#### xaktly | Chemistry

Kinetic-molecular theory

### Developing a model for the behavior of gases

In this section we will develop the ideal gas law from first principles, that is, by considering the behavior of individual particles in an ensemble of gas particles. "Particles" will be our generic term for atoms and/or molecules.

Before we begin, we have to realize that this is a pretty complicated task, so we have to make some approximations. That's the way we often have to move forward in science: We make approximations to come up with a simple model, then ask, if we remove the constraints of our approximations (like adding friction in projectile motion), how does our model change, and can we then add that change to it?

Our assumptions are pretty simple →

1. Gases consist of spherical particles so small the distance between gas particles (atoms or molecules) is always much smaller than the distance between them.

2. The gas particles have the same mass.
1. The number of molecules is so large that an average over, say particle speed, is meaningful.

2. The particles are in constant motion, and undergo perfectly elastic collisions with other particles and with the walls of the container. Interactions, especially attraction, between particles is negligible (i.e. they exert no forces on one another).

3. The temperature of the gas depends on the average kinetic energy of the particles.

4. The time between collisions with the container walls is smaller than the time between collisions with other particles.
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### Ensemble

In chemistry and physics, ensemble means a collection of particles – atoms and/or molecules – large enough that averages of their properties, such as speed or kinetic energy, are meaningful.

An ensemble of particles is a theoretical grouping that helps us think through how individual atoms or molecules contribute to the bulk properties of gases, liquids or solids.

### Force on container walls – pressure

Consider some kind of container holding gas "particles." We'll call them particles, though they could be atoms, like He or Ar, or molecules, like CO2 or O2. The particles are in constant motion in three dimensions. They have a range of velocities, but there are enough of them that an average velocity is a meaningful number. Here we're looking at the container from one side. If we look at just one gas particle with average velocity v, we can decompose that velocity vector into velocity vx, toward the wall, and vy, parallel to it. For the purpose of calculating the force on the wall, we only care about the component of v perpendicular to the wall, vx If we use that velocity to calculate the momentum of the particle in the x-direction, px = mvx, we have this simplified picture. Now we're interested in how much force each particle exerts on the container wall as it hits it, so we'll have to let it bounce off the wall. Remember that one of our approximations is that this will be a perfectly-elastic collision – no attractive forces between particles and walls. In the picture below, the inward and outward momentum vectors are angled just a bit so they won't overlap, but it's easy to see that the change in momentum is the forward momentum minus the backward momentum (a negative number). So the momentum change perpendicular to the wall upon bouncing off of it is Δpx = 2px.

#### Time between collisions with the walls

In order to calculate the total force that any particle exerts on the container walls, it's necessary to know how often it hits a wall. More hits in a given time means more force over that time which means more pressure (remember that pressure is force divided by area). The particle will hit any one side wall, say the left one in the figure above, after traveling a distance of 2L, where L is the distance between walls. Using the velocity definition v = distance/time, we can calculate Δt.

$$\Delta t = \frac{2L}{v_x}$$

#### Force on the walls

Now from our studies of momentum and impulse (Δp), we know that

$$F = \frac{\Delta p}{\Delta t}$$

where Δp is the impulse and Δt is the time between collisions. If px = mvx is the momentum in the x-direction we get the force of a particle on the wall as a function of its velocity, mass and the length of the container:

$$F \ \frac{\Delta p}{\Delta t} = \frac{2 mv_x}{\Delta t} = \frac{mv_x^2}{L}$$

Make sure you work through that last line to see where the 2's come from and why they cancel.

Now if we have an ensemble of N particles, each of which are bouncing off our wall with the average force (because they can all be considered to be moving with the average velocity), the total force on a wall is

$$F = \frac{N m \bar{v}_x^2}{L}$$

Here we've put a bar over the vx to indicate that we're talking about the average velocity in the x-direction.

Now in our container, there's nothing special about the x-direction, and we'd expect that the average velocities in all three dimensions would be equal

$$\bar{v}_x^2 = \bar{v}_y^2 = \bar{v}_z^2$$

Using the squares of the velocities is important because velocity is a vector, and with equal probability of moving in either of two opposite directions, the x-, y- and z-velocities would be zero if we didn't square. We can always take a root when we need to get to the velocity.

Now the Pythagorean theorem for a three-dimensional velocity vector with x-, y- and z-components is

$$\bar{v}^2 = \bar{v}_x^2 + \bar{v}_y^2 + \bar{v}_z^2$$

Now using the equivalence of the squares of the velocities, we can rewrite that as

$$\bar{v}^2 = 3 \bar{v}_x^2$$

And we can solve that for the x-component:

$$\bar{v}_x^2 = \frac{1}{3} \bar{v}^2$$

Now we can plug that into our formula for the force, to express it as a function of the average velocity – forget about components now – of the particles in the container:

$$F = \frac{Nm\bar{v}^2}{3L}$$

#### Finding the pressure

Now pressure is force divided by area: P = F/A, and the area of a wall of our container is A = L2. So we can convert to pressure like this:

$$P = \frac{Nm\bar{v}^2}{3L^3}$$

Now if L3 is the volume we can write the pressure as

$$P = \frac{Nm\bar{v}^2}{3V}$$

The average kinetic energy is in there

Now the average kinetic energy of a particle in our ensemble is

$$KE = \frac{1}{2} mv^2$$

and if we insert that into the pressure equation we get

\begin{align} P &= \left( \frac{N}{3V} \right) \left( \frac{1}{2} mv^2 \right) \cdot 2 \\ \\ &= \frac{2}{3} \frac{n\cdot KE}{V} \end{align}

Now this is a remarkable place to pause. We've just related energy, the average kinetic energy of the particles in our gas, to the pressure and volume of the gas.

That expression is usually written in terms of the pressure-volume product, like this:

$$PV = \frac{2}{3} N\cdot KE$$

#### Relationship between pressure, volume and kinetic energy

The average kinetic energy of the particles of a gas is related to the product of pressure and volume (PV) by Because N is just a number (no units) it follows that the product, PV, has units of energy.

### The ideal gas law & average KE

In the early days of exploring the properties of gases, Boyle, Charles, Gay-Lussac and Clapeyron developed a few important empirical laws of gas behavior, relating pressure, temperature, volume and number of particles. The most important relationship, due to Clapeyron, showed that the product of pressure and volume (PV) is proportional to the number of particles and the temperature:

$$PV = NkT$$

where k is a proportionality constant known as the Boltzmann constant, k = 1.381 × 10-23 J/K

The constant can be viewed merely as being there to get the units right. The product kT has units of energy (Joules, J)

If we equate the theoretical and empirical formulas for PV,

$$PV = \frac{2}{3} N\cdot KE \; \text{ and } \; PV = NkT$$

The transitive property gives us

$$\frac{2}{3} N\cdot KE = NkT$$

Canceling the N's and solving for the average kinetic energy per particle, KE gives us a remarkable result:

$$KE = \frac{3}{2} kT$$

The average kinetic energy of the particles of a gas depends only on the temperature and can be approximated, to very good measure, with this formula.

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### empirical

An empirical rule or law is one that is based on experiment and observation instead of pure mathematical logic. Empirical observations inform theoretical investigations, and theory invites experiment in order to confirm, disprove or improve the theory.

#### Translational kinetic energy of gas particles

The average translational kinetic energy of the atoms or molecules of a gas is where k = 1.381 × 10-23 J/K is the Boltzmann constant, and T is the temperature in Kelvin.

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### Translational KE

The translational kinetic energy is the energy of movement in 3D space, along the x-, y- and z-directions. It does not include the kinetic energy of rotation of molecules or vibration of molecular bonds, neither of which is a factor for an ideal gas of microscopic particles.

### Connection between the microscopic and macroscopic worlds

The empirically-derived ideal gas law, PV = NkT and the kinetic molecular theory of an ideal gas give us a connection between the measurable properties of gases and the average kinetic energy of the gas particles.

If we let R be the Boltzmann constant multiplied by Avogadro's number,

\begin{align} \left(1.38065 \times 10^{-23} \, \frac{J}{K}\right)&(6.02215 \times 10^23/mol) \\ &= 8.314 \frac{J}{mol\cdot K} \end{align}

We get the molar gas constant R = 8.314 J/mol·K or 8.314 J·mol-1·K-1.

Then the ideal gas law can be rewritten,

$$PV = nRT$$

where n is the number of moles instead of the number of particles (N).

#### The ideal gas law

The ideal gas law has two forms: where N is the number of gas particles (atoms or molecules), n is the number of moles, k = 1.381 × 10-23 J/K is the Boltzmann constant and R = 8.314 J/mol·K is the molar gas constant.  xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.