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 q(,) V T What are N, V, and T? QNVT(,,)= 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 low pressure).fssumption omolecules (a 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: q V(,)(,)(,) T q= trans V Telec q V T Total atomic Translational atomic Electronic atomic partition function partition function partition function We’ll consider both separately… Translations of Ideal Gas:qtrans(,) V T PFIG-3 −βε trans General form of partition function:qtrans = ∑ e states z QM slides…Recall from c b x a 2 h 2 2 2 ε = n( + n + ) n,n ,x n y 1 nz= , 2∞ ,..., trans 8ma2 x y z So what is qtrans? Let’s simplify qtrans … PFIG-4 ∞ ∞ ∞ ∞ 2 −βε nx,,ny nz ⎡ βh 2 2 2⎤ qtrans = e = exp⎢− n2 ()x+ n y + nz⎥ ∑ ∑ ∑ ∑8ma nx, n y , n= z 1 nx1= n y= 1 n = z1 ⎣ ⎦ a b+ c + a b c Recall:e = e e e ∞ ⎛ βh2 n 2⎞ ∞ ⎛ βh2 n 2⎞ ∞ ⎛ βh2 n 2⎞ 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, ⎡ ∞ ⎛ βh2 n 2⎞⎤ q(trans , V )= T ⎢ 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 ∞ ⎡ ⎛ βh2 n 2⎞⎤ Work the q( V , ) T= dn exp⎜− ⎟ trans ⎢∫ ⎜ 2 ⎟⎥ integral ⎣ 0 ⎝ 8ma ⎠⎦ 3 / 2 Note limit change … ⎛ 2mkπ B ⎞ T only way to solve but adds qtrans(,) V T= ⎜ 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 ⎞ ⎛ 2mkπ B ⎞ T ε trans =kB T⎜ ⎟ here…qtrans(,) V T= ⎜ 2 ⎟ V ⎝ ∂T ⎠V ⎝ h ⎠ ⎛ ⎡ 2/3 ⎤⎞ 2 ⎜ ∂ 2/3 ⎛ 2πmkB ⎞ ⎟ ε trans =kB T ln⎢T ⎜ ⎟ V ⎥ ⎜ ∂T ⎝ h2 ⎠ ⎟ ⎝ ⎣⎢ ⎦⎥⎠V As we found in BZ notes! (Recall: this is per atom.) Ideal monatomic gas:q elec(,) V T PFIG-7 Next consider the electronic contribution to q: qelec Again, start from the general −βε ei form of q, but this time sum qelec = ∑ gei e 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) q Telec()= ge1 + g e 2 e +e3 g... e + terms are getting small rapidly… The electronic energy levels are spaced far apart, and therefore we typically only need to consider the first or two in the series…term 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 q Telec()= ge1 + g e 2 e +e3 g... e + General trends, 1. Nobel gas atoms: 5 -1 εe2 = 10 cm (at 300K, keep ___ term(s)) 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 qelec T()≈ ge1 + e g 2 e Howevr, you should always keeep 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: q( (,)(,) V T q V)N T QNVT(,,)= trans elec N! with 3 / 2 ⎛ 2mkπ B ⎞ T qtrans(,) V T= ⎜ ⎟ V ⎝ h2 ⎠ −βε e 2 qelec T()≈ ge1 + e g 2 e Finding thermodynamic parameters… U PFIG-12 We can now calculate the average energy,UE = 2 ⎛ ∂ lnQ ⎞ 2 ⎛ ∂ ln q ⎞ 2 ⎛ ∂ lnqtrans q elec⎞ U= E = B k⎜ T ⎟ =NkB T⎜ ⎟ =NkB T⎜ ⎟ ⎝ ∂T ⎠V ⎝ ∂T ⎠V ⎝ ∂T ⎠V Plug in qtrans and qelec … 3 / 2 ⎛ ∂ ⎛⎛ 2mkπ ⎞ T ⎞⎞ 2 ⎜ ⎜ B −βε e 2 ⎟⎟ U= NkB T ln ⎜ V⎟ g()e1+ g e 2 e ⎜ ∂T ⎜⎝ h2 ⎠ ⎟⎟ ⎝ ⎝ ⎠⎠V 3 Ngε −βε ee 2 Electronic e2 e 2 contribution U= NkB + T 2 q y small lltypica elec (i.e., negligible) So … U ≈ or U = molar energy Finding thermodynamic parameters… CV PFIG-13 Molar heat capacity for a monatomic ideal gas: d⎛ U⎞ CV = ⎜ ⎟ ⎝ dT ⎠NV, ⎛ ⎛ 3 ⎞ ⎞ d⎜ ⎜ RT⎟ ⎟ 2 3 C = ⎜ ⎝ ⎠ ⎟ = R V ⎜ dT ⎟ 2 ⎜ ⎟ ⎝ ⎠NV, 3 Could als find heat capacity:o C= Nk V 2 B Finding thermodynamic parameters… P PFIG-14 ⎛ ∂ lnQ ⎞ ⎛ ∂ ln q ⎞ ⎛ ∂ln(qtrans q elec⎞ ) P= kB ⎜ T ⎟ =NkB ⎜ T ⎟ =NkB ⎜ T ⎟ ⎝ ∂V ⎠NT, ⎝ ∂V ⎠T ⎝ ∂V ⎠T Plug in qtrans and qelec … 3 / 2 ⎛ ∂ ⎛⎛ 2mkπ ⎞ T ⎞⎞ ⎜ ⎜ B −βε e 2 ⎟⎟ P= NkB T ln ⎜ V⎟ g()e1+ g e 2 e ⎜ ∂V ⎜⎝ h2 ⎠ ⎟⎟ ⎝ ⎝ ⎠⎠T Only function of V So… P = or P = Look familiar?? molar pressure Ideal Monatomic Gas: A Summary PFIG-15 q(,) VN T rtition Function:Pa QNVT(,,)= From N! 3 / 2 −βε ⎛ 2mkπ B ⎞ T e 2 q(,) V T= Vqelec T()≈ ge1 + e g2 e Q trans ⎜ ⎟ ⎝ h2 ⎠ get (molar) 3 3 U≈ Nk TU= RT 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 +rotε + ε vib+ 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 +rotε + ε vib+ 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!.