CHM 3410 – Problem Set 1
Due date: Wednesday, August 31st
Do all of the following problems. Show your work.
“Physical chemist (n.) – Someone who studies anything that is interesting.” – Larry Bartell
1) Consider 1.0000 mol of xenon (Xe) 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 1.19b)
Give your values for pressure (to four significant figures) in units of bar. Values for the van der Waals a and b
coefficients for xenon are given in Table 1.6, p 916 of Atkins. The value for the virial coefficient B for xenon is
given in Table 1.4, p 915 of Atkins.
2) Nitrogen dioxide (NO2, M = 46.006 g/mol) will dimerize in the gas phase to form dinitrogen tetroxide (N 2O4, M =
92.011 g/mol). The reaction may be written as follows:
2 NO2(g) ⇄ N2O4(g)
K=
p(N2O4)
[p(NO2)]2
Note that the "pressures" appearing in the expression for the equilibrium constant are actually the partial 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). Ideal behavior is also assumed.
One interesting method for finding the equilibrium constant for the above reaction is to simultaneously
measure the pressure and mass of a sample of NO2 and N2O4 at equilibrium at known temperature and volume. This
can be done by measuring the mass of an evacuated glass bulb of known volume, and then remeasuring the mass of
the bulb when filled with an NO2-N2O4 gas mixture at equilibrium at a known temperature and pressure. If ideal gas
behavior is assumed a numerical value for K can be found.
In a particular experiment the following data were obtained:
T = 32.7 C
V(bulb) = 1057. mL
ptotal = 851. torr
m(bulb + gas) = 59.165 g
m(bulb) = 55.653 g
Based on these data find the numerical value for K for the dimerization reaction at 32.7 C.
3) 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.
4) Plot Z (the compressibility factor) vs p for oxygen (O 2) at T = 300.0 K. Include in your plot pressures in the range
p = 0 – 1000 atm. Also find the value of p corresponding to the minimum in Z, and the value for Z at this minimum.
Assume that oxygen obeys the van der Waals equation of state. An example of a plot of Z vs p for a real gas is given
in Fig. 1.14, p 31 of Atkins. [Hint: It is easiest to solve this problem by picking values for V and then calculating the
corresponding value for p.]
Also do the following from Atkins:
Exercises
1.8b At 100.0 C and 16.0 kPa the mass density of phosphorus vapor is 0.6388 kg/m3.
What is the molecular formula of phosphorus under these conditions?
Problems
1.2 Deduce the relationship between pressure (p) and mass density () of a perfect gas of
molar mass M. Confirm graphically, using the following data on dimethyl ether (C2H6O) at 25.0 C, that perfect
behavior is reached at low pressure and find the molar mass of dimethyl ether.
p (kPa)
 (kg/m3)
12.223 25.20 36.97 60.37 85.23 101.3
0.225 0.456 0.664 1.062 1.468 1.734
[Hint: There are many ways to do this problem. I would suggest the following. 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, for dimethyl ether at T = 25. C. 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 = (1.0000 mol) (0.083145 L.bar/mol.K) (273.00 K) = 22.70 bar
V
(1.0000 L)
b) From the van der Waals equation ( a = 4.137 L2.atm/mol2 ; b = 0.0516 L/mol )
p =
nRT - an2
(V - nb)
V2
= (1.000 mol) (0.083145 L.bar/mol.K) (273.00 K) - (4.137 L2.atm/mol2) (1.0000 mol)2 1.01325 bar
[1.0000 L - (1.000 mol) (0.0516 L/mol)]
(1.0000 L)2
1. atm
= (23.934 – 4.192) bar = 19.74 bar
c) From the virial equation ( B = - 153.7 x 10-3 L/mol )
p = nRT { 1 + (B/Vm) }
Vm = V/n = (1.0000 L)/(1.0000 mol) = 1.0000 L/mol
= (1.0000 mol) (0.083145 L.bar/mol.K) (273.00 K) { 1 + (- 153.7 x 10-3 L/mol) }
(1.0000 L/mol)
= (22.699 bar) (1.0000 – 0.1537) = 19.21 bar
Notice that there are significant (~ 10 - 15%) differences among the three results. The van der Waals result is likely
to be closest to the true value for pressure for these conditions.
2) Our strategy will be to first find the number of moles of NO 2 and N2O4 at equilibrium, and then find the partial
pressures, from which the equilibrium constant can be calculated.
The moles of gas at equilibrium, assuming ideal behavior, is
n = pV = (851. torr) (1 atm/760 torr) (1.057 L) = 0.04717 mol
RT
(0.082057 L.atm/mol.K) (305.8 K)
The grams of gas is m = 59.165 g - 55.653 g = 3.512 g
Now we have the following two relationships involving nNO2 and nN2O4.
n = nNO2 + nN2O4
m = nNO2 MNO2 + nN2O4 MN2O4
From the first expression we may say nN2O4 = n - nNO2. If we substitute into the second expression, we get
m = nNO2 MNO2 + (n - nNO2) MN2O4
m = nNO2 (MNO2 - MN2O4) + n MN2O4
or, solving for nNO2,
nNO2 =
(m - n MN2O4) = (3.512 g - (0.04717 mol) (92.011 g/mol)) = 0.01800 mol NO2.
(MNO2 - MN2O4)
((46.006 g/mol) - (92.011 g/mol))
The moles of N2O4 is then
nN2O4 = 0.04717 mol - 0.01800 mol = 0.02917 mol
From the ideal gas law
p(NO2) = (0.01800 mol) (0.083145 L.bar/mol.K) (305.8 K) = 0.4330 bar
1.057 L
p(N2O4) = (0.02917 mol) (0.083145 L.bar/mol.K) (305.8 K) = 0.7017 bar
1.057 L
And so K = (0.7017) = 3.74
(0.4330)2
3) The van der Waals constants for CO2 are a = 3.610 L2.atm/mol2, b = 0.0429 L/mol. Using the equations in Table
1.7, we may say (experimental values are in parentheses)
pc = a/27b2 = (3.610 L2.atm/mol2)/27(0.0429 L/mol)2 = 72.6 atm (72.85 atm)
Vm,c = 3b = 3(0.0429 L/mol) = 0.1287 L/mol (0.0940 L/mol)
Tc = 8a/27bR = 8(3.610 L2.atm/mol2)/27(0.0429 L/mol)(0.082057 L.atm/mol.K) = 303.9 K (304.2 K)
There is good agreement between the van der Waals values for p c and Tc and the experimental results. The van der
Waals value for Vm,c is about 35% higher than the actual result. This can be attributed to the way the van der Waals
equation corrects for the volume occupied by molecules (treating them under dilute conditions, where it is pairs of
molecules that interact, instead of the multimolecule interactions near the point where condensation occurs).
4) The van der Waals coefficients for oxygen are a = 1.364 L2.atm/mol2, b = 0.0319 L/mol. The equations for p
(from the van der Waals equation) and Z (compressibility factor) are as follows
p=
RT
- a
(Vm – b) Vm2
Z = pVm
RT
As discussed in the hint, it is easiest to pick values for Vm and then find the corresponding values for p and Z. Data
are given below and plotted on the next page
Vm(L/mol)

2.000
1.000
0.600
0.400
0.300
0.200
0.150
0.140
0.130
0.120
p(atm)
0.0
12.2
24.1
39.5
58.4
76.7
112.3
147.8
158.1
170.2
184.7
Z
1.000
0.989
0.978
0.964
0.948
0.934
0.913
0.901
0.899
0.899
0.900
Vm(L/mol)
0.100
0.090
0.080
0.070
0.060
0.055
0.052
0.050
0.048
0.047
p(atm)
225.1
255.3
298.9
367.8
497.2
614.8
720.3
814.5
937.0
1012.8
Z
0.914
0.933
0.971
1.046
1.212
1.374
1.522
1.654
1.827
1.934
Plot of Z vs p
2
1.8
Z
1.6
1.4
1.2
1
0.8
0
200
400
600
800
1000
p (atm)
There is a shallow minimum at p  165. atm.
Exercise 1.8b
If we assume that the vapor obeys the ideal gas law, then
n = p =
16.0 x 103 N/m2
5.156 mol/m3
V RT (8.3145 J/mol.K) (373.2 K)
The molecular mass is
M = (mass density) = 0.6388 x 103 g/m3 = 123.9 g/mol
(molar density)
5.156 mol/m3
If we assume the vapor molecules have the formula Pj, then (since M(P) = 30.97 g/mol)
j = 123.9 g/mol = 4.00  4
30.97 g/mol
The formula for the vapor molecules is P4.
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
V RT
If we multiply both sides by M (molecular mass) and recall that m = nM, then
nM = m =  = pM
V
V
RT
Solving for M gives
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)
RT/p
(g/mol)
12.233
45.63
25.20
44.85
36.97
44.52
60.67
43.61
85.23
42.70
101.3
42.43
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
M (g/mol)
46
45
44
43
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
pVm = RT { 1 + (B/Vm) + (C/Vm2) + ... }
Dividing by RT gives
pVm = Z = 1 + B + C + ...
RT
Vm
Vm2
If we take the derivative of the above equation with respect to (1/V m) 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)
Z
0.00488
1.0096
0.00990
1.0270
0.01441
1.0347
0.02305
1.0564
0.03186
1.0790
0.03764
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
a nice result, since it gives Z  1 in the limit 1/Vm = 0, as expected. Therefore, B = 2.339 L/mol = 2339. cm3/mol.
If this value is compared to those given in Table 1.4, page 915 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
Z
1.06
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.