Gases and the Virial Expansion

Transcription

1 Gases and the irial Expansion February 7, 3 First task is to examine what ensemble theory tells us about simple systems via the thermodynamic connection Calculate thermodynamic quantities: average energy, pressure, etc.. Expressions can be evaluated if partition function can be calculated. Consider a system of N identical point particles with mass m occupying a volume. General Hamiltonian Hx N = pn p N m + Ur N. The canonical partition function is: Q N T, = = N!h 3N N!h 3N Integral over the momenta can be done explicitly: dp N pn p N β e m = dx N e βhxn = N!h 3N dr N dp N e β p N p N +Ur m N dp N pn p N β e m dr N e βurn [ ] 3N p β dp e m = Defining the configurational part of the partition function as Z n = dr N e βurn, we have Q N = N!h 3N mπ β mπ 3N/ = πmkt 3N/. 3N/ Z N = πmkt 3N/ N! h Z N. β

2 Ideal Gas System: Ur N = Simplest possible system: particles do not interact but are confined to volume. Z N = dr N e βurn = dr N = N so Q N = N! πmkt h 3N/ N = qn N! πmkt 3/ q. h Note that q = Q is the partition function for a -particles system. Recall that: E = ln Q N β and ln QN P = kt Now for the ideal gas system, qe ln Q N = N ln q ln N! N ln q N ln N + N lnπn / N ln [ N πmkt ] [ 3/ πm ] e 3/ e = N ln h = N ln N βh N = N ln 3 N ln β + N ln fm, N T.. The pressure is therefore: ln QN P = kt T = kt N ln = nkt Ideal gas equation of state. The energy and heat capacity at constant volume are E = ln Q N = 3 β N ln β β E C v = = 3 T Nk. = 3 nkt. Note that E and C v scale with N as assumed earlier.

3 Non-ideal gas Must evaluate Z N for each model. -particle system: external potential such as gravitational: Ur N = φr leads to a non-uniform system where properties depend on location in volume ex. barometric pressure formula. -particle system in absence of external potential: particles interact via a pair potential: Ur, r = U r r = Ur. Example: Lennard-Jones potential: Ur = Ar + Br 6. 3-particle system in absence of external potential: Ur, r, r 3 = Ur + Ur 3 + Ur 3 + ωr, r, r 3 ωr, r, r 3 = 3 body interaction potential Typically, ωr, r, r 3 is usually small and neglected hence the interactions are considered to be pairwise-additive: Ur, r, r 3 3 i= 3 Ur ij = j>i 3 i= 3 Ur ij j i For an N-particle system, with pairwise additive potential: Ur, r,..., r N N i= N Ur ij = j>i N i= N Ur ij j i N Ur ij. i<j For non-electrostatic potentials, pair interactions Ur ij are usually short-ranged so that Ur ij if r ij ξ, where typically ξ 8 m. Particles separated by long distances do not interact. Define the Ursell-Mayer function fr = exp{ βur} so that fr for r ξ. In terms of the Mayer function, Z N is: Z N = dr N e β i<j Urij = dr N N i<j e β i<j Urij + = dr N N i<j f ij +, where the short hand notation f ij = fr ij has been used.

4 The irial Expansion For systems of very low density, the ideal gas equation of state is approximately correct. On average, particles are far apart an non-interacting. What about correcting this approximation by expanding the partition function in powers of the density ρ = N/? Can we correct the ideal gas equation of state in this fashion? P = NkT [ + ρb T + ρ B 3 T + ] P kt = ρ + ρ B T + ρ 3 B 3 T + Can relate B i T to integrals of Ursell-Mayer functions It is rather difficult to obtain these results in the canonical ensemble. Is it simpler to use other ensembles? irial expansion using the grand canonical ensemble Goal: write pressure P in a series in density ρ = N /. P = kt ρ + ρ j B j+ T Recall that the grand canonical partition function is: Ξµ,, T = Q N T, λ N λ = e βµ N= and P = kt ln Ξ = kt ln + Q N λ N. N= Also recall that Q N = πmkt 3N/ N! h Z N Z N = dr dr N e βurn. Note that in the absence of an external potential, πmkt Q = h 3/ so Q N = N! Q N Z N.

5 The pressure can now be written as: P = kt ln Ξ = kt ln + Z N z N N= where z = λ Q = λa. Formally, we can expand this to get a series expression for P in terms of z homework problem: P = kt b j z j with b = Z = b =! Z Z b 3 = 3! Z3 3Z Z + Z 3 and so on. Strategy is to express this series in terms of the density by expressing the density in a power series of z and inverting. The density is given by ρ = N = ln Ξ βµ,t Since λ = exp{βµ}, βµ = λ βµ λ = λ λ, and since P = kt ln Ξ, ρ = λ ln Ξ λ,t = λ P kt λ = λ P kt λ. Recall that z = aλ, where a = Q /, so ρ = λ kt a P z = z P/kT z = z jb j z j = jb j z j. We wish to invert this relationship to write z in terms of ρ. The procedure is to solve for the coefficients a i where: ρ = b z + b z + 3b 3 z 3 + z = a ρ + a ρ + a 3 ρ 3 +.

6 Inserting expression for z into the expansion of ρ and collecting like powers of ρ: so ρ = ρb a + ρ b a + b a + ρ 3 b a 3 + 4b a a + 3b 3 a 3 + b a = a = /b = b a = b a a = b b = a / b a 3 = 4b a a 3b 3 a 3 a 3 = 8b 3b 3 b 3 = 3 Thus the density expansion for z is: z = ρ b ρ + 8b 3b 3 ρ Inserting this expression of z is the expansion of P/kT, a a 3 P kt = b z + b z + b 3 z 3 + = z a z + a 3 a 3 z = a ρ + a ρ + a 3 ρ a a ρ + a ρ + a 3 ρ a a 3 a ρ + a ρ + a 3 ρ = a ρ + ρ a a a + ρ 3 a 3 a a + 3 a a 3 a = ρ + ρ b + ρ 3 8b 3b 3 4b 3 = ρ + ρ b + ρ 3 4b b Comparing this with the virial expansion, we have: B T = b = Z Z B 3 T = 4b b 3 = Z Z 3 Z3 3Z Z + Z 3 We must now show that these results are equivalent to the Mayer expansion results obtained in the canonical ensemble. Evaluation of irial Coefficients Must express configurational partion functions Z i in terms of Mayer functions. If no external potenial, U = j<i Ur ij. Now: Z = Z = dr = dr dr e βur = dr dr + f = + dr dr f

7 We introduce the new graphical notation: Each graph has a factor of /. Each line represents a factor of f connecting the vertices. Each darkened dot indicates the argument is integrated over. For example: Thus: f = 3 3 = = dr dr f dr dr dr 3 f f 3 dr dr dr 3 f f 3 f 3 Z = + Now Z 3 is given by: Z 3 = dr dr dr 3 + f + f 3 + f 3 = dr dr dr 3 [ + f + f 3 + f 3 + f f 3 + f f 3 + f 3 f 3 + f f 3 f 3 ] = The diagram is reducible since = dr dr dr 3 f f 3 = dr dr dr 3 f f 3 = dr f dr 3 f 3 =. Thus Z 3 =

8 The virial coefficients are therefore: B T = and = B 3 T = 4B T 3 Z Z = dr dr fr = + = Z3 + 3Z Z + Z 3 = = 3 dr e βur dr dr dr 3 fr fr 3 f r + r 3. as obtained with the Mayer expansion method. In general, B n T are expressed in terms of doubly connected irreducible graphs: B n+ T = n n + γ n+ γ n+ = dr dr n+ S,,...,n+ n! S,,...,n+ = Sum of all topologically distinct doubly-connected graphs with n + vertices. For example: S,,3,4 = Coefficients are number of topologically distinct ways of permuting indices on graph. Can generalize this to multi-component systems homework?. Limits of validity of virial expansion. Large systems i.e. N, rigorous in thermodynamic limit N, with fixed ρ = N /.. Short-ranged potentials with at least Ur r 3+ɛ for large r. B T = dr e βur = π dr r e βur. For large r, βur so exp{ βur} βur. Now the integral for B T will converge at upper limit in thermodynamic limit only if converges i.e. ɛ >. a dr r Ur a dr r r 3+ɛ

9 3. Convergence of power series in ρ requires ρ not too big: Not valid for condensed systems liquids. Evaluation of virial coefficients Obviously model dependent since depends on interaction potential Ur. We will look at specific cases:. Hard spheres: No attractive interaction but short-ranged repulsion. { r σ Ur = r > σ This gives: B T = = π dr σ e βur = π dr r = π 3 σ3 P NkT = + ρπ 3 σ3. dr r e βur Note that no temperature dependence of B j T : All isotherms of plot of P versus ρ or are identical. No condensation or other phase transition for such a system.. Square well potential B T = π = π 3 [ σ 3 + Ur = r σ ɛ σ < r < σ r σ dr r e βur [ σ = π e βɛ ] σ 3 σ 3. Isotherms of P vs. now differ and condensation possible. σ drr + dr r e βɛ] σ Perturbation Theory and the an-der-waals Equation Recall that the configurational partition function is: Z N = dr dr N e βur,...,r N

10 Suppose that the potential can be written as the sum of two components: Ur N = U r N + U r N where U is small in some sense compared to U. Let Z N = dr dr N e βu r,...,r N n r N = e βu r N Z N Note that n r N is the configurational part of the probability density for a system with potential U r N. Ideally, one would like to be able to evaluate Z N analytically. We may express Z N as: Z N = Z N ZN = ZN dr N e βu r N e βu r N dr N n r N e βu r N = Z N e βu r N. Notation means average with respect to probability density for the unperturbed system. Implications: the contribution of the configurational part of the partition function to the Helmholtz free energy is [ πmkt AN,, T = ] 3N/ β ln Z N h N! [ πmkt = ] 3N/ β ln ZN h N! β ln e βu r N = A + A = A β j C j β j! The coefficients C j are called cumulants and are defined by β j exp C j j! = e βx β k X k β j = exp C j k! j! = + β j C j + β j C j +... j! j! k= where X denotes the average of variable X with respect to a probability density fx.

11 Solving for C j homework gives: General properties of cumulants: C = X C = X X C 3 = X 3 3 X X + X 3.. Typically, if the property X N, then C j N. This makes them much more useful in expansions than moment expansions.. Quite often, C j > C j+ so truncations of the expansions are possible. In terms of the cumulants, we have the perturbation expansion of the Helmholtz free energy A = A + C β β! C β + β 3! C 3β + = A + U β U! U + β U 3 3! 3 U U + U 3 + Derivation of the an der Waals Equation We define the pairwise additive reference potential to be with U r N = u r ij i j { if r > σ u r = if r σ where σ is the diameter of the spherical particles. The form of the attractive potential U is not particularly important According to the perturbation scheme above, AN, T, = A N, T, + U +... We define the an der Waals parameter a so that a = U Nρ AN,, T = A N,, T an + Evaluation of a can be done once the form of U is specified. If we take σ =, then Z N = N is the ideal gas result.

12 Due to hard core repulsions, particles cannot overlap, so volume is restricted. Excluded volume per pair of particles is 4π/3σ 3, and hence excluded volume per particle is π/3σ 3 b. The configurational partition function is therefore Z N = Nb N Nb is the excluded volume for each particle, Nb is the total free volume for each particle. The Helmholtz free energy is therefore: [ πmkt AN, T, ] 3N/ β ln Nb N h an N! [ πmkt ] 3N/ = kt ln h kt ln Nb N an N! From the thermodynamic relations, the energy and pressure are therefore E = ln Q N = 3 kt + anρ β A N P = = kt an Nb N,T = kt ρ ρb aρ This is the an der Waals equation of state. In the low density limit, ρb << so ρb + ρb and hence P kt ρ + ρ b a kt The second virial coefficient is therefore approximately B T b a/kt. Plot of P- isotherms of an der Waals gas show interesting behavior: At high temperatures, we have P/ < as expected of a gas: decreasing the volume increases the pressure. At a particular called critical temperature T c, there is a region in the curve at a critical volume c where P/ = and there is no curvature, P/ =. For all isotherms below T c, there are unphysical regions where P/ >. For these regions, see a volume discontinuity as pressure is lowered: liquid-gas transition.

13 Correct isotherm curves determined by the equal areas approach that follows from the equality of chemical potentials of the liquid and gase phases. What are these critical points? Determined by the conditions P = P = Evaluation of these conditions gives: c = 3Nb kt c = 8a 7b P c = a 7b Near the critical point, we can expand the pressure around the critical temperature and density and find that: P kt = P c + P kt c kt c ρ ρ ρ c + P ρ=ρc kt c ρ ρ ρ c + 3 P ρ=ρc 6kT c ρ 3 ρ ρ c 3 + ρ=ρc = P c + [ P kt c kt c ρ ρ ρ c + ] P ρ=ρc kt c + P ρ ρ ρ ρ c ρ=ρ c + 3 P 6kT c ρ 3 ρ ρ c 3 + ρ=ρc = P c + 3 P kt c 6kT c ρ 3 ρ ρ c 3 + = ρ=ρc 8b b ρ ρ c 3 + Thus, we expect that near the critical point, with the critical exponent δ = 3. Experimentally, it is found that δ 4.. P kt constant + Cρ ρ c δ Improvements: higher order in perturbation theory or better zeroth order potential