From #PV = 1/3 mN<< upsilon^2 >>#, show that the kinetic energy for a mole of an ideal gas is #E_k = 3/2 RT#?

Depends on the units you want... I'm guessing your #E_k# is the average kinetic energy per mol. This of course assumes the molecule is not subject to an external potential.

From Levine's "Physical Chemistry" (pg. 446, Ch. 14), from a derivation based on the kinetic theory of gases, an ideal gas has:

#PV = 1/3 mN << upsilon^2 >>#

where #m# is the mass in #"kg"# and #N# is the number of molecules.

Note that this asks for the average squared speed, NOT the average speed squared, i.e. #<< upsilon^2 >> ne << upsilon >>^2# unless the variance is zero. Furthermore, #upsilon_(RMS) = << upsilon^2 >>^(1//2) ne << upsilon >>#.

We know that the translational energy of a free particle is given by:

#E_(tr) = K_(tr) = 1/2 m upsilon^2#

So, dividing by the number of molecules #N# gives the average translational energy:

#<< epsilon >>_(tr) -= E_(tr)/N = 1/2 m << upsilon^2 >>#

As a result:

#PV = 1/3 mN (2/m << epsilon >>_(tr))#

#= 2/3 N << epsilon >>_(tr)#

Therefore, for an ideal gas, #PV = nRT#, with the usual energy units of #R#, giving:

#nRT = 2/3 N << epsilon >>_(tr) = 2/3 E_(tr)#

#=> barul|stackrel(" ")(" "color(blue)(E_(tr) = 3/2nRT)" ")|# #" ""J"#

Or since #nR = Nk_B#,

#=> barul|stackrel(" ")(" "color(blue)(E_(tr) = 3/2Nk_BT)" ")|# #" ""J"#

Here we have defined the energy to be in #"J"#... and this is just what we know from the equipartition theorem for ideal gas molecules in the absence of rotation and vibration... 3 degrees of freedom for translation.

In terms of molar energy:

#barul|stackrel(" ")(" "color(blue)(barE_(tr) = E_(tr)/n = 3/2RT)" ")|# #" ""J/mol"#

In terms of molecular energy:

#barul|stackrel(" ")(" "color(blue)(<< epsilon >>_(tr) = E_(tr)/N = 3/2k_BT)" ")|# #" ""J/molecule"#