Question

How do you calculate the average kinetic energy of the `CH_4` molecules in a sample of `CH_4` gas at 253 K and at 545 K?

Answer 1

The average kinetic energy `K_"avg"` is based off of the equipartition theorem. I have derived it here from the statistical mechanics standpoint if you wish to see it.

Suppose `"CH"_4` was an ideal, polyatomic molecule. Then, the only degrees of freedom it has are due to linear motion, that is, `x,y,z`, and rotational motion (`R_x, R_y, R_z`).

For methane this is a decent approximation that omits vibrational degrees of freedom; I'll talk about it at the bottom of the answer.

For the equipartition theorem, we have:

`\mathbf(K_"avg" = N/2nRT)`

where:

• `N` is the number of degrees of freedom.
• `n` is the number of `\mathbf("mol")`s.
• `R` is the universal gas constant, `"8.314472 J/mol"cdot"K"`.
• `T` is the temperature in `"K"`.

NOTE: If you include the vibrational contribution to the degrees of freedom according to the equipartition theorem (`"+9"`), you will be WAY, WAY off!

Given `"1 mol"` of `"CH"_4`, we then get three linear and three rotational degrees of freedom, each of which contribute `1` to `"N"`.

Thus, we have

`color(blue)(K_"1,avg") = 6/2RT_1`

`= 3*8.314472*253`

`= "6310.68 J" ~~ color(blue)(6.31xx10^3 "J")`

`color(blue)(K_"2,avg") = 6/2RT_2`

`= 6/2*8.314472*545`

`= "13594.2 J" ~~ color(blue)(1.36xx10^4 "J")`

rounded to three sig figs.

NOTE: This corresponds to an estimated constant-volume molar heat capacity of `barC_"V,tot" ~~ 3R ~~ "24.9 J/mol"cdot"K"`, which is about `8.9%` below the true value.

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This is a decent approximation but could be better because methane has four vibrational modes in its IR spectrum (`A_1 + E + 2T_2`),.

These contribute a total of `0.2823R` to its constant-volume molar heat capacity `barC_"V,tot"` as follows:

`barC_"V,tot" = barC_"V,trans" + barC_"V,rot" + barC_"V,vib"`

`= 3/2R + 3/2R + 0.2823R`

`~~ color(blue)(3.2823R)`

whereas the equipartition theorem predicts:

`barC_"V,tot" = barC_"V,trans" + barC_"V,rot" + barC_"V,vib"`

`= 3/2R + 3/2R + stackrel("omit this for ideality")(overbrace(9R)`

`= color(red)(stackrel("'real'")(overbrace(12R))`, or `color(red)(stackrel("ideal")(overbrace("3R"))`

which is fairly off from the true `barC_("V,tot")`!

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And just so you know I'm not making these numbers up, I got this information while doing a speed-of-sound in methane lab last year:

`barC_(V_"vib") = Rsum_(i)^(3N-6) [g_i((theta_i)/T)^2 e^(-theta_i"/"T)/(1-e^(-theta_i"/"T))^2]`
(from my lab handout)

where:

• `R` is the universal gas constant.
• `g_i` is the degeneracy of mode `i`. `E` is doubly degenerate and `T` is triply degenerate (`A` is unique, so `g_A = 1`).
• `N` here is the number of atoms on `"CH"_4`: `5`.
• `theta_i` is the vibrational temperature in `"K"` of vibrational mode `i`. Those are given in Table 1 below.
• `T` is temperature in `"K"` as usual.

`barC_("V,vib")` and `barC_("V,tot")` were calculated in Excel below at `"295.15 K"` (`2.3469/8.314472 ~~ 0.2823`).

And a summary of `barC_("V,vib")` from various approaches is shown below:

The takeaway is that the statistical mechanics approach was the most accurate approach.