Kinetic-molecular theory of gases, Distribution of molecular Speeds in gas kinetics. Diffusion and applications, Effusion. Using the van der Waals equation to calculate the pressure of a gas.

1. Kinetics of gases
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.

2. Kinetic-Molecular Theory of Gases
It is based on the model outlined as follows.
A gas is composed of a very large number of extremely small particles (molecules or, in
some cases, atoms) in constant, random, straight-line motion.
Molecules of a gas are separated by great distances. The gas is mostly empty space. (The
molecules are treated as so-called point masses, as though they have mass but no volume.)
Molecules collide only fleetingly with one another and with the walls of their container,
and most of the time molecules are not colliding.
There are assumed to be no forces between molecules except very briefly during collisions.
That is, each molecule acts independently of all the others and is unaffected by their
presence, except during collisions.
Individual molecules may gain or lose energy as a result of collisions. In a collection of
molecules at constant temperature, however, the total energy remains constant.

3. Derivation of Boyle s Law
Boyle s law was stated mathematically in equation 𝑃𝑉 = 𝑎
The value of a depends on the number, N, of molecules in the sample and the temperature, T.
Let’s focus on a molecule traveling along the x direction toward a wall perpendicular to its
path. The speed of the molecule is denoted by ux. The force exerted on the wall by the
molecule depends on the following factors.
The frequency of molecular collisions
𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 ∝ 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑 𝑥 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒
𝑐𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 ∝ 𝑢 𝑥 𝑥
𝑁
𝑉
The momentum transfer, or impulse.
𝑖𝑚𝑝𝑢𝑙𝑠𝑒 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 ∝ 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑥 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑
𝑖𝑚𝑝𝑢𝑙𝑠𝑒 ∝ 𝑚𝑢 𝑥

4. The pressure of a gas (P) is the product of impulse and collision frequency. Thus, the
complete proportionality expression for factors that affect pressure is
𝑃 ∝ 𝑚𝑢 𝑥 𝑥 𝑢 𝑥 𝑥
𝑁
𝑉
∝
𝑁
𝑉
𝑚 𝑢 𝑥
2
At any instant, however, the molecules in a gas sample are traveling at different
speeds. Therefore, we must replace 𝑢 𝑥
2 in the expression above with the average
value of which is denoted by (𝑈 𝑥
2
)
𝑃 ∝
𝑁
𝑉
𝑢 𝑥
2
we should expect that 𝑈 𝑥
2
= 𝑈 𝑦
2
= 𝑈𝑧
2
=
1
3
𝑈2, where the quantity is the average
value of U2. Thus
𝑃 =
1
3
𝑁 𝑚
𝑉
𝑢 𝑥
2
𝑃𝑉 =
1
3
𝑁 𝑚 𝑢 𝑥
2

5. Distribution of Molecular Speeds
The Maxwell equation for the distribution of
speeds gives us fraction of molecules that
have speed ‘u’ is
𝐹(𝑢) = 4𝜋
𝑀
2𝜋𝑅𝑇
3/2
𝑢3
𝑒−(
𝑀𝑢2
2𝑅𝑇)
The distribution depends on the product of two opposing factors: a factor that is
proportional to U and an exponential factor, 𝑒−(
𝑀𝑢2
2𝑅𝑇
)
.
As u increases, the factor increases from a value of zero, while the exponential factor
decreases from a value of one. The factor favors the presence of molecules with high
speeds and is responsible for there being few molecules with speeds near zero

6. The distributions for O2 (g) at 273 K and at 273
K and H2 (g) at 273 K reveal that the lighter the
gas, the broader the range of speeds.
The most probable speed, or modal speed, is
denoted by um , more molecules have this speed
than any other speed. The average speed is
denoted by Uav. The root-mean-square speed,
Urms, is obtained from the average of u2
𝑢 𝑟𝑚𝑠 = 𝑢2 . The root-mean-square, speed is
of particular interest because we can use the
value of urms to calculate the average kinetic
energy, 𝑒 𝑘 of a collection of molecules.
𝑒 𝑘 =
1
2
𝑚 𝑢2
=
1
2
𝑚 𝑢 𝑟𝑚𝑠
2

7. PV = nRT
We also know that
𝑃𝑉 =
1
3
𝑁𝐴 𝑚 𝑢2
3 𝑃𝑉 = 𝑁𝐴 𝑚 𝑢2
3 𝑅𝑇 = 𝑁𝐴 𝑚 𝑢2
the product NAm represents the mass of 1 mol of molecules, the molar mass, M.
3 𝑅𝑇 = 𝑀𝑢2
𝑢 𝑟𝑚𝑠 = 𝑢2 =
3𝑅𝑇
𝑀
6.20

8. Equation (6.20) shows that urms of a gas is directly proportional
to the square root of its Kelvin temperature and inversely
proportional to the square root of its molar mass.
From this we infer that, on average,
1. molecular speeds increase as the temperature increases, and
2. lighter gas molecules have greater speeds than do heavier
ones

9. The Meaning of Temperature
The basic equation of the kinetic-molecular theory
𝑃𝑉 =
1
3
𝑁𝐴 𝑚 𝑢2
modify it slightly 𝑃𝑉 =
2
3
𝑥
1
2
𝑁𝐴 𝑚 𝑢2
The quantity represents the average translational kinetic energy, 𝑒 𝑘 of a
collection of molecules.
1
2
𝑚 𝑢2
𝑃𝑉 =
2
3
𝑁𝐴
1
2
𝑚 𝑢2 =
2
3
𝑁𝐴 𝑒 𝑘

10. 𝑅𝑇 =
2
3
𝑁𝐴 𝑒 𝑘
𝑒 𝑘 =
3
2
𝑅𝑇
𝑁𝐴
Because R and are constants, equation (6.21) simply states that
𝑒 𝑘 = constant * T … 6.22
The Kelvin temperature (T)of a gas is directly proportional to the average translational
kinetic energy 𝑒 𝑘 of its molecules
Molecules in the hotter object have, on average, higher kinetic energies than do the
molecules in the colder object. When the two objects are placed into contact with each
other, molecules in the hotter object give up some of their kinetic energy through
collisions with molecules in the colder object until the temperatures become equalized
the absolute zero of temperature, is the temperature at which translational molecular
motion should cease.

11. Gas Properties Relating to the Kinetic-Molecular Theory
Diffusion is the migration of molecules as a result of random molecular motion
The diffusion of two or more gases results in an intermingling of the molecules and, in a
closed container, soon produces a homogeneous mixture.
Effusion, is the escape of gas molecules from their container through a tiny orifice or pinhole
The rate at which effusion occurs is directly proportional to molecular speeds. That is,
molecules with high speeds effuse faster than molecules with low speeds.
Let us consider the effusion of two different gases at the same temperature and pressure
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐴
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝐵
=
𝑈𝑟𝑚𝑠 𝐴
𝑈𝑟𝑚𝑠 𝐴
=
3𝑅𝑇/𝑀𝐴
3𝑅𝑇/𝑀 𝐵
=
𝑀𝐴
𝑀 𝐵

12. The rate of effusion of a gas is inversely proportional to the square root of
its molar mass … Graham’s law
Graham s law has serious limitations :
It can be used to describe effusion only for gases at very low pressures, so that
molecules escape through an orifice individually, not as a jet of gas.
Also, the orifice must be tiny so that no collisions occur as molecules pass through.
When effusion takes place under the restrictions described above, equation (6.23) can be
used to determine which of two gases effuses faster,
𝑟𝑎𝑡𝑖𝑜 𝑜𝑓
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑𝑠
𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒𝑠
𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑡𝑖𝑚𝑒𝑠
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑏𝑦 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑔𝑎𝑠 𝑒𝑓𝑓𝑢𝑠𝑒𝑑
= 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡𝑤𝑜 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠𝑒𝑠 6.25

13. Applications of Diffusion
Addition of mercaptants (CH3SH) to liquefied petroleum gas
(LPG) to detect leaks. Liquefied petroleum gas (LPG) are odorless.
The mercaptan has an odor that can be detected in parts per billion
(ppb) or less. When a leak occurs, which can lead to asphyxiation
or an explosion, we rely on the diffusion of this odorous compound
for a warning.
Separation of 235U from 238 U. the Manhattan Project. The method
is based on the fact that uranium hexafluoride is one of the few
compounds of uranium that can be obtained as a gas at moderate
temperatures

14. Nonideal (Real) Gases
The measure of how much a gas deviates
from ideal gas behavior is found in its
compressibility factor.
The compressibility factor of a gas is the
Ratio PV/nRT. For an ideal gas it is = 1
At 300 K and 10 bar, He, H2, CO, N2 and O2
and behave almost ideally PV/nRT ≈ 1
But NH3 and SF6 do not PV/nRT ≈ 0.88
Conclusion from this plot is that all gases behave ideally at sufficiently low pressures,
say, below 1 atm, but that deviations set in at increased pressures.

15. To explain the significance of the term an2/V2 in the modified pressure factor,
Provided V is not too small, the first term in the equation above is approximately equal to
the pressure exerted by an ideal gas,
𝑛𝑅𝑇
𝑉 − 𝑛𝑏
≈
𝑛𝑅𝑇
𝑉
= 𝑃𝑖𝑑𝑒𝑎𝑙
𝑃 =
𝑛𝑅𝑇
𝑉 − 𝑛𝑏
−
𝑎𝑛2
𝑉2
The equation given above for P predicts that the pressure exerted by a real gas will be
less than that of an ideal gas.
The term an2/V2 takes into account the decrease in pressure caused by intermolecular
attractions

16. The van der Waals Equation
For modeling the behavior of real gases, the van der Waals equation, equation is the
simplest to use and interpret
𝑃 +
𝑎𝑛2
𝑉2
𝑉 − 𝑛𝑏 = 𝑛𝑅𝑇 … 6.26
The equation incorporates two molecular parameters, a and b, whose values vary from
molecule to molecule
𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑓𝑐𝑡𝑜𝑟 𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑛𝑅𝑇
The van der Waals equation uses
A modified pressure factor P + an2/V2
A modified volume factor V-nb

17. The proportionality constant, a, provides a measure of how strongly the molecules attract
each other.
the values of both a and b increase as the sizes of the molecules increase.
The smaller the values of a and b, the more closely the gas resembles an ideal gas.
Deviations from ideality, as measured by the compressibility factor, become more
pronounced as the values of a and b increase.

18. Using the van der Waals Equation to Calculate the
Pressure of a Gas
Example 6.17
Use the van der Waals equation to calculate the pressure exerted by 1.00 mol Cl2 (g)
confined to a volume of 2.00 L at 273 K. The value of and that of a = 6.49 L2 atm mol-2
and that of b= 0.0562 L mol-1
𝑃 =
𝑛𝑅𝑇
𝑉 − 𝑛𝑏
−
𝑎𝑛2
𝑉2
Then substitute the following values into the equation.
Solution
n = 1.00 mol; V = 2.00 L; T = 273 K; R = 0.08206 atm L mol-1 K-1
n2a = (1.00)2 mol2 * 6.49 L2 atm /mol2 = 6.49 L2 atm

19. nb = 1.00 mol * 0.0562 L mol-1 = 0.0562 L
𝑃 =
1.00 𝑚𝑜𝑙 𝑥 0.8206 𝑎𝑡𝑚 𝐿 𝑚𝑜𝑙
− 1 𝐾
− 1 𝑥 273 𝐾
2.00 − 0.0562 𝐿
−
6.49 𝐿2 𝑎𝑡𝑚
(2.00)2 𝐿2
P = 11.5 atm - 1.62 atm = 9.9 atm