solved problems in physical chem

Chem 120B

Problem Set 1

Due: September 10, 2014

1. (i) Show that the mean square fluctuation in any quantity A can be written in terms of its mean and mean square values, h(δA)2 i = hA2 i − hAi2 . (Here, as usual, the fluctuation in A is defined as δA = A − hAi.) (ii) Generalize this result to the case of two different fluctuation quantities, i.e., show that hδA δBi = hABi − hAihBi. 2. Consider a system of N molecules contained in a fixed volume V . Imagine dividing this total volume into a large number M of microscopic subvolumes v (so that M = V /v), and let ni be the number of molecules within subvolume i. (This notional division does not prevent molecules from moving among subvolumes; it is just a way to keep track of their spatial distribution.) As molecules move within the container, the values of n1 , n2 , . . . , nM fluctuate, while the total number N=

M X

ni

i=1

remains constant. What is the average number of molecules within a single subvolume, hni i? Make your argument carefully, using the fact that all subvolumes are statistically equivalent, i.e., they undergo the same set of fluctuations. Write your answer for hni i in terms of v and the total density ρ0 = N/V . P 3. The notation 3i=1 xi is shorthand for a sum over the quantities x1 , x2 , and x3 , 3 X

xi = x1 + x2 + x3 .

i=1

(i) Write out the double summation 3 X 3 X

Aij

i=1 j=1

explicitly in terms of the quantities A11 , A12 , etc. P (ii) Let S = 3i=1 xi . Write the quantity S 2 in terms of x1 , x2 , and x3 . (iii) Now consider the case Aij = xi xj . Using your results from parts (i) and (ii), show that 3 X 3 X

Aij = S 2 ,

i=1 j=1

and therefore that

!2 X

xi

i

(iv) Write out all terms in the sum

P3

2 i=1 xi ,

=

XX i

xi xj .

j

and show that it is in general different from (

P3

i=1 xi )

2.

4. Consider a dilute solution with a total number Ntot of solute molecules contained in a total volume Vtot . The total number density of solutes is thus ρ0 = Ntot /Vtot . We will focus on a limited region of the solution, marked by a dashed line in the sketch below. Chem 120B, Fall 2014

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cell i, hi = 1 cell j, hj = 0

(Lattice cells and solute molecules not drawn to scale)

This observation region has a volume V . As solutes move across its boundary, the number N of solute molecules inside the observation region fluctuates about an average value hN i. In order to examine fluctuations in N , it is useful to imagine dividing the observation region into microscopic cells of a cubic lattice, each with volume v. Let ni be the number of solute molecules in cell number i in a given measurement. We will take the solution to be sufficiently dilute that finding two solutes in the same cell is negligibly unlikely. In other words, either ni = 0 or ni = 1. Assume as well that fluctuations in different cells are uncorrelated. (i) Estimate the size of typical fluctuations in N by calculating the root mean square deviation σ = p h(δN )2 i, where δN = N − hN i is the deviation of N from its average value. The correct answer can be written solely in terms of hN i, ρ0 , and v. (Hint: Note that n2i = ni .) (ii) Show that h(δN )2 i = hN i in the limit of very low concentration, ρ0 v  1. (iii) Calculate σ/hN i and comment on the size of fluctuations in N relative to its mean when the observation region is macroscopically large. 5. Consider the dilute solution of question 4 from a slightly different perspective: Any of the M cells in the observation region should be occupied (ni = 1) with probability p1 = ρ0 v. The probability P (n1 , n2 , . . . , nM ) of finding the system in a particular configuration of the observation volume (in M −N . which the values of n1 , n2 , . . . , nM are all specified) is thus pN 1 (1 − p1 ) M −N . (i) Show and/or explain these two facts, p1 = ρ0 v and P (n1 , n2 , . . . , nM ) = pN 1 (1 − p1 )

(ii) Let W (N ) be the number of configurations of the observation volume when N solutes are present. Calculate the probability P (N ) of observing a given value of N , in terms of p1 , W (N ), M , and N . (iii) Later in the course, we will show that, when N and M are large, ln W (N ) ≈ −M [φ ln φ + (1 − φ) ln(1 − φ)] , where φ = N/M is the fraction of occupied cells. Using this fact, write ln P (N )/M solely in terms of φ and p1 . (iv) Using a computer, plot your result for P (N ) as a function of the fraction of occupied lattice cells, φ. On a single graph, show results for p1 = 0.1 and the cases M = 10, M = 100, M = 1, 000, and M = 10, 000. (v) Comment on the trends evident from your graph and your expectations for a macroscopic observation volume (with M ∼ 1024 ).

Chem 120B, Fall 2014

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