CHM 3400 – Problem Set 1
Due date: Wednesday, January 16th
Do all of the following problems. Show your work. (NOTE: Conversion factors between different pressure units are
given in Table 0.1, p 5 of Atkins. Values for R in various units are given in Table 1.1, p 16 of Atkins. Note that 1
dm3 = 1 L.)
1) The density of oxygen gas (O2, M = 32.00 g/mol) in a new oxygen gas cylinder, measured at T = 20.0 C, is
0.1632 g/mL. Find the following:
a) The number density of oxygen molecules in the gas cylinder (in units of molecules/cm3).
b) The calculated pressure of gas in the cylinder, using the ideal gas law (Atkins, eq 1.2). Give your answer
in units of atm, to four significant figures.
c) The calculated pressure of gas in the cylinder, using the van der Waals equation (Atkins, eq 1.23a). Give
your answer in units of atm, to four significant figures. The van der Waals coefficients for O 2 are a = 1.364
L2atm/mol2, b = 0.0319 L/mol.
d) Which calculated value for pressure (ideal or van der Waals) do you expect to be closer to the true value
for gas pressure for the conditions in the problem? Why?
2) 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?
3) The atmosphere of Mars is composed almost entirely of carbon dioxide (CO 2, M = 44.01 g/mol). In this problem
we will assume the atmosphere is pure CO2. At the surface of Mars typical values for pressure and temperature are p
= 4.5 torr, T = 220. K. Find the following for these conditions of pressure and temperature:
a) crms, the rms average speed of a CO2 molcule in the atmosphere at surface level (Atkins, eq 1.15). Give
your answer in units of m/s.
b) , the mean free path of a CO2 molecule in the atmosphere at surface level (Atkins, eq 1.19). Give your
answer in units of nm.
c) z, the collision frequency for a CO2 molecule in the atmosphere at surface level (Atkins, eq 1.19). Give
your answer in units of collisions/s.
The value for  for CO2 is given in Table 1.3, p 28 of Atkins. Note that if all of the quantities appearing in
eq 1.15 and eq 1.19 are given in MKS units the values calculated using those equations will also have MKS units.
Also, c, , and z are related to one another by eq 1.18 of Atkins.
4) Consider a mixture of argon (Ar, M = 39.95 g/mol) and helium (He, M = 4.00 g/mol). The density of the mixture,
measured at T = 22.8 C and p = 791. torr, is  = 0.662 g/L. What is XAr, the mol fraction of argon in the gas
mixture? You may assume that argon and helium behave ideally for the conditions in the problem.
5) A “generic” phase diagram for a typical pure chemical substance is given in Fig 5.12, p 115 of Atkins.
a) What is the significance of the critical point in the phase diagram?
b) The van der Waals equation is the simplest equation of state that can be used to describe critical behavior
for pure substances. Use the values for the van der Waals a and b coefficients for carbon dioxide (Table 1.5, p 34 of
Atkins) to estimate the values for the critical constants for CO2 (pC, VC, and TC, see eq 1.24). Compare your values
to the experimental values (pC = 72.85 atm, VC = 94.0 cm3/mol, and TC = 304.2 K), and briefly discuss the
agreement between the experimental values and those predicted using the van der Waals coefficients.
6) The Maxwell distribution of molecular speeds in an ideal gas is given by eq 1.16 of Atkins. Starting with this
equation find an expression for cmp, the most probable speed of a molecule in a gas. HINT: The most probable speed
corresponds to the peak in a plot of number of molecules vs speed (see Atkins, Fig 1.8). Using the condition for an
extreme point in a function you can find an expression for cmp.
Solutions.
1)
Before we begin, we will convert the density of the gas (in g/mL) to the molar density (in mol/mL and in
mol/L). If we let  = m/V be the density of the gas, and M be the molecular mass of the gas, then
Vm = V = M = (32.00 g/mol) = 196.1 mL/mol = 0.1961 L/mol
n

(0.1632 g/mL)
a) Using the result above, the number density of molecules in the gas is
N =
V
1 mol
196.1cm3
6.022 x 1023 molecule = 3.071 x 1021 molecule/cm3
mole
b) From the ideal gas law
p = nRT = RT = (0.08206 Latm/molK) (293.15 K) = 122.7 atm
V
Vm
(0.1961 L/mol)
c) From the van der Waals equation ( a = 1.364 L2atm/mol2 ; b = 0.0319 L/mol )
p =
RT (Vm - b)
=
a
Vm2
(0.08206 Lbar/molK) (293.15 K) [ (0.1961 L/mol) - (0.0319 L/mol) ]
(1.364 L2atm/mol2)
(0.1961 L/mol) 2
= 146.50 atm - 35.47 atm = 111.0 atm
The percent difference between the value calculated using the ideal gas law and that found using the van der Waals
equation of state is
% difference = | 122.7 – 111.0 | . 100. % = 10.5 %
111.0
which is a significant difference.
d) We would expect the value calculated using the van der Waals equation to be closer to the true value for
pressure than that found using the ideal gas law, because the van der Waals gas contains corrections for the
approximations made in the ideal gas law.
2) If we assume that the vapor obeys the ideal gas law, then
n = p =
16.0 x 103 N/m2
=
V RT (8.3145 J/mol.K) (373.2 K)
5.156 mol/m3
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.
3)
a)
crms = (3RT/M)1/2 = [3(8.3145 J/molK)(220. K)/(44.01 x 10-3 kg/mol) ]1/2 = 353. m/s
b)
NA = 6.022 x 1023 molecule/mol
 = 0.52 nm2 = 0.52 x 10-18 m2
p = 4.5 torr (1 atm/760 torr) (1.01325 x 10 5 Pa/atm) = 600.0 Pa
=
RT
=
(8.3145 J/molK) (220. K)
21/2NAp
21/2 (6.022 x 1023 molecule/mol) (0.52 x 10-18 m2) (600.0Pa)
= 6.88 x 10-6 m = 6880. nm
c)
z = c/ =
(353. m/s)
= 5.13 x 107 collisions/s
-6
(6.88 x 10 m)
4) Assume you have 1.000 L of gas. Based on the density, the mass of gas you have is m = 0.662 g. The moles of
gas may be found using the ideal gas law
n = pV = (791. torr) (1. atm/760. torr) (1.000 L) = 0.04286 mol gas
RT
(0.08206 Latm/molK) (295.95 K)
Let
ntotal = nHe + nAr = 0.04286 mol be the total number of moles of gas. Then
nHe = ntotal – nAr
m = nHeMHe + mAr MAr = (ntotal – nAr)MHe + nArMAr = ntotalMHe + nAr(MAr – MHe)
or, solving for nAr
nAr = m – ntotalMHe = [ (0.662 g – (0.04286 mol) (4.00 g/mol) ] = 0.01365 mol Ar
(MAr – MHe)
[ (39.95 g/mol) – (4.00 g/mol) ]
The mol fraction of argon in the gas is then
XAr = nAr = 0.01365 mol = 0.318
ntotal
0.04286 mol
5)
a) The significance of the critical point is most easily seen by considering the critical temperature, T C. For a
substance initially in the gas phase, the substance can be converted from gas to condensed phase (solid or liquid) by
an isothermal reversible compression if T < T C. If T > TC an isothermal reversible compression will never result in a
phase transition (that is, the properties of the substance will change continuously from those of a gas at low pressure
to those of a liquid at high pressure, but without a phase transition ever taking place). p C and VC represent the values
for pressure and temperature at the critical temperature on the liquid-gas phase boundary.
b)
For CO2, a = 3.610 L2.atm/mol2, and b = 0.0429 L/mol. So (eq 1.24)
VC = 3 (0.0429 L/mol) = 0.1287 L/mol = 128.7 cm3/mol (experimental value is 94.0 cm3/mol)
TC =
8a =
8 (3.610 L2.atm/mol2)
27Rb
27 (0.08206 L.atm/mol.K)(0.0429 L/mol)
= 303.8 K (experimental value is 304.2 K)
pC =
a = (3.610 L2.atm/mol2) = 72.65 atm (experimental value is 72.85 atm)
27b2
27 (0.0429 L/mol)2
The values for the critical temperature and pressure calculated using the van der Waals coefficients are within a few
tenths of a percent of the experimental value. The value for V C calculated from the van der Waals coefficients is
37% higher than the experimental value. The van der Waals equation typically does a good job in predicting T C and
pC, but usually overestimates the value for VC.
6)
The Maxwell distribution for speeds in an ideal gas is given by the expression
f(s) ds = 4 (M/2RT)3/2 s2 exp(-Ms2/2RT) ds
= A s2 exp(-Bs2) ds
where
A = 4 (M/2RT)3/2
B = M/2RT
At an extreme point in f(s) the first derivative will be equal to zero. So
df(s)/ds = 0 = d/ds { A s2 exp(-Bs2) } = A {2s exp(-Bs2) + s2 exp(-Bs2) (- 2Bs) }
= 2As exp(-Bs2) { 1 – Bs2 }
The above will be equal to zero when the term in brackets is equal to zero. Therefore
1 – Bs2 = 0
1 = Bs2
s2 = 1/B, or s = cmp = (1/B)1/2 = (2RT/M)1/2
is the most probable speed for a molecule in an ideal gas.
Note that we have, strictly speaking, only found an extreme point for the function. To show that it is a
maximum we could use the second derivative test (maximum if d2f(s)/ds < 0). However, based on our discussion of
the Maxwell distribution in class we know the extreme point we have found is a maximum.