Chemistry 1000 Lecture 18: The kinetic molecular theory of...

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.


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.


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)]


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


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]


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.


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.


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.


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.


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


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.


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


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


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


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.


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.


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.


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.


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


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


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.


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