Heat energy: statistical mechanics vs atomic orbitals

[Zebulin]

Normally, I prefer to do my own research, but I'm drawing a blank on this one. Any help would be appreciated.

My understanding is that statistical mechanics accounts for all of the heat energy in a gas by the kinetic energy of the molecules. I also understand that atomic orbitals have different energy levels, and because of the links between heat, radiation, and orbitals, I assume that the atomic orbitals must contribute to the heat energy of a gas. But this seems to be a contradiction with statistical mechanics where all of the heat energy is in kinetic energy.

 

[Charles Link]

The thermal excitation of the electrons out of the ground state to higher atomic orbitals is quite minimal at ordinary temperatures for a gas. Thereby, there is not a significant contribution to the specific heat from the electrons being thermally excited into higher orbitals. $\\$ To quantify, a typical electronic transition would be on the order of $\Delta E=2 eV$. Statistically, we can determine the ratio of the probability of an electron being in the excited state (at $\Delta E=2 \, eV$) vs. the ground state at $T=300 K$ : That ratio is $p=e^{-\Delta E/(k_b T)}=e^{-2(1.602E-19)/((1.381 E-23) (300))}$ where $k_b$ is Boltzmann's constant. $\\$ Basically, all of the electrons can be assumed to remain in the ground state in regards to computations of the specific heat. $\\$ If we had one mole of a gas, the energy contribution to the above would be $E=(6.02 E+23)(2)(1.602 E-19)(p) \, Joules$. The factor $p$ is so small that this number for the energy $E$ is miniscule compared to the kinetic energy of the atoms, $E_{kinetic}=(3/2)nRT$, where $R=(4.184)1.987 \, Joules/(mole-degree \, K)$ with $n=1$ mole. $\\$ Editing note: Initially I had the incorrect exponent on Boltzmann's constant.

 

[Drakkith]

Statistically, we can determine the ratio of the probability of an electron being in the excited state (at $\Delta E=2 \, eV$) vs. the ground state at $T=300 K$ : That ratio is $p=e^{-\Delta E/(k_b T)}=e^{-2(1.602E-19)/((1.381 E-23) (300))}$

Whoa, if my calculations are right, that's about 2.6 x 10^^-34. That's really small...

 

[Charles Link]

Whoa, if my calculations are right, that's about 2.6 x 10^^-34. That's really small...

It really is a good question by the OP @Zebulin, but the specific heat contribution from the excited electron states in a monatomic gas is so small that most Statistical Physics textbooks don't even consider it. In Reif's Statistical Physics book, on p.251 (chapter 7.6 Simple Applications), he states "For an ideal monatomic gas the entire energy is kinetic, so that the mean energy per mole of gas is simply $\bar{E}=N_a (\frac{3}{2}k T)=\frac{3}{2}RT$ ". $\\$ It really is a heads-up question to ask "what about the excited states"? In diatomic molecules the rotational modes are excited states that do come into play, and at very high temperatures even the vibrational modes of diatomic molecules can make a contribution to the specific heat. (The energy of a typical excited rotational state is much much less than 2 eV, so that atoms with excited rotational states are quite abundant and the number increases considerably with an increase in temperature.)

 

[Zebulin]

Wow. You guys are great. Thanks.