443, we see that truncation in this instance is not really a very accurate approximation, although it will obviously get much, much better as r increases. If we were to keep the third term in the power series expansion, eq. 2.25 would become
8 B 2V σ ( ) = 2πN A ( 1) r 2 c dr 2πN 6 0 A r + 1 2 c 6 ) ) ( + r σ ( ' 6 2 ' r 6 * + 2 dr * = 2πN Aσ 3 3 = 2πN Aσ 3 3 2πN A c 6 ( ' σ dr r 4 + c 6 2πN c A 6 3 σ 2πN c 2 A 6 3 9k 2 B 2 σ 9 σ dr r 10 ) + * hus, the second virial coefficient becomes inverse quadratic in, as opposed to the van der Waals equation of state, where it simply has inverse dependence. Note that this does not correspond to any of the equations of state that are addressed in Chapter 2, but it can certainly be further developed in analogy to the others. 9. Example multiple choice problem: When a P vs. V isotherm for a substance has a horizontal region of finite length, which of the following is true? (a) = c (d) he compressibility Z = 1 (b) he isotherm passes through (e) Large changes in pressure can the liquid-vapor coexistence occur with infinitesimal curve changes in volume (c) he gaseous substance is (f) (b) and (e) behaving ideally