10.40 Lectures 23 and 24 Computation of the properties of ideal gases

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1 1040 Lectures 3 and 4 Computation of the properties of ideal gases Bernhardt L rout October (In preparation for Lectures 3 and 4 also read &M ) Degrees of freedom Outline Computation of the thermodynamic properties of a monatomic ideal gas Computation of the thermodynamic properties of polyatomic ideal gases including diatomic ideal gases Summary of thermodynamic functions 31 Degrees of freedom We have discussed translational electronic andnuclear degrees of freedom Diatomic and polyatomic molecules have two additional internal degrees of freedom vibrational and rotational degrees of freedom Excluding electronic and nuclear degrees of freedom atoms and molecules have a total of 3N atoms degrees of freedom where N atoms is the number of atoms in a molecule (or atom) Linear molecules have rotational degrees of freedom and non-linear molecules have 3 rotational degrees of freedom he rest of the degrees of freedom 3N atoms -5 for linear molecules and 3N atoms -6 for non-linear molecules are considered to be vibrational degrees of freedom From now on we approximate these vibrational degrees of freedom as normal modes ie modes of simple harmonic oscillators We will also ignore nuclear degrees of freedom from now on 3 Computation of the thermodynamic properties of a monatomic ideal gas We have N independent and indistinguishable particles Let s label these particles a b Particlea cantakeonenergylevelsε aj where as before j is just an index referening each of the possible states Particle b can take on energy levels ε bj etc Designate the partition function of each particle with the symbol q hus q a = j q b = j e βεaj e βε bj etc 1

2 1040: Fall 003 Lectures 3 and 4 hen the energy for the entire system will be hus Q should be of the form E ij = ε ai + ε bj + (1) e βe ij = q N ij where q is the partition function for an individual particle However since each particle is indistinguishable the ε s in equation 1 can be permuted in N! ways hus in order to avoid overweighting each quantum state we much divide q N by 1 N! his leads to Q = 1 N! qn Now we need to compute q for a single atom We know that from quantum mechanics the energy levels of a particle due to their translational degrees of freedom are ε j = ε lx l y l z = h (lx + ly + lz) 8m /3 where h is Planck s constant equal to Js; l x l y l z =1 3 ; m is the mass of the particle; and is the volume occupied by the particle he numbers l x l y l z are the quantum numbers designating the translational quantum level Let R = lx + ly + lz = 8m /3 ε h hen the number of states with energy <εis Φ(ε) = πr3 = π µ 3/ 8mε 6 6 h his result can be visualized by thinking of the quantum energy levels as points on a 3D Cartisian grid in an octet of a sphere Let us remind ourselves of the difference between the designation of states and energy levels as shown in Figure 1 hus q = e βε j = ω i e βε i states j levels i where ω i is the degeneracy of level i (he degeneracy is the number of states with energy ε i ) Write q = e βεlxly lz l x l y l z For states with energies that are very close together we can replace P s with R s How close do they need to be? How close are they? hus q = Z 0 ω(ε)e βε dε where the number of states between ε and ε + dε are ω(ε)dε = dφ dε dε = π µ 3/ 8m 4 h ε 1/ dε

3 1040: Fall 003 Lectures 3 and 4 3 j=4 j=7 j=5 j=3 j= j=1 j=8 states j=6 energy levels i=5 i=4 i=3 i= Figure 1: Difference between states and energy levels hus Plugging this into the integral yields q = π µ 3/ 8m 4 h 0 µ 3/ πmk = h q = Λ 3 Z ε 1/ e βε dε ³ 1/ h where Λ πmk and is called the thermal debroglie wavelength It gives the characteristic wavelength of a gas Note that this q will be designated as q t below for "translational" (see below) From all of this we can determine the partition function of the system Q = 1 N! qn = 1 N! µ N Λ 3 Now recalling the formulas for the thermodynamic quantities in terms of Q yields A = k ln Q = k( N ln N + N + N ln q) " µπmk # 3/ = Nk ln h N e or in intensive form " µπmk 3/ A = k ln e# h 3

4 1040: Fall 003 Lectures 3 and 4 4 and and µ ln Q P = k µ ln q = Nk = Nk N N µ ln Q U = k N 3/ d ln = Nk d = 3 Nk and C v = µ U N = 3 Nk Recalling an expression for S S = U A " µπmk = Nkln h # 3/ N e5/ Also H= U + P = U + Nk and G= A + P = A + Nk Next we may need to consider internal degrees of freedom of the atom such as electronic and nuclear degrees of freedom hus we can write the partition function as Q = 1 N! (q tq e q n ) N where q t is the translational partition function q e is the electronic partition function and q n is the nuclear partition function For most systems the first excited electronic state and nuclear state are at 0 kcal mol 1 and 0000 kcal mol 1 respectively hus they do not play a significant role at temperatures of interest Note that there are important exceptions to this rule of thumb such as with the alkai metal atoms and halogens but we will not concern ourselves with these in this course We often need to take into account degenerate electronic states of atoms that are a result of their spins For example H has a net spin of 1 meaning that it has two spin degrees of freedom at the ground state energy level ω e =and q e = hus an additional factor of Nkln ω e must be added to the formula for the entropy given above and an additional factor of Nk ln ω e must be added to the formula for the Helmholtz free energy given above 4

5 1040: Fall 003 Lectures 3 and Computation of the thermodynamic properties of polyatomic ideal gases including diatomic ideal gases Recall that Q = 1 N! qn First we assume that all of the degrees of freedom are separable hus q( )=q t ( )q r ( )q v ( )q e ( ) where q t is the translational partition function q r is the rotational partition function q v is the vibrational partition function and q e is the electronic partition function We discussed q t and q e lasttime Wenoteherethatifwechoosetheelectronic energy of separated atoms as the electronic reference state then we can define D e as the dissociation energy the energy needed to atomize the molecule Now we need to compute q r and q v 331 ibrational degrees of freedom From quantum mechanics the energy levels of a harmonic oscillator are: ε n = (n + 1 )hν () n = 0 1 where h is Planck s constant and ν is the frequency function is then q v = e Θ v/ 1 e Θ v/ where Θ v = hν/k (You can verify this yourself) hen he vibrational partition A v = Nk ln q v ³ Θv = Nk +ln Θv/ 1 e µ ln U v = Nk qv µ Θv = Nk + Θ v e Θ v/ 1 C v = µ U = Nk µ Θv N e Θ v/ e Θ v / 1 and S v = U v A v µ = Nk/ Θ v e Θv/ 1 5 Nkln ³1 e Θ v/

6 1040: Fall 003 Lectures 3 and 4 6 We also emphasize that the terms Θ v in the expressions for U v and A v above are a result of the fact that the ground state of the harmonic oscillator has a finite energy level as seen in equation he term 1 hν = kθ v is called the zero point energy of the vibration We note that for a diatomic molecule for example the measured bond dissociation energy D 0 is not the energy of atomization described at the beginning of the lecture his is because the zero point energy adds a destabilizing contribution hus D 0 = D e 1 hν his can be easily generalized for polyatomic molecules 34 Rotational degrees of freedom From quantum mechanics the energy levels of a rigid linear rotator are ε j = j(j +1)h 8π I j = 0 1 where I is the moment of inertia Note that for a diatomic ³ molecule consisting of atoms 1 and I = µd whereµ is the reduced mass µ = m 1m m 1+m and d is the bond length his can be generalized for polyatomic molecules (See books on classical mechanics) For diatomic and non-linear polyatomic molecules the rotational partition function is then q r = ω j e βε j = j=0 (j +1)e j(j+1)θr/ j=0 where At high Θ r = h 8π Ik q r Z o (j +1)e j(j+1)θr/ dj = Θ r = 8π Ik h Note that in order to eliminate double counting symmetry must be taken into account he symmetry number is symbolized as σ and for a diatomic molecule σ =1if the molecule is unsymmetrical and σ =ifthemoleculeissymmetrical A non-linear polyatomic molecule will have 3 rotational degrees of freedom and obviously the 3 different moments of inertia will almost always be different For these species q r = π1/ σ µ 8π 1/ µ I A k 8π 1/ µ I B k 8π 1/ I C k h h h 6

7 1040: Fall 003 Lectures 3 and 4 7 where σ is again the symmetry factor which can take on many values up to 1 and I A I B I C are the three principle moments of intertia (We ignore the derivation here because it is somewhat complicated It can be found in a book on classical mechanics) his equation can be written more compactly as µ q r = π1/ 3 1/ From this equation we can derive the thermodynamic functions: " π 1/ µ 3 1/ # A r = Nk ln E r = 3 Nk C r = 3 Nk " π 1/ µ 3 e 3 1/ # S r = Nkln 35 Summary of thermodynamic functions Below is a summary of the thermodynamic functions excluding nuclear and excited electonic degrees of freedom 351 Monatomic ideal gas " µπmk A = Nk ln h U = 3 Nk; # 3/ N e ; C = 3 Nk; " µπmk # 3/ S = Nkln h N e5/ Also H = U + P = U + Nk and G= A + P = A + Nk 35 Diatomic and linear polyatomic ideal gas " µπmk A # 3/ k =ln 8π 3N h N e Ik atoms 5 ³ Θiv +ln σh +ln Θiv/ 1 e + D e k +ln ω e; U k = 3 + 3Natoms 5 + Θiv + Θ iv/ C k = 3 + 3Natoms 5 + " µθiv D e e Θiv/ 1 k ; # e Θ iv/ e Θ iv/ 1 ; 7

8 1040: Fall 003 Lectures 3 and 4 8 " µπmk # 3/ S k =ln h N e5/ +ln σh Note that m is the mass of the molecule G = A + P = A + Nk 8π Ike + 3N atoms Non-linear polyatomic ideal gas " µπmk A # " 3/ k =ln h N e π 1/ µ 3 1/ # +ln Θiv / e Θiv/ 1 ln ³ 1 e Θ iv/ +ln ω e Also H = U + P = U + k and 3N atoms 6 ³ Θiv +ln Θiv/ 1 e + D e k +ln ω e; U k = Natoms 6 + Θiv + Θ iv/ D e e Θiv/ 1 k ; " C k = Natoms 6 + µθiv # e Θ iv/ e Θ iv/ 1 ; " µπmk # " 3/ S k =ln π 1/ e 3/ µ 3 1/ # 3N atoms 6 Θiv / ³ +ln h N e5/ +ln + e Θiv/ 1 ln 1 e Θ iv/ ω e Note that m is the mass of the molecule Also H = U + P = U + k and G = A + P = A + Nk 8