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?

Apr 21, 2016

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

Suppose ${\text{CH}}_{4}$ 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:

$\setminus m a t h b f \left({K}_{\text{avg}} = \frac{N}{2} n R T\right)$

where:

• $N$ is the number of degrees of freedom.
• $n$ is the number of $\setminus m a t h b f \left(\text{mol}\right)$s.
• $R$ is the universal gas constant, $\text{8.314472 J/mol"cdot"K}$.
• $T$ is the temperature in $\text{K}$.

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

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

Thus, we have

$\textcolor{b l u e}{{K}_{\text{1,avg}}} = \frac{6}{2} R {T}_{1}$

$= 3 \cdot 8.314472 \cdot 253$

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

$\textcolor{b l u e}{{K}_{\text{2,avg}}} = \frac{6}{2} R {T}_{2}$

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

= "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 ${\overline{C}}_{\text{V,tot" ~~ 3R ~~ "24.9 J/mol"cdot"K}}$, which is about 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 (${A}_{1} + E + 2 {T}_{2}$),.

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

${\overline{C}}_{\text{V,tot" = barC_"V,trans" + barC_"V,rot" + barC_"V,vib}}$

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

$\approx \textcolor{b l u e}{3.2823 R}$

whereas the equipartition theorem predicts:

${\overline{C}}_{\text{V,tot" = barC_"V,trans" + barC_"V,rot" + barC_"V,vib}}$

= 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 ${\overline{C}}_{\text{V,tot}}$!

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 ${\text{CH}}_{4}$: $5$.
• ${\theta}_{i}$ is the vibrational temperature in $\text{K}$ of vibrational mode $i$. Those are given in Table 1 below.
• $T$ is temperature in $\text{K}$ as usual. ${\overline{C}}_{\text{V,vib}}$ and ${\overline{C}}_{\text{V,tot}}$ were calculated in Excel below at $\text{295.15 K}$ ($\frac{2.3469}{8.314472} \approx 0.2823$). And a summary of ${\overline{C}}_{\text{V,vib}}$ from various approaches is shown below: The takeaway is that the statistical mechanics approach was the most accurate approach.