How many micro states exist for a molecule?
1 Answer
A lot! For example, a single molecule of methane could have over
And thus, a usual sample of
Also, a note about
For most molecules, namely those that are in a scenario where the number of occupied states is much less than the number of available states, one can find the
#ln Omega = sum_i [N_i ln(g_i/N_i) + N_i]# where:
#N_i# is the number of particles in state#i# with energy#epsilon_i# .#g_i# is the degeneracy of the energy#epsilon_i# , i.e. how many#epsilon_i# there are for a given state#i# .#Omega# is the number of microstates.
One can then use this in Boltzmann's expression for the entropy to find:
#S = k_B ln Omega# where
#k_B# is the Boltzmann constant. Often the form#"0.695 cm"^(1)"/K"# of#k_B# is useful in Statistical Mechanics.
The above form of
#S = k_Boverbrace([ln(q_(t ot)/N) + E/(Nk_BT) + 1])^(ln Omega)# where:
 For simple systems,
#q_(t ot) ~~ q_(tr)q_(rot)q_(vib)q_(el ec)# is the partition function for a given molecule, and is based on its translational, rotational, vibrational, and electronic degrees of freedom.#N# is the number of particles.#E# is the internal energy of a set of#N# particles in a given system.#T# is the temperature in#"K"# .
And so, the number of microstates is given by:
#color(blue)(barulstackrel(" ")(" "Omega = "exp"(ln(q_(t ot)/N) + E/(Nk_BT) + 1)" "))#
Take methane,
At

#q_(tr) = 0.02560M^(3//2)T^(5//2) ~~ ul(2.525 xx 10^6)# ,#M = 16.0426# is the molar mass in#"amu"# 
#q_(rot) = 1/12 (0.014837I^(3//2)T^(3//2)) ~~ ul(36.72)# ,#I ~~ 3.216# is moment of inertia in#"amu" cdot Å^2# 
#q_(vib) = prod_(i=1)^(4) q_(vib,i) ~~ ul(6.373 xx 10^(10))# ,
#q_(vib,i) = e^(Theta_(vib,i)//2T)/(1  e^(Theta_(vib,i)//T))#
#Theta_(vib,i) = omega_i/k_B# is the vibrational temperature in#"K"# , where#omega_i# is the fundamental vibrational frequency in#"cm"^(1)# for a given molecular motion, and#k_B = "0.695 cm"^(1)"/K"# .
#ul(q_(el ec) ~~ 1)# for any closedshell molecule (zero unpaired electrons), since the next electronic energy level is usually naturally inaccessible except at temperatures in the tens of thousands of#"K"# .
However, if the molecule has unpaired electrons, that will make
#q_(el ec) = n + 1# , where#n# is the number of unpaired electrons.
As a result, we get that:
#q_(t ot) = overbrace((2.525 xx 10^6))^"translational"overbrace((36.72))^"rotational"overbrace((6.373 xx 10^(10)))^"vibrational"overbrace((1))^"electronic" ~~ 0.0591#
To make this simple, I have already calculated that the internal energy
As a result, for
#ln Omega = ln(0.0591) + ("5012.18 cm"^(1)"/molecule")/(1("0.695 cm"^(1)"/molecule"cdot"K")("298.15 K")) + 1#
#= 22.36#
And so, the number of microstates a single methane molecule has is:
#color(blue)(Omega) = e^(22.36)#
#~~# #ulcolor(blue)(5.14 xx 10^9 "molecule"^(1))#
A more usual quantity is for a mol of methane molecules, which would then possess about
SHOWING THAT THE NUMBER OF MICROSTATES IS CORRECT
From here one could also calculate the standard molar entropy
#S^@ = k_BlnOmega#
#= "0.695 cm"^(1)"/K" xx 22.36#
#= "15.54 cm"^(1)"/molecule"cdot"K"#
Or, in more familiar units, multiply by
#S^@ = (6.626 xx 10^(34) "J"cdotcancel"s") cdot (2.998 xx 10^(10) cancel"cm""/"cancel"s") cdot 6.0221413 xx 10^(23) cancel"molecules""/mol" cdot (15.54 cancel("cm"^(1)))/(cancel"molecule" cdot "K")#
#=# #ul("185.93 J/mol"cdot"K")#
whereas the literature value was