# How do I calculate the average kinetic energy using the equipartition theorem?

##### 1 Answer

**DISCLAIMER:** *Kind of long answer, but the bottom half simply reconciles the discrepancy in the answer we get in the example.*

The **equipartition theorem** (in the context of Physical Chemistry and Physics) is a way to assign degrees of freedom (DOFs) to a gas, assuming ideality, meaning that we assume each DOF *equally* contributes to the average energy.

A **degree of freedom** is simply a dimension in which the gas can move in the specified way. So...

- Since we evidently live in a 3-dimensional world, the
DOFs total*translational*#\mathbf(3)# :#x# ,#y# , and#z# . - The
DOFs total*rotational*#\mathbf(2)# for linear molecules (a third rotation axis would be the internuclear axis, which returns the same molecule back), or#\mathbf(3)# for nonlinear molecules (#x,y,z# rotational axes). - The number of
modes available are*vibrational*#\mathbf(3N - 5)# for*linear*polyatomic molecules or#\mathbf(3N - 6)# for*nonlinear*polyatomic molecules, where#N# is the number of atoms in the molecule, but#N = 1# maximum for vibrations. Note that these may include degenerate vibrations. - The
DOFs are generally considered negligible and so it's usually fine to omit them.*electronic*

Perhaps a bit confusing, but the equipartition theorem as it relates to **internal energy** is often written as:

#\mathbf(stackrel("Molar Internal Energy")overbrace(barU) = N/2RT)# ,where:

#N# is thenumber of DOFs(not the number of atoms in the molecule... bad coincidence).#R# is the universal gas constant,#"8.314472 J/mol"cdot"K"# .#T# is the temperature at which you observe the gas, in#"K"# .

**EXAMPLE (MOLAR CONSTANT-VOLUME HEAT CAPACITY)**

So, for our example, let's say we looked at **molar constant-volume heat capacity**,

According to the equipartition theorem, it has:

#3# translationalDOFs#2# rotationalDOFs#1# vibrationalDOF (corresponding to#4# vibrational modes, i.e. stretching/bending motions) is presumed to be accounted or, but two aredegenerate---same in energy.

Therefore, **DOFs**; the equipartition theorem tells us that

#\mathbf(barC_V = barC_(V,"trans") + barC_(V,"rot") + barC_(V,"vib"))#

#~~ \mathbf(N/2R)# , where#R# is the universal gas constant.

#=> 3/2R + 2/2R + R = \mathbf(7/2R) ~~ color(green)("29.099 J/mol"cdot"K")# .

The actual *overestimate* of the true answer.

**CAVEAT ABOUT EQUIPARTITION THEOREM ON POLYATOMIC MOLECULAR GASES**

It's worth mentioning that the equipartition theorem is *not particularly accurate on polyatomic molecular gases* when it comes to their vibrational DOF.

It often **overestimates** on

To get **linear polyatomic molecules**, I have this equation from my lab manual:

#\mathbf(barC_(V,"vib") = Rsum_(i = 1)^(3N - 5)[g_i((theta_i)/T)^2 e^(-theta_i"/"T)/(1-e^(-theta_i"/"T))^2])# where:

#g_i# is thedegeneracyof vibrational mode#i# .#theta_i# is thecharacteristic vibrational temperatureof vibrational mode#i# in#"K"# .- Other variables have been defined previously.

For nonlinear polyatomics, just replace

#3N - 5 = 4# .- The symmetric stretch (mode
#1# ) has#g_i = 1# . - The asymmetric stretch (mode
#2# ) has#g_i = 1# . - The bends have
#g_i = 2# . #theta_1 = "1890 K"# #theta_2 = "3360 K"# #theta_(3,4) = "954 K"# (doubly-degenerate)

So at

#barC_(V,"vib") = R[stackrel("Symmetric Stretch")overbrace(((("1890 K")/("298.15 K"))^2(e^(-1890"/"298.15)/(1-e^(-1890"/"298.15))^2))) + stackrel("Asymmetric Stretch")overbrace(((("3360 K")/("298.15 K"))^2(e^(-3360"/"298.15)/(1-e^(-3360"/"298.15))^2))) + stackrel("Two Degenerate Bends")overbrace(2*((("954 K")/("298.15 K"))^2(e^(-954"/"298.15)/(1-e^(-954"/"298.15))^2)))]#

Eventually, you should get

So, the corrected value of

#color(blue)(barC_V) = 3/2R + 2/2R + 0.9822R = 3.4822R#

#= color(blue)("28.953 J/mol"cdot"K")#

And indeed, even closer! (less than