10.40 Lectures 23 and 24: Computation of the Properties of Ideal Gases

10.40 Lectures 23 and 24 Computation of the properties of ideal gases October 16, 2003 (In preparation for Lectures 23 and 24, also read T&M, 10.1.5-10.1.7). Outline • Degrees of freedom • 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 23.1 Degrees of freedom We have discussed translational, electronic, and nuclear 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 3Natoms degrees of freedom, where Natoms is the number of atoms in a molecule (or atom). Linear molecules have 2 rotational degrees of freedom and non-linear molecules have 3 rotational degrees of freedom. The rest of the degrees of freedom, 3Natoms -5 for linear molecules and 3Natoms -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, i.e. modes of simple harmonic oscillators. We will also ignore nuclear degrees of freedom from now on. 23.2 Computation of the thermodynamic properties of a monatomic ideal gas We have N independent and indistinguishable particles. Let’s label these particles, a, b, .... Particle a can take on energy levels εaj , where as before, j is just an index referening each of the possible states. Particle b can take on energy levels εbj , etc. Designate the partition function of each particle with the symbol q. Thus, X qa = e−βεaj , j qb = X j etc. 1 e−βεbj , 10.40: Fall 2003 Lectures 23 and 24 2 Then, the energy for the entire system, will be Ei,j,... = εai + εbj + .... (1) Thus, Q should be of the form X e−βEi,j,... = q N , i,j,... 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. Thus, in order to avoid overweighting each quantum state, we much divide q N by N1 ! . This leads to Q= 1 N N! q . 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 h2 (lx2 + ly2 + lz2 ) εj = εlx ly lz = , 8mV 2/3 where h is Planck’s constant, equal to 6.626×10−34 J s; lx , ly , lz = 1, 2, 3, ...; m is the mass of the particle; and V is the volume occupied by the particle. The numbers lx , ly , lz are the quantum numbers designating the translational quantum level. Let R2 = lx2 + ly2 + lz2 = 8mV 2/3 ε . h2 Then, the number of states with energy < ε is π πR3 = Φ(ε) = 6 6 µ 8mε h2 ¶3/2 V. This 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. Thus, X X q= e−βεj = ω i e−βεi , states j levels i where ω i is the degeneracy of level i. (The degeneracy is the number of states with energy εi .) Write ∞ X ∞ X ∞ X q= e−βεlx ly lz . lx ly lz For states with energies that are very close together, we can replace How close do they need to be? How close are they? Thus, Z ∞ ω(ε)e−βε dε, q= 0 where the number of states between ε and ε + dε are ω(ε)dε = = dΦ dε dε ¶3/2 µ π 8m V ε1/2 dε. 4 h2 2 R P ’s with ’s. 10.40: Fall 2003 Lectures 23 and 24 3 . . . j=7 j=4 . . . j=8 j=5 j=3 i=5 i=4 j=6 i=3 i=2 j=2 i=1 j=1 states energy levels Figure 1: Difference between states and energy levels. Plugging this into the integral yields q = = π 4 µ µ 8m h2 ¶3/2 2πmkT h2 V ¶3/2 Z ∞ ε1/2 e−βε dε 0 V. Thus, q= V Λ3 , ³ 2 ´1/2 h where Λ ≡ 2πmkT and is called the thermal deBroglie wavelength. It gives the characteristic wavelength of a gas. Note that this q will be designated as qt below for "translational" (see below). From all of this, we can determine the partition function of the system, 1 1 N Q= q = N! N! µ V Λ3 ¶N . Now, recalling the formulas for the thermodynamic quantities in terms of Q yields A = −kT ln Q = −kT (−N ln N + N + N ln q) "µ # ¶3/2 2πmkT V = −N kT ln e h2 N or in intensive form, A = −kT ln "µ 2πmkT h2 3 ¶3/2 # Ve , 10.40: Fall 2003 Lectures 23 and 24 4 and P µ ¶ ∂ ln Q = kT ∂V T,N µ ¶ ∂ ln q = N kT ∂V T,N = N kT , V and U = kT 2 µ ∂ ln Q ∂T ¶ V ,N d ln T 3/2 = N kT 2 dT 3 = N kT, 2 and, Cv = = µ ∂U ∂T ¶ V ,N 3 N k. 2 Recalling an expression for S, S U −A T " # µ ¶3/2 V 5/2 2πmkT = N k ln . e h2 N = Also, H= U + P V = U + N kT and G= A + P V = A + N kT. Next, we may need to consider internal degrees of freedom of the atom, such as electronic and nuclear degrees of freedom. Thus, we can write the partition function as 1 Q= (qt qe qn )N , N! where qt is the translational partition function, qe is the electronic partition function, and qn is the nuclear partition function. For most systems, the first excited electronic state and nuclear state are at ≈ 20 kcal mol−1 and 20,000 kcal mol−1 respectively. Thus, 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 12 , meaning that it has two spin degrees of freedom at the ground state energy level, ω e = 2 and qe = 2. Thus an additional factor of N k ln ω e must be added to the formula for the entropy given above, and an additional factor of N kT ln ω e must be added to the formula for the Helmholtz free energy given above. 4 10.40: Fall 2003 23.3 Lectures 23 and 24 5 Computation of the thermodynamic properties of polyatomic ideal gases, including diatomic ideal gases Recall that 1 N q . N! First, we assume that all of the degrees of freedom are separable. Thus, Q= q(V , T ) = qt (V , T )qr (T )qv (T )qe (T ). where qt is the translational partition function, qr is the rotational partition function, qv is the vibrational partition function, and qe is the electronic partition function. We discussed qt and qe last time. We note here that if we choose the electronic energy of separated atoms as the electronic reference state, then we can define De as the dissociation energy, the energy needed to atomize the molecule. Now, we need to compute qr and qv . 23.3.1 Vibrational degrees of freedom From quantum mechanics, the energy levels of a harmonic oscillator are: 1 = (n + )hν, 2 n = 0, 1, 2, ..., εn where h is Planck’s constant and ν is the frequency. function is then e−Θv /2T , qv = 1 − e−Θv /T where Θv = hν/k. (You can verify this yourself.) Then, Av CV v Sv = The vibrational partition = −N kT ln qv · ´¸ ³ Θv = N kT + ln 1 − e−Θv /T , 2T Uv and, (2) µ ¶ ∂ ln qv = N kT 2 ∂T ¶ µ Θv Θv , = Nk + Θ /T 2 e v −1 µ ¶ ∂U = ∂T V,N µ ¶2 eΘv /T Θv = Nk ¡ ¢2 , T eΘv /T − 1 U v − Av T µ = N k/T Θv eΘv /T − 1 5 ¶ ³ ´ − N k ln 1 − e−Θv /T . 10.40: Fall 2003 Lectures 23 and 24 6 We also emphasize that the terms Θ2v in the expressions for U v and Av above are a result of the fact that the ground state of the harmonic ¡ v ¢ oscillator has a finite energy level as seen in equation 2. The term 12 hν = kΘ is called the zero point energy 2 of the vibration. We note that for a diatomic molecule, for example, the measured bond dissociation energy, D0 , is not the energy of atomization described at the beginning of the lecture. This is because the zero point energy adds a destabilizing contribution. Thus, 1 D0 = De − hν. 2 This can be easily generalized for polyatomic molecules. 23.4 Rotational degrees of freedom From quantum mechanics, the energy levels of a rigid, linear rotator are j(j + 1)h2 , 8π 2 I = 0, 1, 2, ..., = εj j where I is the moment of inertia. Note that for a diatomic molecule ³ ´ consisting of m2 atoms 1 and 2, I = µd2 , where µ is the reduced mass µ = mm11+m and d is the 2 bond length. This can be generalized for polyatomic molecules. (See books on classical mechanics.) For diatomic and non-linear polyatomic molecules, the rotational partition function is then qr ∞ X = ω j e−βεj j=0 ∞ X = (2j + 1)e−j(j+1)Θr /T , j=0 where Θr = h2 8π 2 Ik At high T , qr → Z = ∞ (2j + 1)e−j(j+1)Θr /T dj o T 8π2 IkT = . Θr h2 Note that in order to eliminate double counting, symmetry must be taken into account. The symmetry number is symbolized as σ, and for a diatomic molecule, σ = 1 if the molecule is unsymmetrical and σ = 2 if the molecule is symmetrical. 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, qr = π1/2 σ µ 8π2 IA kT h2 ¶1/2 µ 8π 2 IB kT h2 6 ¶1/2 µ 8π 2 IC kT h2 ¶1/2 , 10.40: Fall 2003 Lectures 23 and 24 7 where σ is again the symmetry factor, which can take on many values, up to 12, and IA , IB , IC 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.) This equation can be written more compactly as π1/2 qr = σ µ T3 ΘA ΘB ΘC ¶1/2 . From this equation, we can derive the thermodynamic functions: " ¶1/2 # µ π 1/2 T3 Ar = −N kT ln , σ ΘA ΘB ΘC 3 N kT, 2 3 C V r = N k, 2 " ¶1/2 # µ 1/2 π T 3 e3 . S r = N k ln σ ΘA ΘB ΘC Er = 23.5 Summary of thermodynamic functions Below is a summary of the thermodynamic functions, excluding nuclear and excited electonic degrees of freedom. 23.5.1 Monatomic ideal gas A = −N kT ln "µ 2πmkT h2 ¶3/2 # V e ; N 3 N kT ; 2 3 C V = N k; 2 "µ # ¶3/2 2πmkT V 5/2 . e S = N k ln h2 N U= Also, H = U + P V = U + N kT and G= A + P V = A + N kT. 23.5.2 Diatomic and linear polyatomic ideal gas "µ # ¶3/2 ¸ 3NaX · 2 t o m s −5 · ´¸ D ³ Θi,v A 2πmkT 8π IkT V e −Θi,v /T − + − = ln e +ln + ln 1 − e +ln ω e ; kT h2 N σh2 2T kT i=1 ¸ De Θi,v Θi,v /T − + Θ /T ; i,v 2T kT e −1 i=1 "µ # ¶2 3NaX t o m s −5 Θi,v CV eΘi,v /T 3 2 = + + ¡ ¢2 ; k 2 2 T eΘi,v /T − 1 i=1 U 3 2 = + + kT 2 2 3NaX t o m s −5 · 7 10.40: Fall 2003 S = ln k "µ 2πmkT h2 Lectures 23 and 24 ¶3/2 8 # · 2 ¸ 3NaX t o m s −5 · ³ ´¸ V 5/2 8π IkT e Θi,v /T −Θi,v /T − ln 1 − e +ln + +ln ω e . e N σh2 eΘi,v /T − 1 i=1 Note that m is the mass of the molecule. G = A + P V = A + N kT. Also, H = U + P V = U + kT and 23.5.3 Non-linear polyatomic ideal gas "µ # " ¶3/2 ¶1/2 # 3NaX µ t o m s −6 · ´¸ D ³ Θi,v A 2πmkT π 1/2 T3 V e −Θi,v /T − + − = ln e +ln + ln 1 − e +ln ω e ; kT h2 N σ ΘA ΘB ΘC 2T kT i=1 ¸ De Θi,v Θi,v /T − + Θ /T ; i,v 2T kT e −1 i=1 "µ # ¶2 3NaX t o m s −6 Θi,v eΘi,v /T CV 3 3 = + + ¡ ¢2 ; k 2 2 T eΘi,v /T − 1 i=1 # " "µ ¶3/2 ¶1/2 # 3NaX µ t o m s −6 · ³ ´¸ π1/2 e3/2 Θi,v /T S 2πmkT T3 V 5/2 −Θi,v /T − ln 1 − e +ln + +ln ω e . = ln e k h2 N σ ΘA ΘB ΘC eΘi,v /T − 1 i=1 U 3 3 = + + kT 2 2 3NaX t o m s −6 · Note that m is the mass of the molecule. G = A + P V = A + N kT. 8 Also, H = U + P V = U + kT and