University of Groningen Permeation of small molecules across a polymer membrane Sok, Robert Martin IMPORTANT NOTE: You are advised to consult the publisher's version (publisher's PDF) if you wish to cite from it. Please check the document version below. Document ersion Publisher's PDF, also known as ersion of record Publication date: 994 Link to publication in University of Groningen/UMCG research database Citation for published version (APA): Sok, R. M. (994). Permeation of small molecules across a polymer membrane: a computer simulation study s.n. Copyright Other than for strictly personal use, it is not permitted to download or to forward/distribute the text or part of it without the consent of the author(s) and/or copyright holder(s), unless the work is under an open content license (like Creative Commons). Take-down policy If you believe that this document breaches copyright please contact us providing details, and we will remove access to the work immediately and investigate your claim. Downloaded from the University of Groningen/UMCG research database (Pure): http://www.rug.nl/research/portal. For technical reasons the number of authors shown on this cover page is limited to maximum. Download date: 5--9
Appendix A Reprint Molecular dynamics simulation of the transport of small molecules across a polymer membrane R.M. Sok, H.J.C. Berendsen and W.F. van Gunsteren Reprinter from J. Chem. Phys. 96, 4699 (99). 9
Appendix B The error in the diffusion coefficient Let r(t) be the displacement of a diffusing particle and r = jrj. The probability distribution of r is gaussian, and the probability distribution of r = x + y + z is givenbya distribution for 3 degrees of freedom. If we write s = r,wehave f (s)ds = p s 4 p exp(,s=4dt) ds (B:) (Dt) 3= where D is the diffusion constant and t the time. hsi = = sf(s) ds 4 p (Dt) 3= s 3= exp(,s=4dt) ds (B.) on substituting u = s=(4dt),wefind = (4Dt)5= 4 p u 3= exp(,u) du (B:3) The integral is the error function Γ( 5 ) and is equal to 3 4p so: hsi = 45= Dt 4 p 3p (Dt) 3= 4 (B:4) If we work this out we obtain the well known relation: hsi = 6Dt: (B:5) 7
8 The error in the diffusion coefficient Let us now concentrate on the error in s, again we use and using equation B. we find: D (s,hsi) E = D (s,hsi) E = (s,hsi) f (s)ds (B:6) 4 p (s, (Dt) 3= 6Dt) s = exp(,s=4dt)ds (B:7) Again take u = s=(4dt) D(s E,hsi) = 4(Dt) (4Dt) 3= 4 p (4u, u (Dt) 3= + 9) u = exp(,u)du (B:8) the integral is a combination of error functions and has a value of 3 p,so D (s,hsi) E = 8(Dt) 3 p p = (6Dt) 3 (B.9) Thus the average value of s with its standard error is : rd s,hsi E hsi p n ; (B:) so, hsi = 6Dt q 3 p p A ; n (B:) where n is the number of independent samples. The choice for n in a real system is not so obvious. In this thesis i have taken the total simulation time divided by the residence time t :5 (page 4). This residence time is approximately the time a penetrant resides in a certain hole.
Appendix C Deriviation of enthalpy expression The excess enthalpy of insertion h ex can be evaluated by use of an enhanced version of Widom s particle insertion method. The deriviation of the expression used for this is given in this appendix. Let us denote the inserted particle with subscript and the system particles with subscript i, so particle : inserted particle particle i =:::N : system particles The energy of interaction between the inserted particle and the system particles is given by: E i = NX i i (r i ) (C:) and the internal system energy is: E ij = NX i<j ij (r ij ) (C:) where ab (r) is the interatomic interaction potential of particles a and b at distance r. The expression for the excess chemical potential of the inserted particle ex in Widom s method is given as : ex =,kt ln hexp(,e i )i ; (C:3) where 9
Deriviation of enthalpy expression hexp(,e i )i = dr exp(,e i ) exp(,e ij ) R : (C:4) dr exp(,eij ) Where we used the shorthand notation dr = dr :::dr N. The excess partial molar enthalpy h ex associated with the process of solvation can also be calculated in a way similar to that of the chemical potential. It can be derived using : h ex =! T (=T ) (C:5) and using = k we find (=T ) h ex =, ln hexp(,e i)i : (C:6) This evaluates to: h ex =, hexp(,e i )i hexp(,e i)i (C:7) Let us first concentrate on the derivative:, hexp(,e i)i =, dr exp(,e i ) exp(,e ij ) (C:8) = dr exp(,e i ) exp(,e ij ) dr(,e ij ) exp(,e ij ),
Deriviation of enthalpy expression dr(,e i, E ij ) exp(,e i ) exp(,e ij ) : (C:9) We can also write this using the <> notation as: =,he ij ihexp(,e i )i + he i exp(,e i )i + he ij exp(,e i )i (C:) So the final expression for the excess enthalpy of the inserted particle is: h ex = he i exp(,e i )i hexp(,e i )i + he ij exp(,e i )i hexp(,e i )i,he ij i : (C:) The first term in this expression simply is the energy of the inserted particle weighed with its Boltzmann weight (and normalized). The second and third term can be thought of as being a correction term to the first. They will cancel each other if there is no correlation between the system energy and the energy of the inserted penetrant. In practice the correction term will show poor convergence. A similar expression for the patial enthalpy in the constant-npt ensemble is given in [9].