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

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

#"N"_2# and#"O"_2# , but overestimates for more complicated molecules like#"CH"_4# and#"NH"_3# .

THE EQUIPARTITION THEOREM AT THE CLASSICAL LIMIT

Well, the equipartition theorem for the

free particleathigh enough temperaturesis:

#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

#<< kappa >> = K_(avg)/(nN_A)# as the average molecular kinetic energy, then:

#color(blue)(<< kappa >> = N/2 k_B T)# EXAMPLE USING NITROGEN MOLECULE

For instance, let's say we were looking at

#"N"_2# , a linear molecule. Here is how we could determine its degrees of freedom:

#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

#"N"_2# , we expect its average kinetic energy to be:

#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

#5# degrees of freedom we found, we can approximate:

#barC_P ~~ (5+2)/2 R = 7/2 R ~~ ul("29.10 J/mol"cdot "K")# From NIST, we have the graph for

#barC_P# :

which shows

#ul("29.124 J/mol"cdot"K")# at#"299 K"# , around#0.08%# error.

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The equipartition theorem states that, in thermal equilibrium, each degree of freedom of a system has an average energy of (1/2) kT, where k is the Boltzmann constant and T is the temperature in Kelvin. For a molecule, each degree of freedom corresponds to a way in which the molecule can store energy, such as translational motion (three degrees of freedom for a molecule in a gas), rotational motion (two additional degrees of freedom for linear molecules, three for nonlinear molecules), and vibrational motion (which varies depending on the molecule). By knowing the number of degrees of freedom and the temperature of the system, the equipartition theorem can be used to estimate the average kinetic energy of molecules.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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