How do you calculate the average kinetic energy of the $C H_{4}$ $CH_4$ molecules in a sample of $C H_{4}$ $CH_4$ gas at 253 K and at 545 K?

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How do you calculate the average kinetic energy of the $C H_{4}$ $CH_4$ molecules in a sample of $C H_{4}$ $CH_4$ gas at 253 K and at 545 K?

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Answer 1 · 2016-04-21T07:07:45

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

Suppose ${CH}_{4}$ $"CH"_4$ was an ideal, polyatomic molecule. Then, the only degrees of freedom it has are due to linear motion, that is, $x, y, z$ $x,y,z$ , and rotational motion ( $R_{x}, R_{y}, R_{z}$ $R_x, R_y, R_z$ ).

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:

$K_{avg} = \frac{N}{2} n R T$ $\mathbf(K_"avg" = N/2nRT)$

where:

$N$ $N$ is the number of degrees of freedom.

$n$ $n$ is the number of $mol$ $\mathbf("mol")$ s.

$R$ $R$ is the universal gas constant, $8.314472 J/mol \cdot K$ $"8.314472 J/mol"cdot"K"$ .

$T$ $T$ is the temperature in $K$ $"K"$ .

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

Given $1 mol$ $"1 mol"$ of ${CH}_{4}$ $"CH"_4$ , we then get three linear and three rotational degrees of freedom, each of which contribute $1$ $1$ to $N$ $"N"$ .

Thus, we have

$K_{1,avg} = \frac{6}{2} R T_{1}$ $color(blue)(K_"1,avg") = 6/2RT_1$

$= 3 \cdot 8.314472 \cdot 253$ $= 3*8.314472*253$

$= 6310.68 J \approx 6.31 \times 10^{3} J$ $= "6310.68 J" ~~ color(blue)(6.31xx10^3 "J")$

$K_{2,avg} = \frac{6}{2} R T_{2}$ $color(blue)(K_"2,avg") = 6/2RT_2$

$= \frac{6}{2} \cdot 8.314472 \cdot 545$ $= 6/2*8.314472*545$

$= 13594.2 J \approx 1.36 \times 10^{4} J$ $= "13594.2 J" ~~ color(blue)(1.36xx10^4 "J")$

rounded to three sig figs.

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

This is a decent approximation but could be better because methane has four vibrational modes in its IR spectrum ( $A_{1} + E + 2 T_{2}$ $A_1 + E + 2T_2$ ),.

These contribute a total of $0.2823 R$ $0.2823R$ to its constant-volume molar heat capacity ${¯ ¯ ¯ C}_{V,tot}$ $barC_"V,tot"$ as follows:

${¯ ¯ ¯ C}_{V,tot} = {¯ ¯ ¯ C}_{V,trans} + {¯ ¯ ¯ C}_{V,rot} + {¯ ¯ ¯ C}_{V,vib}$ $barC_"V,tot" = barC_"V,trans" + barC_"V,rot" + barC_"V,vib"$

$= \frac{3}{2} R + \frac{3}{2} R + 0.2823 R$ $= 3/2R + 3/2R + 0.2823R$

$\approx 3.2823 R$ $~~ color(blue)(3.2823R)$

whereas the equipartition theorem predicts:

${¯ ¯ ¯ C}_{V,tot} = {¯ ¯ ¯ C}_{V,trans} + {¯ ¯ ¯ C}_{V,rot} + {¯ ¯ ¯ C}_{V,vib}$ $barC_"V,tot" = barC_"V,trans" + barC_"V,rot" + barC_"V,vib"$

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

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

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

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.

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How do you calculate the average kinetic energy of the $C H_{4}$ $CH_4$ molecules in a sample of $C H_{4}$ $CH_4$ gas at 253 K and at 545 K?

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