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