Physics

The Physics Philes, lesson 127: Kinetic-Molecular Model of Ideal Gases, Part 2

It’s been a couple of weeks, but now it’s time to continue our discussion of the kinetic-molecular model of ideal gases. Let’s have a little discussion on pressure and the kinetic energies of molecules.

Let’s think about the absolute value of the velocity in the x-direction from last time. That value is not actually the same for all molecules, as you may have guessed. However, we could separate all the different velocities and determine their effect on the pressure. But in doing so, we’re basically replacing the square of the x-velocity in the pressure equation with the square of the average x-velocity. This average value is related to the speeds of the molecules. We know that

Screen shot 2015-01-18 at 10.34.10 AM

We can find the average of this relationship over all molecules in a similar way:

Screen shot 2015-01-18 at 10.37.50 AM

It turns out, though, that in our model there is no difference between the x-, y-, and z-directions. So the square of the velocity is can be represented as three times the velocity in the x-direction. Or, if we move the terms around a little bit:

Screen shot 2015-01-18 at 10.45.10 AM

So, instead of using the square of the velocity in our equation for pressure, we can use this. The equation becomes

Screen shot 2015-01-18 at 11.00.11 AM

Notice that term in the brackets. Look familiar? Of course it does! That’s the average translational kinetic energy! This term, multiplied by the total number of molecules N, is the total random kinetic energy of the the molecules’ translational motion. That means that we can write the above equation in an even simpler way:

Screen shot 2015-01-18 at 11.13.21 AM

That’s not all! Remember the ideal gas equation? You know, pV = nRT? If we set these two pV equations equal to each other and solve for the kinetic energy, we get

Screen shot 2015-01-18 at 11.22.24 AM

And just like that, we have the equation for the average translational kinetic energy for a given number of moles of an ideal gas. What this equation shows us that the kinetic energy is directly proportional to the temperature T.

We’re still not done, because we can look at this whole kinetic energy situation in a slightly different way. If the last equation shows the total translational kinetic energy per mole, we can do some mathematical manipulation to get an equation that shows the translational kinetic energy per molecule. Let’s get to work.

To find the average translational kinetic energy of a single molecule, we need to divide the total translational kinetic energy by the number of molecules N. In math, that looks like this:

Screen shot 2015-01-18 at 11.35.53 AM

Remember that we can find the total number of molecules N by multiplying the number of moles n by Avagadro’s number:

Screen shot 2015-01-18 at 11.38.59 AM

We can work this into our equation and get

Screen shot 2015-01-18 at 11.43.18 AM

Stay with me because this is about to get much simpler. The ratio R/N_A is often referred to as the Boltzmann constant k. It’s equal to 1.38 x 10^-23 J/molecule · K. That means that we can write this equation like this:

Screen shot 2015-01-18 at 11.46.31 AM

This is the average translational kinetic energy for a single molecule. Like the per mole situation, the translational kinetic energy is directly proportional to the temperature. Neat, right?

One last thing: We can use the Boltzmann constant k to write the ideal gas law in an alternative form. We know that

Screen shot 2015-01-18 at 11.50.15 AM

and

Screen shot 2015-01-18 at 11.50.34 AM

We can use these to right the ideal gas equation in a different way:

Screen shot 2015-01-18 at 11.51.39 AM

That’s all for now. More on the kinetic-molecular model of ideal gases next week!

Featured image credit: Jeff Turner via Flickr

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Mindy

Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

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