Florida International University s6

CHM 3410 – Problem Set 1 Due date: Friday, September 4th Do all of the following problems. Show your work.

"Only he (or she) who is not frightened away by a simple mathematical formula can expect to derive any real and practical benefit from physical chemistry." (Arnold Eucken, Fundamentals of Physical Chemistry, 1922).

1) Consider 2.0000 mol of argon gas at a temperature T = 273.0 K and confined in a volume V = 1.0000 L. Find the pressure of the gas using a) the ideal gas law (eq 1.8) b) the van der Waals equation (eq 1.21a or 1.21b) c) the virial equation (eq 1.19)

Give your values for pressure (to four significant figures) in units of bar. Values for the van der Waals a and b coefficients for argon are given in Table 1.6, p 992 of Atkins. Values for the virial coefficients B and C are given in Table 1.4, p 991 of Atkins.

2) Use the values for the van der Waals a and b coefficients for carbon dioxide (Table 1.6, p 992 of Atkins) to estimate the values for the critical constants for CO2 (pC, VC, and TC). Compare your values to the experimental values (Table 1.5, p 991 of Atkins), and briefly discuss the level of agreement.

3) Nitrogen dioxide will dimerize in the gas phase to form dinitrogen tetroxide. The reaction may be written as follows:

2 NO2(g) ⇄ N2O4(g) K = p(N2O4) 2 [p(NO2)]

Note that the "pressures" appearing in the expression for the equilibrium constant are actually the pressure of the gas divided by standard pressure, p = 1.000 bar = 0.9869 atm. K therefore has no units (this will be discussed in detail when we cover Chapter 7 of Atkins).

One interesting method for finding the equilibrium constant for the above reaction is to simultaneously measure the pressure and density of an NO2 - N2O4 gas mixture at equilibrium at some temperature T. If ideal gas behavior is assumed a numerical value for K can be found.

In a particular experiment the following data were obtained:

T = 33.8 C ptotal = 857. torr  = 3.323 g/L

Based on these data find the numerical value for K for the dimerization reaction at 33.8 C.

4) One form of the virial equation is

p = RT + B + C 2 3 Vm Vm Vm where Vm = V/n is the molar volume of the gas, and B and C are constants. Show that this equation of leads to critical behavior, and find values for pC, VC, and TC (the critical pressure, volume, and temperature) in terms of B, C, and/or other constants. Also do the following from Atkins:

Exercises 1.8a (assume ideal gas behavior)

Problems 1.2 (Starting with the ideal gas law, find an expression for M, the molecular mass, in terms of  (density), p, and T. Based on your result, find an appropriate way of plotting the data. Based on the plot find the value for M for dimethyl ether, and compare your value to that found from the molecular formula, C2H6O).

EXTRA CREDIT - Use the data in problem 1.2 to find an experimental value for B, the first virial coefficient in eq 1.19, at T = 25. C. Use the value for M for dimethyl ether calculated from the molecular formula in your work. Do the data justify finding C, the second virial coefficient? Briefly justify your answer (but you do not have to find a value for C, even if the data are sufficient to do so).

Solutions 1) a) From the ideal gas law

p = nRT = (2.0000 mol) (0.083145 L. bar/mol .K) (273.00 K) = 45.40 bar V (1.0000 L)

b) From the van der Waals equation ( a = 1.337 L2.atm/mol2 ; b = 0.0320 L/mol )

p = nRT - an2 (V - nb) V2

= (2.000 mol) (0.083145 L.bar/mol.K) (273.00 K) - (1.337 L2.atm/mol2) (2.0000 mol)2 1.01325 bar [1.0000 L - (2.000 mol) (0.0320 L/mol)] (1.0000 L) 1. atm

= (48.501 - 5.419) bar = 43.08 bar

c) From the virial equation ( B = - 21.7 x 10-3 L/mol ; C = 1200. x 10-6 L2/mol2 )

2 p = nRT { 1 + (B/Vm) + (C/Vm ) } Vm = V/n = (1.0000 L)/(2.0000 mol) = 0.5000 L/mol

= (2.0000 mol) (0.083145 L.bar/mol.K) (273.00 K) { 1 + (- 21.7 x 10-3 L/mol) + (1200. x 10-6 L2/mol2) } (0.5000 L/mol) (0.5000 L/mol)2

= (45.394 bar) (1.0000 - 0.0434 + 0.0048) = 43.64 bar

2) The critical constants derived from the van der Waals equation (Table 1.7, p 19) are

2 pC = a/27b VC = 3b TC = 8a/27Rb

2. 2 For CO2, a = 3.610 L atm/mol ; b = 0.0429 L/mol . The calculated values for the critical constants for CO 2, and the experimental values (in brackets) are

2. 2 pC = (3.610 L atm/mol ) = 72.6 atm [experimental = 72.85 atm] 27 (0.0429 L/mol)2

VC = 3 (0.0429 L/mol) = 0.129 L/mol [experimental = 0.0940 L/mol]

2. 2 TC = 8 (3.610 L atm/mol ) = 303.9 K [experimental = 304.2 K] 27 (0.082057 L.atm/mol.K) (0.0429 L/mol)

The values for pC and TC calculated from the van der Waals coefficients are close (within 1%) to the experimental values, but the value for VC calculated from the van der Waals coefficients is almost 40% larger than the experimental value. The reason for this is that the van der Waals b coefficient is obtained for conditions of low density, and so assumes pairwise interactions between molecules. At the critical point the density is sufficiently large that interactions among 3, 4, 5, ... molecules must be considered. The value of VC obtained from the b coefficient therefore overestimates the critical volume.

It is also interesting that the critical temperature for carbon dioxide is close to room temperature, therefore making it relatively easy to work with carbon dioxide in the vicinity of its critical point. This is the basis for a large number of techniques in chemistry, including supercritical fluid chromatography, as well as processing techniques in the food and pharmaceutical industry. For example, supercritical CO2 extraction is used to remove caffeine from (decaffinate) coffee (Atkins, pp 119-120). 3) To determine the value for the K, the equilibrium constant, requires that we find values for the partial pressure for

NO2 and N2O4. If we assume the gases behave ideally (Dalton's law) this can be done.

The following will be of use in doing this problem

M(NO2) = 46.01 g/mol M(N2O4) = 92.02 g/mol p = 857. torr 1 atm 1 bar = 1.1426 bar 760. torr 0.9868 atm

T = 307.0 K

For convenience, let us assume 1.0000 L of gas. Then from the ideal gas law

n = pV = (1.1426 bar) (1.0000 L) = 0.04476 mol RT (0.083145 L.bar/mol.K) (307.0 K)

Now, the above is the total number of moles of gas. Let x = n(N 2O4) be the number of moles of N2O4, then it follows that n(NO2) = (0.04476 - x).

Based on the value for density we know that 1.0000 L of the gas mixture has a mass m = 3.323 g.

m = m(N2O4) + m(NO2) = m(N2O4) n(N2O4) + m(NO2) n(NO2). Using this (and ignoring, for convenience, units)

3.323 = x (92.02) + (0.4476 - x) (46.01)

3.323 = 46.01 x + 2.059

x = (3.323 - 2.059) = 0.02747 mol 46.01

So n(N2O4) = 0.02747 mol ; n(NO2) = 0.01729 mol

From Dalton's law

p(N2O4) = X(N2O4) p p(NO2) = X(NO2) p Xi = ni/n

p(N2O4) = (0.02747 mol/0.04476 mol) (1.1426 bar) = 0.7012 bar

p(NO2) = (0.01729 mol/0.04476 mol) (1.1426 bar) = 0.4414 bar

K = p(N2O4) = (0.7012) = 3.60 2 2 [p(NO2)] (0.4414)

As we will discuss in Chapter 7, what actually appears in the expression for K is a, the activity, of the reactants and products, raised to their appropriate powers. For an ideal gas

a = p/p where p is standard pressure (now taken as 1.0000 bar). As can be seen from this expression, activity (and therefore K) does not have units. 4) The gas law we are considering is (for convenience, we will use V to represent Vm = V/n).

p = RT + B + C V V2 V3

In a plot of pressure vs volume for the critical isotherm the first and second derivatives are equal to zero (a condition for a critical point that is a saddle point), that is

2 2 (p/V)T = 0 ( p/V )T = 0

2 3 4 (p/V)T = 0 = - (RT/V ) - (2B/V ) - (3C/V )

2 2 3 4 5 ( p/V )T = 0 = (2RT/V ) + (6B/V ) + (12C/V )

4 2 2 5 If we multiply (p/V)T by ( - V ), and ( p/V )T by (V /2), we get

0 = RTV2 + 2BV + 3C

0 = RTV2 + 3BV + 6C

If we subtract the first equation above from the second, we get

0 = BV + 3C

Solving for V gives VC = - 3C/B

Now, returning to the expression

0 = RTV2 + 2BV + 3C

and solving for T, we get

2 T = TC = - (2BV + 3C) = - [ (2B) (- 3C/B) + (3C) ] = B RV2 R (- 3C/B)2 3RC

To find pC we now substitute our values for VC and TC into the original equation.

2 3 3 3 3 pC = R (B /3RC) + B + C = - B + B - B = - B (- 3C/B) (- 3C/B)2 (- 3C/B)3 9C2 9C2 27C2 27C2

Although not part of the problem, it is interesting to test the above results. If we compare the equation of state in this problem to the virial equation (eq 1.19) we can find values for B and C above in terms of the virial coefficients. The results are B = BVRT, C = CVRT, where BV and CV are the virial coefficients in eq 1.19. Using the 2. 2 values for BV and CV from Table 1.4, p 991 of Atkins (at T = 273.0 K) we get B = - 0.4926 L bar/mol , C = 0.02724 3. 3 L bar/mol , which gives pC = 6.0 bar, VC = 0.166 L/mol, and TC = 35.7 K. Compared to the good agreement in the values for the critical constants found from the van der Waals coefficients (problem 2, above), this is terrible agreement, additional proof that methods based on physical insight (like the van der Waals equation) are often superior to brute force methods based solely on mathematics.

In fact, we might have suspected there would be poor agreement, since at the Boyle temperature (Atkins, p

16) the above relationships would predict pC = 0, TC = 0, and VC =  !

Exercise 1.8a Assume a volume V = 1.0000 m3. From the ideal gas law (and noting that 1 kPa = 0.01 bar, 1. m3 = 1000. L)

n = pV = (93.2 x 10-2 bar) (1000.0 L) = 14.50 mol RT (0.083145 L.bar/mol.K) (773.2 K)

Since  = 3.710 kg/m3 = 3710. g/m3, the mass of 1.000 m3 of vapor is 3710.g. Since M = m/n, it follows that

M = (3710. g) = 255.9 g/mol (14.50 mol)

Since the atomic mass of sulfur is M(S) = 32.06 g/mol, the number of sulfer atoms per gas particle is

N = (255.9 g/mol) = 7.98  8 The formula is S8 . (32.06 g/mol)

Problem 1.2

There are several ways in which this problem can be done. One way is to find an expression for M based on the ideal gas law. Since

pV = nRT

n = p If we multiply both sides by M (molecular mass) and recall that m = nM, then V RT

nM = m =  = pM Solving for M gives V V RT

M =  RT p

Since all gases obey the ideal gas law in the limit of pressure approaching zero, the above suggests that we plot (RT/p) vs p, and extrapolate to p  0.

The data are tabulated below, and plotted on the next page. p (kPa) 12.233 25.20 36.97 60.67 85.23 101.3  RT/p 45.63 44.85 44.52 43.61 42.70 42.43 (g/mol)

The data show sufficient curvature to justify a fit to a second order polynomial rather than to a line. If the data are fit to a second order polynomial by the method of least squares, we get

RT/p = (46.22 g/mol) - (0.535 g/mol.kPa) p + (0.000155 g/mol.kPa2) p2

which leads to a value M = 46.22 g/mol. This is close to the value calculated from the atomic masses and molecular formula, M = 46.07 g/mol. Plot of M vs p

47 46 ) l 45 o m

/ 44 g ( 43 M 42 41 0 20 40 60 80 100 120 p (kPa)

EXTRA CREDIT

The virial equation, written as an expansion in terms of 1/Vm, is

2 pVm = RT { 1 + (B/Vm) + (C/Vm ) + ... } Dividing by RT gives

pVm = Z = 1 + B + C + ... 2 RT Vm Vm

If we take the derivative of the above equation with respect to (1/Vm) we get

(Z/(1/Vm))T = B + 2C + ... Vm

This means that if we plot Z vs (1/Vm) and fit the data to an expansion in powers of (1/Vm) we can relate the terms in the expansion to the virial coefficients B, C, ...

Now, since

 = m/V = nM/V = M/Vm , it follows that 1/Vm = /M . We use M = 46.07 g/mol for dimethyl ether, as calculated from the molecular formula and atomic masses.

The data are tabulated below and plotted on the next page

1/Vm (mol/L) 0.00488 0.00990 0.01441 0.02305 0.03186 0.03764 Z 1.0096 1.0270 1.0347 1.0564 1.0790 1.0857

The data do not show sufficient curvature to justify a fit to a second order polynomial (which means a value for the virial coefficient C cannot be obtained from the data). If the data are fit to a first order polynomial by the method of least squares, we get

Z = 1.001 + (2.339 L/mol) Vm

3 a nice result, since it gives Z  1 in the limit 1/Vm = 0, as expected. Therefore, B = 2.339 L/mol = 2339. cm /mol. If this value is compared to those given in Table 1.4, page 991 of Atkins, it can be seen that this is much larger than values given there for other gases. However, this is not surprising, since the gases in Table 1.4 are all small nonpolar molecules and so are expected to behave more like ideal gases than dimethyl ether.

Plot of Z vs 1/Vm

1.1 1.08 1.06

Z 1.04 1.02 1 0.98 0 0.01 0.02 0.03 0.04 0.05 1/Vm (mol/L)

The above method shows how experimental data can be fit to the virial equation and how the virial coefficients can be obtained. A similar method can be used to obtain values for the a and b coefficients for the van der Waals equation. Such methods are in fact now routine.