5.62 Physical Chemistry II Spring 2008

Transcription

1 MIT OpenCourseWare Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit:

2 5.62 Lecture #9: CALCULATION OF MACROSCOPIC PROPERTIES FROM MICROSCOPIC ENERGY LEVELS: q trans The macroscopic thermodynamic properties are written in terms of Q. Q is related to the single-molecule partition function q, which is the sum over the molecular energy levels or states. Atoms and molecules have different kinds of states or energy levels. Each type of state makes its contribution, through q, to the macroscopic property under consideration. TRANSLATIONAL CONTRIBUTION TO MACROSCOPIC PROPERTIES Single-Molecule Translational Partition Function 2πmkT 3/2 qtrans = V 2 h Canonical Translational Partition Function Q trans q N trans 1 2πmkT = = N! N! h 2 3/2 N V 2πmkT lnq trans = ln N! + N ln h 2 3/2 V 3 N ln 2πmk 3 = N ln N + N + N lnt + N lnv 2 h (ln N! = N ln N N is Stirling s Approximation) Calculate the translational contribution to the average energy (one of the contributions to U, internal energy) E = kt 2 lnq trans lnq trans T N,V 3 N = T N,V 2 T E = kt 2 3 N 2 T 3 3 = NkT = nrt 2 2

3 5.62 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 C V C V U T N,V trans = E trans T 3 = NkT N,V T 2 = 3 Nk = 3 nr 2 2 Recall from 5.60 that C V for an ideal monatomic gas was often observed to be 3 C V = C V / n = R 2 There are no other important places other than translation for an ideal monatomic gas to put internal energy. Calculate translational contribution to pressure : p = V A N,T = kt lnq V N,T (recall da = pdv SdT + µdn) because A = kt ln Q 1 + N ln 2πmkT lnq trans = N lnv + ln N! h (only the first term is V dependent) So lnq trans N V = N,T V p = NkT/V pv = NkT = nrt IDEAL GAS LAW Calculate translational contribution to entropy S = k ln Q + E /T

4 5.62 Spring 2008 Lecture 9, Page 3 (2πk) lnq trans = N ln 3 h 3/ N ln m + N ln T + N lnv ln N! 2 2 Again ln N! ~ N ln N N Stirling's approximation, which is valid for large N = N ln (2πk)3/2 h So lnq ln T + ln V trans ln m N Now V kt ln V ln k + ln T ln p N = p N = So lnq trans = N ln (2π)3/2 k 5 2 h ln m + ln T ln p 2 2 S = k lnq + E T = k lnq + (3 / 2) Nk S = Nk ln (2π)3/2 k 5/ h ln m + ln T ln p S ln T + Nk = 2 2 ln m ln p + + ln (constants) 2 S 5 ln T + 3 ln m ln p Nk = 2 2 [T] = K; [m] = g mol 1 ; [p] = atm (not S. I.)

5 5.62 Spring 2008 Lecture 9, Page 4 P(ε) FOR TRANSLATION S Nk = 5 2 ln T ln m ln p [p] = bar SACKUR-TETRODE EQUATION [Sackur and Tetrode were people, not equipment!] We know P i for translation P i = e ε(l,m,n)/kt where ε(l,m,n) = h2 L 2 M 2 N 2 = P L,M,N q trans 8m a b 2 c 2 quantum state The lowest possible energy is for L = M = N = 1 = ε x (L) + ε y (M) + ε z (N) probability of molecule in translational state with the three quantum numbers # L, M, N But P(ε) is a more useful form than P i. Rewrite P L,M,N as P(ε) P(ε) = g(ε)e ε/kt / q trans prob. of molecule with energy ε degeneracy (# of molecular states with energy ε) NEED: to calculate g(ε) Consider energy of N 2 molecule in state L = 1, M = 1, N = 1 in 10 cm cube. ε(1,1,1) = kcal/mol (3 = )

6 5.62 Spring 2008 Lecture 9, Page 5 Next higher energy state is ε(2,1,1) = kcal/mol (6 = ) States are very close in energy: Δε kcal/mol Because the allowed energies of a molecule are so closely spaced, the discrete P(ε) can be approximated by a continuous P(ε) dε. If P(ε) is treated as continuous, then P L,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(ε x )d(ε x ) P(L)dL represents probability of finding the molecule with energy between ε x and ε x + dε x due to translation in the x direction; we want to determine this continuous distribution represents probability of finding the molecule with quantum number L between L and L + dl for motion in x-direction; we know this distribution by virtue of knowing the discrete distribution PL dl dε x = # states between L and L + dl ε x (L + dl) ε x (L) Often described as # states per unit ε x. quantum number, not length P(L) = e ε x (L)/kT 2πmkTa a is length of container in where q x = h 2 x direction q x because 2πmkTa πmkTb πmkTc q trans = q x q y q z = h 2 h 2 h 2 Now P(ε x ) and P(L) are two continuous distribution functions which must map onto each other. In essence, they represent the same distribution but the variable has changed. Problem is to relate P(L) to P(ε x ). This can be done by

7 5.62 Spring 2008 Lecture 9, Page 6 P(ε x )dε x = P(L)dL P(ε x ) = P(L) dl dε x Jacobian of the transformation (see pages 9-7, 9-8 about the change 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 ε x ( density of states, dn/de). [In quantum mechanics the density of states is also important. In one dimensional systems, it is proportional to the period of motion in a potential.] dl Calculate dεx so that P(ε x ) can be calculated L 2 h 2 8ma 2 ε x 1/2 ε x = 8ma 2 or L = h 2 so dl d 8ma 2 ε = x 1/2 2ma 2 1/2 1/2 dε x dε x h 2 = h 2 εx so ( )/kt 2ma 2 P (ε x ) = P(L) dl = e ε x L dε x q x h 2 1/2 ε 1/2 x = e ε x /kt 2ma 2 1/2 2πmkTa 2 1/2 h 2 ε h 2 1/2 x P (ε x ) = (πkt) 1/2 ε 1/2 x e ε x Check normalization (require P (ε x )dε x = 1) 0 P (ε x ) dε x = kt 1 1/2 ε 1/2 x e ε x πkt π 1/2 kt dε x P(ε x )dε x = 1 1/2 πkt = 1 1/2 (1 / kt) correctly normalized. Therefore:

8 5.62 Spring 2008 Lecture 9, Page 7 P (ε x ) = (πkt) 1/2 ε 1/2 x e ε x kt BOLTZMANN KINETIC ENERGY DISTRIBUTION FUNCTION IN 1D Probability of finding a free molecule in a gas with kinetic energy between ε x and ε x + dε x due to translation in x-direction. For functions of one variable Note on Change of Variable(s) We know f(x) but we want to take the information contained in f(x) and re-express it in terms of a different variable, y. f(x)dx g(y)dy We want to know g(y) and know how to use it in place of f(x). dy dy = dx dx f (x)dx = g(y) dy dx thus f (x) = g[ y(x) ] dy dx dx and g(y) = f [x(y)] dx dy g[y(x)] means: evaluate the function g(y) at the value of y determined by the value of the function y(x) evaluated at x.

9 5.62 Spring 2008 Lecture 9, Page 8 For functions of two or more variables f (x, y, z, )dxdydz = g( r, s,t, )drdsdt f (x, y, z ) = (r, s,t, ) g( r ( x, y, z, ), s( x, y, z, ),t ( x, y, z, )) (x, y, z, ) Jacobian of transformation ( r, s,t, ) x, y, z, ( ) = r x r y s x s y t x t y r s t z z z where the single derivative for one variable is replaced by an N N determinant of derivatives for N variables. Example: x, y, z, r, θ, φ x = r sin θ cos φ y = r sin θ sin φ z = r cos θ 3 x 3 determinant dxdydz = x y z r r r sθcφ sθsφ cφ x y z drdθdφ = rcθcφ rcθsφ rsθ θ θ θ rsθsφ rsθcφ 0 x y z φ φ φ drdθdφ which reduces to the familiar result: dxdydz = r 2 sinθ drdθdφ.

10 5.62 Spring 2008 Lecture 9, Page 9 A Note About Negative Temperature We saw that the molecular partition function is q = e ε i i and that the fractional population of the i-th level in one molecule is kt f i = e ε i kt q and the population of that level in a sample containing N molecules is The population ratio for levels i and j n i = Nf i. For T > 0. f i n i ( εi ε j ) kt = = e f j n j When ε i < ε j, f j < f i For T 0K f j f i For T and ε i < ε j f j f i But if f j > f i (as occurs in a laser gain medium) that would imply T < 0. There is no possibility of equilibrium with T < 0K, but one often uses a population ratio as a thermometer. Thus a steady state (but not equilibrium) situation is often described by a negative absolute temperature. Note that, if you started with a steady state sample at T < 0K, and then turned off the source of energy input that sustains the nonequilibrium steady state, that T (defined by a population ratio) would become increasingly negative, pass through directly to +, and then decrease to the equilibrium T of the surroundings.