Partition Functions and Ideal Gases PFIG-1

You’ve learned about partition functions and some uses, now we’ll explore them in more depth using ideal monatomic, diatomic and polyatomic gases!

Before we start, remember:

N TVq ),( What are N, V, and T? TVNQ ),,( = N!

We now apply this to the ideal gas where:

1. The molecules are independent. 2. The number of states greatly exceeds the number of

molecules (assumption of low pressure).

Ideal monatomic gases PFIG-2

Where can we put energy into a monatomic gas?

ε atomic = ε trans + ε elec

Only into translational and electronic modes! ☺

The total partition function is the product of the partition functions from each degree of freedom:

= trans elec TVqTVqTVq ),(),(),(

Total atomic Translational atomic Electronic atomic partition function partition function partition function We’ll consider both separately… Translations of Ideal Gas: trans TVq ),( PFIG-3

−βε trans General form of partition function: trans = ∑eq states z Recall from QM slides…

c b x a 2 h 222 ε = ( ++ nnn ) nnn zyx = ,...,2,1,, ∞ trans 8ma2 zyx

So what is qtrans? Let’s simplify qtrans … PFIG-4 ∞ ∞ ∞ ∞ 2 −βε nx ny ,, nz ⎡ βh 222 ⎤ qtrans = e = exp⎢− 2 ()++ nnn zyx ⎥ ∑∑∑∑ 8ma nnn zyx =1,, = nn =11 nzyx =1 ⎣ ⎦

++ cbacba Recall: = eeee

∞ ⎛ β nh 22 ⎞ ∞ ⎛ β nh 22 ⎞ ∞ ⎛ β nh 22 ⎞ q exp⎜−= x ⎟ exp⎜− y ⎟ exp⎜− z ⎟ trans ∑ ⎜ 8ma2 ⎟∑ ⎜ 8ma2 ⎟∑ ⎜ 8ma2 ⎟ nx =1 ⎝ ⎠ny =1 ⎝ ⎠nz =1 ⎝ ⎠

All three sums are the same because nx, ny, nz have same form! We can simplify expression to:

qtrans is nearly continuous PFIG-5

3 We’d like to solve this expression, ⎡ ∞ ⎛ β nh 22 ⎞⎤ trans TVq ⎢ exp),( ⎜−= ⎟⎥ but there is no analytical solution ∑ ⎜ 8ma2 ⎟ for the sum! ⎣ n=1 ⎝ ⎠⎦ No fears… there is something we can do!

Since translational energy levels are spaced very close together, the sum is nearly continuous function and we can approximate the sum as an integral… which we can solve!

3 ∞ ⎡ ⎛ β nh 22 ⎞⎤ Work the = dnTVq exp),( ⎜− ⎟ trans ⎢∫ ⎜ 2 ⎟⎥ integral ⎣ 0 ⎝ 8ma ⎠⎦

2/3 Note limit change … ⎛ 2π BTmk ⎞ only way to solve but adds trans TVq ),( = ⎜ 2 ⎟ V very little error to result ⎝ h ⎠ a3 Translational energy, ε trans PFIG-6

With q we can calculate any thermodynamic quantity!!

In Ch 17 (BZ notes) we showed this for the average energy … 2/3 2 ⎛ ∂ ln qtrans ⎞ ⎛ 2π BTmk ⎞ ε trans = BTk ⎜ ⎟ here… trans TVq ),( = ⎜ 2 ⎟ V ⎝ ∂T ⎠V ⎝ h ⎠

⎛ ⎡ 2/3 ⎤⎞ 2 ⎜ ∂ 2/3 ⎛ 2πmkB ⎞ ⎟ ε trans = BTk ln⎢T ⎜ ⎟ V ⎥ ⎜ ∂T ⎝ h2 ⎠ ⎟ ⎝ ⎣⎢ ⎦⎥⎠V

As we found in BZ notes! (Recall: this is per atom.) Ideal monatomic gas: elec TVq ),( PFIG-7

Next consider the electronic contribution to q: qelec

Again, start from the general −βε ei form of q, but this time sum elec = ∑ eiegq over levels rather than states: levels Energy of level i Degeneracy of level i

We choose to set the lowest electronic energy state at zero, such that all higher energy states are relative to the ground state.

ε1 = 0 For monatomic gases! Can we simplify qelec? PFIG-8

−βε e 2 −βε e3 (18.1) elec )( ee 21 e3egeggTq +++= ... terms are getting small rapidly…

The electronic energy levels are spaced far apart, and therefore we typically only need to consider the first term or two in the series…

General rule of thumb: At 300 K, you only need to keep terms 3 -1 -βε where εej < 10 cm (e > 0.008)

A closer look at electronic levels… PFIG-9

(18.1) −βε e 2 −βε e3 elec )( ee 21 e3egeggTq +++= ...

General trends,

1. Nobel gas atoms: ε = 105 cm-1 (at 300K, keep ___ term(s)) e2

2. Alkali metal gas atoms: 4 -1 εe2 = 10 cm (at 300K, keep ___ term(s))

3. Halogen gas atoms: 2 -1 εe2 = 10 cm (at 300K, keep ___ term(s))

Final look at qelec PFIG-10

In general it is sufficient to keep only the first two terms for qelec

−βε e 2 elec )( +≈ ee 21 eggTq

However, you should always keep in mind that for very high

temperatures (like on the sun) or smaller values of εej (like in F) that additional terms may contribute.

If you find that the second term is of reasonable magnitude (>1% of first term), then you must check to see that the third term can be

neglected. Finally… we can solve for Q! PFIG-11

For a monatomic ideal gas we have:

( TVqTVq ),(),( )N TVNQ ),,( = trans elec N!

with 2/3 ⎛ 2π BTmk ⎞ trans TVq ),( = ⎜ ⎟ V ⎝ h2 ⎠

−βε e 2 elec )( +≈ ee 21 eggTq Finding thermodynamic parameters… U PFIG-12

We can now calculate the average energy, = EU

2 ⎛ ∂ lnQ ⎞ 2 ⎛ ∂ ln q ⎞ 2 ⎛ ∂ ln qq electrans ⎞ == BTkEU ⎜ ⎟ = BTNk ⎜ ⎟ = BTNk ⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠V ⎝ ∂T ⎠V

Plug in qtrans and qelec … 2/3 ⎛ ∂ ⎛⎛ 2π Tmk ⎞ ⎞⎞ 2 ⎜ ⎜ B −βε e 2 ⎟⎟ = BTNkU ln ⎜ ⎟ ()+ ee 21 eggV ⎜ ∂T ⎜⎝ h2 ⎠ ⎟⎟ ⎝ ⎝ ⎠⎠V 3 ε eNg −βε e 2 Electronic ee 22 contribution BTNkU += 2 q typically small elec (i.e., negligible)

So … U ≈ or U = molar energy Finding thermodynamic parameters… CV PFIG-13

Molar heat capacity for a monatomic ideal gas: ⎛ Ud ⎞ CV = ⎜ ⎟ ⎝ dT ⎠ ,VN

⎛ ⎛ 3 ⎞ ⎞ ⎜ ⎜ RTd ⎟ ⎟ 2 3 C = ⎜ ⎝ ⎠ ⎟ = R V ⎜ dT ⎟ 2 ⎜ ⎟ ⎝ ⎠ ,VN

3 Could also find heat capacity: = NkC V 2 B Finding thermodynamic parameters… P PFIG-14

⎛ ∂ lnQ ⎞ ⎛ ∂ ln q ⎞ ⎛ ∂ qq electrans )ln( ⎞ = BTkP ⎜ ⎟ = BTNk ⎜ ⎟ = BTNk ⎜ ⎟ ⎝ ∂V ⎠ ,TN ⎝ ∂V ⎠T ⎝ ∂V ⎠T

Plug in qtrans and qelec … 2/3 ⎛ ∂ ⎛⎛ 2π Tmk ⎞ ⎞⎞ ⎜ ⎜ B −βε e 2 ⎟⎟ = BTNkP ln ⎜ ⎟ ()+ ee 21 eggV ⎜ ∂V ⎜⎝ h2 ⎠ ⎟⎟ ⎝ ⎝ ⎠⎠T

Only function of V

So… P = or P =

Look familiar?? molar pressure Ideal Monatomic Gas: A Summary PFIG-15 TVq ),( N Partition Function: TVNQ ),,( = From N! 2/3 −βε ⎛ 2π BTmk ⎞ e 2 TVq ),( = V elec )( +≈ ee 21 eggTq Q trans ⎜ ⎟ ⎝ h2 ⎠ get

(molar) 3 3 ≈ TNkU = RTU Energy 2 B 2

Heat Capacity

Pressure Adding complexity… diatomic molecules PFIG-16

In addition to trans. and elec. degrees of freedom, we need to consider: 1. Rotations 2. Vibrations

ε diatomic = ε trans + ε + ε vibrot + ε elec

Electronic Translational Rotational Vibrational Total Energy Energy

z Energy Energy Energy (( ((

c b x a Diatomic Partition Function PFIG-17

Q. What will the form of the molecular diatomic partition function be given: ε diatomic = ε trans + ε + ε vibrot + ε elec ?

Ans.

Q. How will this give us the diatomic partition function?

Ans.

Now all we need to know is the form of qtrans , qrot , qelec , and qvib . q Start with trans: This is the same as in the monatomic case but with m = m1+m2!

Diatomic Gases: qelec PFIG-18

We define the zero of the electronic energy to be separated atoms at rest in their ground electronic energy states.

With this definition, ε −= D e1 e

And…

Note the slight difference in qelec between monatomic and diatomic Figure 18.2 gases!