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Chemistry 1000 Lecture 18:The kinetic molecular theory of gases
Marc R. Roussel
October 11, 2018
Marc R. Roussel Kinetic molecular theory October 11, 2018 1 / 22
Ideal gases
The kinetic molecular theory of gases
Matter is in constant movement and, as we have seen, subject to avariety of intermolecular forces.
Can we use basic ideas from physics to connect the microscopic forcesacting on molecules to our everyday (macroscopic) world?
Yes, if we take a statistical approach.
This is made possible because of the very large size of Avogadro’snumber and with the help of the law of large numbers.
In this context, the law of large numbers says that the behavior of asystem containing many molecules is unlikely to deviate significantlyfrom the statistical average of the properties of the individualmolecules.
Marc R. Roussel Kinetic molecular theory October 11, 2018 2 / 22
Ideal gases
The Maxwell-Boltzmann distribution
One of the results of the kinetic molecular theory is theMaxwell-Boltzmann distribution of molecular speeds.
This is the probability distribution for the speeds (v) of molecules in agas:
f (v) = 4π
(M
2πRT
)3/2
exp
(−Mv2
2RT
)v2
where M is the molar mass of an isotopomer, R is the ideal gasconstant, and T is the absolute temperature.
Marc R. Roussel Kinetic molecular theory October 11, 2018 3 / 22
Ideal gases
Typical speeds and the Maxwell-Boltzmann distribution
0
0.0005
0.001
0.0015
0.002
0.0025
0.003
0 200 400 600 800 1000
f(v)
v/m s-1
most probable
average
root mean square
[16O2 at 298.15 K (25 ◦C)]
Marc R. Roussel Kinetic molecular theory October 11, 2018 4 / 22
Ideal gases
Maxwell-Boltzmann distribution for 16O2 at differenttemperatures
0
0.0005
0.001
0.0015
0.002
0.0025
0.003
0.0035
0.004
0 200 400 600 800 1000 1200 1400
f(v)
v/m s-1
100 K300 K500 K
Marc R. Roussel Kinetic molecular theory October 11, 2018 5 / 22
Ideal gases
Maxwell-Boltzmann distribution: Effect of mass
0
0.0005
0.001
0.0015
0.002
0.0025
0.003
0.0035
0 500 1000 1500 2000 2500 3000 3500 4000
f(v)
v/m s-1
1H2 (0.002 kg/mol)2H2 (0.004 kg/mol)
16O2 (0.032 kg/mol)35Cl2 (0.070 kg/mol)37Cl2 (0.074 kg/mol)
[T = 300 K]
Marc R. Roussel Kinetic molecular theory October 11, 2018 6 / 22
Ideal gases
Assumptions of the kinetic molecular theoryfor ideal gases
The particles (molecules or atoms) of the gas are small compared tothe average distance between them.
Corollary: The particles occupy a negligible fraction of the volume.
These particles are in constant motion.
There are no intermolecular forces acting between them, exceptduring collisions.
a good approximation for real gases provided the gas is at a sufficientlylow pressure so that the distance between the molecules is very large.
At constant temperature, the energy of the gas is constant.
Marc R. Roussel Kinetic molecular theory October 11, 2018 7 / 22
Ideal gases
Pressure of an ideal gas
Basic bits of physics we need:
Pressure: p = F/A
Newton’s second law: F = ma = m∆v∆t
Newton’s third law: For every action there is an equal and oppositereaction.
The pressure on the wall of a container will be the force exerted on itdue to collisions of molecules with the wall divided by the area of thewall.
This force will be the negative of the sum of the average forcesexperienced by all the molecules.
Marc R. Roussel Kinetic molecular theory October 11, 2018 8 / 22
Ideal gases
For simplicity, imagine a rectangular container containing an ideal gas.
Consider a single particle impacting the wall:
x
y
z
−vx
vx
We choose the coordinate system so that the x axis is perpendicularto the wall.
The y and z components of the velocity won’t affect the pressure onthis wall.
If the total energy is conserved, then on average, the x component ofthe velocity after collision is just the negative of this componentbefore collision.
Marc R. Roussel Kinetic molecular theory October 11, 2018 9 / 22
Ideal gases
∆vx = vx ,after − vx ,before = −vx − vx = −2vx
How often do collisions with this wall occur?
If Lx is the x dimension of the container, then the particle travels 2Lxbefore returning, so the time between collisions is ∆t = 2Lx/vx .
The average force experienced by one particle over time due tocollisions with this wall is therefore
Fx = m∆vx∆t
= −mv2x
Lx
If v2x is the average value of v2
x for all the molecules in the gas, thenthe force on the wall is
F =Nmv2
x
Lx
where N is the total number of molecules of gas.
Marc R. Roussel Kinetic molecular theory October 11, 2018 10 / 22
Ideal gases
The mean squared speed is
v2 = v2x + v2
y + v2z
There is no physical difference between the three directions in space,so v2
x = v2y = v2
z , from which we conclude that v2x = 1
3v2.
F =Nmv2
3Lx
p = F/A, so
p =Nmv2
3ALx=
Nmv2
3V
using the fact that the area of the wall times the distance betweenthe walls is the volume of the container.
pV =1
3Nmv2
Marc R. Roussel Kinetic molecular theory October 11, 2018 11 / 22
Ideal gases
Root mean squared speed and temperature
pV =1
3Nmv2
In this equation, m is the mass of one molecule and N is the numberof molecules.
We have N = nNA and m = M/NA.
∴ pV =1
3nMv2
Experimentally, we know that pV = nRT for dilute (ideal) gases.Combining the two, we get
1
3Mv2 = RT =⇒ v2 =
3RT
M=⇒
√v2 =
√3RT
M√v2 is the root mean squared (rms) speed.
Marc R. Roussel Kinetic molecular theory October 11, 2018 12 / 22
Ideal gases
Example: rms speed of N2
The calculation of rms speeds is straightforward, provided we use SIunits consistently.
The SI unit of mass is the kg.
For N2, M = 2(14.0067× 10−3 kg/mol) = 2.801 34× 10−2 kg/mol.
At room temperature, we would have
√v2 =
√3RT
M=
√3(8.314 472 J K−1mol−1)(293 K)
2.801 34× 10−2 kg/mol
= 511 m/s
Marc R. Roussel Kinetic molecular theory October 11, 2018 13 / 22
Ideal gases
The Boltzmann constant
Recall the ideal gas equation
pV = nRT
If we want to rewrite the ideal gas equation in terms of the number ofmolecules (rather than the number of moles of molecules), we usen = N/NA:
pV = (N/NA)RT = N(R/NA)T
R/NA ≡ kB is Boltzmann’s constant.It is the ideal gas constant on a per molecule basis.
pV = NkBT
kB = 1.380 649× 10−23 J K−1
Marc R. Roussel Kinetic molecular theory October 11, 2018 14 / 22
Ideal gases
Average kinetic energy
pV =1
3Nmv2
The average kinetic energy is K = 12mv2, so
pV =2
3NK
Since pV = NkBT , equating the two expressions for pV gives
K = 32kBT
Marc R. Roussel Kinetic molecular theory October 11, 2018 15 / 22
Nonideal gases
Why and when the ideal gas law breaks down
Not all gases behave ideally under all conditions.
Intermolecular forces can be significant.
The volume taken up by the molecules can be a significant fraction ofthe total volume of the container.
Both of these effects become more important as the density of thegas increases.
The density is proportional to
n
V=
p
RT
so nonideal effects should become important at high pressures or atlow temperatures.
Marc R. Roussel Kinetic molecular theory October 11, 2018 16 / 22
Nonideal gases
Excluded volume
The molecules occupy some of the volume of the container.
The volume available for them to move in is therefore less than thetotal volume of the container.
We can correct for this by subtracting the excluded volume, whichwill be proportional to the number of molecules, from the totalvolume in the ideal gas law:
p(V − nb) = nRT
b is a constant determined experimentally and is about four times thevolume of a molecule times Avogadro’s constant.
Marc R. Roussel Kinetic molecular theory October 11, 2018 17 / 22
Nonideal gases
Intermolecular forces
Provided the density isn’t too high, intermolecular forces are primarilyattractive, as discussed in a previous lecture.
Attractive forces will tend to decrease the pressure:As a molecule approaches the container wall, there is an imbalancebetween the number of molecules ahead of it and the number behind.The force of attraction from molecules behind provide a braking forcewhich slows the approach of a molecule to the wall, and thusdecreases the force of impact.
Marc R. Roussel Kinetic molecular theory October 11, 2018 18 / 22
Nonideal gases
The attractive force is found to depend on the square of the density.
Including this correction in the equation of state for a gas would give(p +
an2
V 2
)V = nRT
where a is a constant determined experimentally that depends on thestrength of the intermolecular forces, and thus on the particular gaswe are studying.
Marc R. Roussel Kinetic molecular theory October 11, 2018 19 / 22
Nonideal gases
van der Waals equation
Putting both corrections together, we get the van der Waals equation:(p +
an2
V 2
)(V − nb) = nRT
Solving the vdW equation for p isn’t too difficult:
p =nRT
V − nb− an2
V 2
Marc R. Roussel Kinetic molecular theory October 11, 2018 20 / 22
Nonideal gases
Example
For N2, a = 0.1408 Pa m6mol−2 and b = 3.91× 10−5 m3/mol.If we have 40 mol of N2 in 1.0 m3 at 298 K, then
p =nRT
V − nb− an2
V 2
p =(40 mol)(8.314 472 J K−1mol−1)(298 K)
1.0 m3 − (40 mol)(3.91× 10−5 m3/mol)
− (0.1408 Pa m6mol−2)(40 mol)2
(1.0 m3)2
=(40 mol)(8.314 472 J K−1mol−1)(298 K)
0.9984 m3
− (0.1408 Pa m6mol−2)(40 mol)2
(1.0 m3)2
= 99 264− 225 Pa = 99 kPa
Marc R. Roussel Kinetic molecular theory October 11, 2018 21 / 22
Nonideal gases
Example
Suppose we have 4000 mol of N2 in 1.0 m3 at 298 K:
p =(4000 mol)(8.314 472 J K−1mol−1)(298 K)
1.0 m3 − (4000 mol)(3.91× 10−5 m3/mol)
− (0.1408 Pa m6mol−2)(4000 mol)2
(1.0 m3)2
=(4000 mol)(8.314 472 J K−1mol−1)(298 K)
0.8436 m3
− (0.1408 Pa m6mol−2)(4000 mol)2
(1.0 m3)2
= 1.175× 107 − 2.25× 106 Pa = 9.5 MPa
Using the ideal gas law, we would have predicted 9.9 MPa.
Marc R. Roussel Kinetic molecular theory October 11, 2018 22 / 22