Question

What equation relates average kinetic energy to temperature in the high-temperature limit?

Answer 1

`KE=(3RT)/2`

Explanation:

The relationship between the kinetic energy (`KE`) and temperature (`T`) is the following:

`KE=(3RT)/2`

`R=8.3145J/(mol*K)` and is the universal gas constant.

Which implies that the kinetic energy is independent of the nature of the gas, it only depends on the temperature at which the gas exists.

Answer 2

In short, the observable kinetic energy (the kind we know in everyday chemistry and physics)---which is the same as the ensemble average of the kinetic energy---for a monatomic ideal gas, is:

`\mathbf(<< barE >> = U = K = 3/2 RT)`

but the average kinetic energy for a single system of monatomic ideal gases is:

`color(blue)(<< E >> = 3/2 nRT)`

So we see that the kinetic energy here depends on the number of `\mathbf("mol")`s of gas.

But as we should know, `"1 mol"` of a monatomic gas is equal to `"1 mol"` of any other monatomic gas when defined in terms of `"mol"`s. The masses in `"g"` are different, though that doesn't matter because this equation asks for `"mol"`s of gas, not `"g"` of gas.

Furthermore, `R`, the universal gas constant, never changes.

Therefore, the identity of a monatomic ideal gas doesn't matter when determining its average kinetic energy. Only its temperature.

You can read below for an interesting derivation.

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A SINGLE SYSTEM OF MONATOMIC IDEAL GASES

We have what's called a single system of gases, and then we have what's called an ensemble of systems of gases. We're focusing on a single system for now.

The average energy of a single system is defined in Statistical Mechanics as:

`\mathbf(<< E >> = -(del ln Q)/(del beta))`

where, for a monatomic ideal gas:

• `Q = [((2pim)/(h^2beta))^("3/2")V]^N/(N!)`
• `beta = 1/(k_BT)`
• `k_B` is the Boltzmann constant, `1.38064852 xx 10^(-23) "J/K"`
• `T` is temperature in `"K"`
• `m` is the mass of the gas
• `h` is Planck's constant, `6.626xx10^(-34) "J"*"s"`
• `N` is the number of gas particles in the system
• `V` is the volume of the system

The `del` means "partial derivative", which is just a fancy way of saying "let's focus on only this function when finding its slope at all points on the function."

If we work with the first equation for a bit, we can figure out why the identity of the monatomic ideal gas doesn't matter.

SIMPLIFYING THE FUNCTION THAT WILL BE DIFFERENTIATED

First, let's figure out `ln Q` and simplify it:

`color(green)(lnQ) = ln[[((2pim)/(h^2beta))^("3/2")V]^N/(N!)]`

`= ln[((2pim)/(h^2beta))^("3/2")V]^N - lnN!`

`= Nln[((2pim)/(h^2beta))^("3/2")V] - lnN!`

`= -Nln(beta)^("3/2") + Nln[((2pim)/(h^2))^("3/2")] + NlnV - lnN!`

`= color(green)(-(3N)/2lnbeta + (3N)/2ln((2pim)/(h^2)) + NlnV - lnN!)`

AVERAGE ENERGY OF A SINGLE SYSTEM OF MONATOMIC IDEAL GASES

And now when we take the partial derivative with respect to `beta`, we ignore everything other than `beta` to get:

`color(green)(<< E >> = -(del lnQ)/(delbeta)) =`

`= -del/(delbeta)[-(3N)/2lnbeta + cancel((3N)/2ln((2pim)/(h^2)) + NlnV - lnN!)^("not " beta)]`

`= -(-(3N)/(2beta))`

`= color(green)(3/2 Nk_BT)`

Something cool is that `Nk_B = nR`, where `n` is the number of `"mol"`s and `R` is the universal gas constant. So, we really just derived:

`color(blue)(<< E >> = 3/2 nRT)`

which shows that the average kinetic energy of a single system of monatomic ideal gases is only dependent on the number of `"mol"`s (which is the same across all monatomic ideal gases), and the temperature (which we assumed was the same for all systems in question).

THE ENERGY WE KNOW AND LOVE??

Okay, now we can proceed to the ensemble of systems of monatomic ideal gases. The idea is, we can say that a certain number of `"mol"`s of systems of gases is an ensemble:

`<< barE >> = << E >> / n`

This is the equivalent of us, observing kinetic energy in real life.

In other words, this is the kinetic and potential energy we observe in physics and chemistry classes:

`\mathbf(<< barE >> = U = K = 3/2 RT)`