MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.5.62 Lecture #9: CALCULATION OF MACROSCOPIC PROPERTIES FROM MICROSCOPIC ENERGY LEVELS: qtrans The macroscopic thermodynamic properties are written in terms of Q. Q isrelated to the single-molecule partition function q, which is the sum over the molecularenergy levels or states. Atoms and molecules have different kinds of states or energylevels. Each type of state makes its contribution, through q, to the macroscopic propertyunder consideration. TRANSLATIONAL CONTRIBUTION TO MACROSCOPIC PROPERTIES Vq⎜ ⎟2trans ⎝ ⎠h⎞⎠Single-Molecule Translational Partition Function ⎛ 2πmkT ⎞ 3 /2 = Canonical Translational Partition Function ⎟ N3 /2 qNtrans 1 2πmkT ⎡ ⎤ ⎛⎝⎜2πmkT Qtrans V ⎢⎣ ⎥⎦ = = h2N! N! 3/2 ⎡ ⎤ ⎛⎜⎝⎢⎣ ⎞⎟⎠ lnQtrans = – ln N! + N ln V ⎥⎦ h2 3 2 ⎛⎜⎝N ln 2πmk h2 3 ⎞⎟⎠ = −N ln N + N + N lnT + N lnV+ 2 (ln N! = N ln N – N is Stirling’s Approximation) Calculate the translational contribution to the average energy (one of the contributions toU, internal energy) ∂ln Qtrans E = kT2⎛⎜⎝ ⎞⎟⎠ ∂T N,V 3 N = ∂ln Qtrans ⎛⎜⎝ ⎞⎟⎠ ∂T N,V 2 T E = kT2 ⎝⎜⎛ 23 NT ⎠⎟⎞ = 23NkT = 23nRT5.62 Spring 2008 Lecture 9, Page 2 Average translational energy of N molecules in a gas What about the contribution of translational energy to the heat capacity CV ≡ ⎜⎛∂U ⎞⎟⎝∂T ⎠N ,V CV trans = ⎝⎜⎛∂E∂Ttrans ⎞⎠⎟ N ,V = ∂∂ T ⎡⎣⎢23 NkT ⎤⎦⎥ = 3 Nk = 3 nR2 2 Recall from 5.60 that CV for an ideal monatomic gas was often observed to be 3R2 CV = CV / n = There are no other important places other than translation for an ideal monatomic gas toput internal energy. Calculate translational contribution to pressure : ∂A ⎞⎟⎠N,T ⎛⎜⎝ = kT ∂lnQ ∂V⎛⎜⎝ ⎞⎟⎠N,T (recall dA = –pdV – SdT + µdN) p = − ∂V because A = –kT ln Q 3 21 ⎛⎝+ N ln 2πmkT ⎜h2 ⎞⎠⎟ is V dependent) ⎛∂ ln Qtrans ⎞ N ⎝⎜ ∂V ⎠⎟ N,T = V So p = NkT/V pV = NkT = nRT IDEAL GAS LAW Calculate translational contribution to entropy S = k ln Q + E/T ⎛⎝ ⎞⎠ ln Qtrans = N lnV + ln (only the first termN! revised 2/27/08 9:48 AM5.62 Spring 2008 Lecture 9, Page 3 ⎡ln Qtrans = N ln ⎢(2πhk)3 3 /2 ⎦⎥⎤ + 23N ln m + 23N ln T + N ln V − ln N! ⎣Again ln N! ~ N ln N – N Stirling's approximation, which is valid for large N = N ln (2πk)3 /2 h3 ⎡ ⎢⎣ ⎧⎪⎨ ⎫⎪⎬⎞⎟⎠ ⎤ ⎥⎦ + 1+ 3 3 ⎛⎜⎝ln T + ln V NSo ln Qtrans ln m + 2 2 ⎪⎩ ⎪⎭ Now V kT ∴ln V ln k + ln T − ln p N = pN = ln Qtrans = N ln (2π)3/2 k5 2 h3 ⎡ ⎢⎣ ⎡⎢⎣ ⎤ ⎥⎦ + 1+ ⎤ ⎥⎦ 3 5 ln m + ln T − ln p So 2 2 S = k ln Q + ET = k ln Q +(3 / 2) Nk ⎡ ⎢⎣ ⎡⎢⎣ S = Nk ln (2π)3 /2 k5/2 h3 ⎤ ⎥⎦ 5 3 5+ 2 + 2 ln m + 2 ln T − ln p ⎤ ⎥⎦ S 5ln T + 3 5 Nk = 2 2ln m − ln p + 2 + ln (constants) S 5ln T + 3ln m − ln p − 1.164871Nk = 2 2 [T] = K; [m] = g mol–1; [p] = atm (not S. I.) revised 2/27/08 9:48 AM 5.62 Spring 2008 Lecture 9, Page 4 S Nk = 5 2 ln T + 3 2 ln m − ln p − 1.15170 [p] = bar SACKUR-TETRODE EQUATION 1911-13 [Sackur and Tetrode were people, not equipment!] P(ε) FOR TRANSLATION We know Pi for translation Pi = e− ε(L,M,N)/kT where ε(L,M, N) = h2 ⎛ L2 M2 N2 ⎞ = PL,M,N qtrans 8m ⎝⎜ a2 + b2 + c2 ⎠⎟ quantum= εx(L) + εy(M) + εz(N) state The lowest possible energy is forL = M = N = 1 probability of molecule in translational state with the three quantum numbers # L, M, N But P(ε) is a more useful form than Pi. Rewrite PL,M,N as P(ε) P(ε) = g(ε)e− ε/kT / q trans NEED: to calculate g(ε) Consider energy of N2 molecule in state L = 1, M = 1, N = 1 in 10 cm cube. ε(1,1,1) = 3 × 1.695 × 10–20 kcal/mol (3 = 1 + 1 + 1) prob. ofmolecule with energy ε degeneracy (# of molecularstates with energy ε) revised 2/27/08 9:48 AM5.62 Spring 2008 Lecture 9, Page 5 Next higher energy state is ε(2,1,1) = 6 × 1.695 ×10–20 kcal/mol (6 = 4 + 1 + 1) States are very close in energy: Δε ≈ 10–20 kcal/mol Because the allowed energies of a molecule are so closely spaced, the discrete P(ε) can beapproximated by a continuous P(ε) dε. If P(ε) is treated as continuous, then PL,M,N must also be treated continuous because these distributions must map onto each other. The problem is more easily solved for 1 dimension at a time. Consider the x-dimension only P(εx)d(εx) represents probability of finding the molecule with energy betweenεx and εx + dεx due to translation in the x direction; we want todetermine this continuous distribution P(L)dL represents probability of finding the molecule with quantum numberL between L and L + dL for motion in x-direction; we know thisdistribution by virtue of knowing the discrete distribution PL dL # states between L and L + dL = dεx εx (L + dL) − εx(L) Often described as “# states per unit εx.” quantum number, not length   ↓ e−εx(L)/kT ⎛ 2πmkTa2 ⎞1 2 a is length of container inP(L) = qx where qx = ⎝⎜ h2 ⎠⎟ x direction because ⎛ 2πmkTa2 ⎞1 2 1 2 1 2 ⎛ 2πmkTb2 ⎞ ⎛ 2πmkTc2 ⎞ qtrans = qxqyqz = ⎝⎜ h2 ⎠⎟ ⎝⎜ h2 ⎠⎟ ⎝⎜ h2 ⎠⎟ Now P(εx) and P(L) are two continuous distribution functions which must map onto eachother. In essence, they represent the same distribution but the variable has changed.Problem is to relate P(L) to P(εx). This can be done by revised 2/27/08 9:48 AM5.62 Spring 2008 Lecture 9, Page 6 P(εx)dεx = P(L)dL P(εx) = P(L) dL dεx Jacobian of the transformation (see pages 9-7, 9-8 about thechange of variables) The Jacobian for this change of variables is essentially the degeneracy. It tells us how many states there are within a small interval in εx (“density of states”, dn/dE). [Inquantum mechanics the density of states is also important. In one dimensional systems, itis proportional to the period of motion in a potential.] dLCalculate dεx so that P(εx) can be calculated L2h2 ⎛ 8ma2 εx ⎞1/2 εx = 8ma2 or L = ⎝⎜ h2 ⎠⎟ dL = d ⎛ 8ma2 εx ⎞1/2 ⎛ 2ma2 ⎞1/2 −1/ 2 so dεxdεx ⎝⎜ h2 ⎠⎟ ⎝⎜ h2 ⎠⎟ ε= x ( )/kT …