# Calculate the ratio of the average kinetic energies of #"8 g"# #"H"_2# compared to #"8 g"# of #"O"_2# at #27^@ "C"#?

##### 1 Answer

The average kinetic energy *at a high enough temperature* is given by:

#n<< kappa >> -= << K >> ~~ 5/2 nRT# in#"J"# ,

and we got that:

#(<< K >>_(H_2))/(<< K >>_(O_2)) ~~ (9.922n_(H_2)cdotRT)/(0.625n_(O_2)cdotRT) ~~ 15.87#

How to use the equipartition theorem is given below.

**DISCLAIMER:** *LONG ANSWER!*

The **equipartition theorem** can be best used to approximate the average kinetic energy in the *high temperature limit* (also called the "classical" limit), where rotational and vibrational states are not accessed.

#<< kappa >> = << K >>/n = N/2 RT# ,where:

#<< kappa >># is the average kinetic energy in#"J/mol"# , while#<< K >># is in#"J"# .#N# is the number of degrees of freedom, which we can consider the number of dimensions of each type of motion that should be considered.#R = "8.314472 J/mol"cdotK"# is the universal gas constant#n# is the mols of gas.#T# is temperature in#"K"# .

In considering whether we are at a high enough temperature, it turns out that:

- For pretty much all molecules, the high temperature limit applies for
**translational**motion (#x,y,z# ) and**rotational**motion. This includes#"O"_2# and#"H"_2# .

So, we can assume that equipartition applies for the translational and rotational kinetic energy,

#<< K >>_(tr)# and#<< K >>_(rot)# .

- For
#T = 27 + 273.15 = "300.15 K"# , both#"O"_2# and#"H"_2# have#Theta_(vib) -= (N_Ahnu)/(R)# #">>"# #T# , where:

#nu# is the fundamental vibrational frequency of#"O"_2# or#"H"_2# .#N_A# is Avogadro's number.#h# is Planck's constant.#Theta_(vib)# is the vibrational temperature, the temperatureabove whichvibrational motions become important.We actually have that:

#Theta_(vib,O_2) = "2256 K"#

#Theta_(vib,H_2) ~~ "6323 K"# So,

#Theta_(vib)# is much greater than#"300.15 K"# for both molecules. This allows us to assume these molecules are in their ground vibrational state.In other words, for

#"O"_2# and#"H"_2# ,it is better to just not invoke equipartition for vibrational motion, so that we don't overestimate#<< K >># .

In considering the degrees of freedom,

- For
**each**dimension of translational motion (#x,y,z# ), we assign#1# to#N# . - For
**each**dimension of rotational motion (#theta, phi# ), we assign#1# to#N# . - We
**don't**assign a value for#N# for vibrational motion because the high temperature limit doesn't apply for the vibrations of#"O"_2# and#"H"_2# .

So far, we then have from equipartition, a *general expression* for

#color(green)(<< kappa >>) ~~ << kappa >>_(tr) + << kappa >>_(rot) + cancel(<< kappa >>_(vib))^("small")#

#= 3/2 RT + 2/2 RT#

#= color(green)(5/2 RT)#

So, their **average kinetic energy** is predicted to be:

#<< K >>_(O_2) = n<< kappa >>_(O_2) ~~ 5/2 nRT#

#<< K >>_(H_2) = n<< kappa >>_(H_2) ~~ 5/2 nRT#

Thus, with

#<< K >>_(O_2) ~~ 5/2 (8 cancel("g") xx "1 mol O"_2/(31.998 cancel"g"))RT ~~ 0.625RT#

#~~# #"1559.8 J"#

#<< K >>_(H_2) ~~ 5/2 (8 cancel("g") xx "1 mol H"_2/(2.0158 cancel"g"))RT ~~ 9.922RT#

#~~# #"24760.3 J"#

Therefore, their **kinetic energy ratio** at

#color(blue)((<< K >>_(H_2))/(<< K >>_(O_2))) ~~ 9.922/0.625 = 24760.3/1559.8#

#~~ color(blue)(15.87)#

Thus,