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Problem 3.1 Reif (a) At equilibrium, the temperature in the larger (L) side and smaller (S) side of the box will be equal, TL = TS . With a hole in the partition, it should be intuitively clear that the pressures, and, therefore, the number densities of particles should be equal on each side of the partition. Thus we have NS NL = , VS VL (1) for each species of particle. Since we are given that VL = 3VS , it follows immediately that NL = 3NS and, thus, that 3/4 of the particles of each species are in the large side and 1/4 in the small side. In terms of numbers, we have 750 Ne particles and 75 He particles on the large side and 250 Ne and 25 He particles on the small side. Because intuition in physics is a developed skill, the above solution may be demonstrated in a systematic approach based on our knowledge of effusion. At steady state, the two-way fluxes of each species through the hole will be equal and given by the expression nv/4 for the total flux. Thus we have (nv)L = (nv)S , (2) for each species. Since TL = TS , we see that (v)L = (v)S for each species, and thus we have (n)L = (n)S , (3) which is equivalent to Eq.(1). The only problem with this approach is that knowledge of effusion and the total molecular flux is covered in Ch. 7, and would not be available for use in Ch. 3 if we were covering the chapters in numerical order. So there should be a way to solve this problem using only information covered in Ch. 3. This is called "making an easy problem into a difficult one." The Ch. 3 method is to find the value of NS (or NL ) that maximizes the total number of accessible states for the composite system, Ω(0) , where Ω(0) = ΩL ΩS , (4) and ΩL and ΩS are the numbers of accessible states for the large and small subsystems, respectively. Assuming the gases are ideal, we can write (for each species) 3N /2 B 3Nα /2 VαNα Eα α Ωα = , α = L, S (5) Nα !(3Nα /2)! where B is the constant, B= 2πm . h2 (6) The total energy is constant E (0) = EL + ES = N ε, 1 (7) where ε is the average energy per particle and ES = NS ε, EL = NL ε. (8) After substituting the above expressions and using Stirling’s approximation for ln(n!), we find ln Ω(0) 3NL 3NL 3NL ln ε − NL ln NL + NL + + ln B + 2 2 2 3NS 3NS 3NS NS ln VS + ln ε − NS ln NS + NS + + ln B. (9) 2 2 2 = NL ln VL + Remembering that N = NL + NS , we can further simplify this expression to read 3N 5N 3N ln ε + + ln B. 2 2 2 (10) Now we differentiate Eq.(10) with respect to NS , while holding N , ε, VS , and VL constant and realizing that ∂NL /∂NS = −1, to find ln Ω(0) = NL ln VL + NS ln VS − NL ln NL − NS ln NS + VS NL ∂ ln Ω(0) = − ln VL + ln VS + ln NL + 1 − ln NS − 1 = ln . ∂NS VL NS (11) The maximum value of Ω(0) occurs when ∂ ln Ω(0) /∂NS = 0, and thus when VS NL = 1, VL NS (12) which is equivalent to the intuitive result found above in Eq.(1). (b) Think of this problem as two simultaneous random walks, one with N = 1000 and the other with N = 100. For each walk the probability of a "step" into the small side is VS 1 p= = , (13) V 4 and the probability for a "step" into the large side is q =1−p= VL 3 = . V 4 (14) The probability of simultaneously finding 100 He atoms in the small side and 1000 Ne atoms in the large side is P = W100 (100)W1000 (0), (15) where WN (n) is the binomial probability distribution for achieving n successes in N trials. Using results from Ch. 1, we have µ ¶100 µ ¶1000 1 3 P = ≈ 10−185 . 4 4 This is a vanishingly small number even for such relatively small values of N . 2