# How would the equipartition theorem be used to estimate the average kinetic energy of molecules?

##### 1 Answer

At high enough temperatures,

#<< kappa >> -= K_(avg)/(nN_A) ~~ N/2 k_B T# where

#<< kappa >># is the average molecular kinetic energy in#"J/molecule"cdot"K"# ,#n# is the mols,#N_A# is Avogadro's number in#"mol"^(-1)# ,#N# is the number of degrees of freedom,#k_B# is Boltzmann's constant in#"J/K"# , and#T# is temperature in#"K"# .

At **room temperature**, this holds for simple molecules such as

**THE EQUIPARTITION THEOREM AT THE CLASSICAL LIMIT**

Well, the **equipartition theorem** for the *free particle* at *high enough temperatures* is:

#bb(E_(avg) ~~ K_(avg) = N/2nRT)# ,

or

#K_(avg) = N/2 cdot stackrel("Number of Molecules")overbrace(nN_A) cdot k_B T# where:

#N# is the number of degrees of freedom in the molecule.#n# is the#"mols"# of substance.#N_A = 6.0221413 xx 10^(23) "mol"^(-1)# is Avogadro's number.#k_B = 1.38065 xx 10^(-23) "J/K"# is the Boltzmann constant.#T# is the temperature in#"K"# .

If we define

#color(blue)(<< kappa >> = N/2 k_B T)#

**EXAMPLE USING NITROGEN MOLECULE**

For instance, let's say we were looking at

#3# translational dimensions (#x,y,z# )

#-># #3# translational degrees of freedom#2# rotational angles (#theta, phi# in spherical coordinates) for linear molecules

#-># #2# rotational degrees of freedom

(it would have been

#3# for nonlinear molecules)

- Hardly any vibrational degrees of freedom, because of a very stiff triple bond.

So, for

#color(blue)(<< kappa >>) ~~ (3+2+0)/2 k_B T#

#= color(blue)(5/2k_B T)# #color(blue)("J/molecule"cdot"K")# or,

#= color(blue)(5/2R T)# #color(blue)("J/mol"cdot"K")#

**CHECKING LITERATURE VALUES**

We can't really look up the average molecular kinetic energy, but we can check by looking at the **constant-pressure molecular heat capacity**:

#CC_P -= C_P/(nN_A)#

#~~ (N+2)/2k_B# #"J/molecule"cdot"K"# ,

or the **constant-pressure molar heat capacity**:

#barC_P -= C_P/n#

#~~ (N+2)/2 R# #"J/mol"cdot"K"# .

From the

#barC_P ~~ (5+2)/2 R = 7/2 R ~~ ul("29.10 J/mol"cdot "K")#

From NIST, we have the graph for

which shows