3.10: Problems
- Page ID
- 206325
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1. Consider a system with three non-degenerate quantum states having energies ϵ1=0.9 kT, ϵ2=1.0 kT, and ϵ3=1.1 kT. The system contains N=3×1010 molecules. Calculate the partition function and the number of molecules in each quantum state when the system is at equilibrium. This is the equilibrium population set {N⦁1,N⦁2,N⦁3}. Let Wmp be the number of microstates associated with the equilibrium population set. Consider the population set when 10−5 of the molecules in ϵ2 are moved to each of ϵ1 and ϵ3. This is the population set {N⦁1+10−5N⦁2, N⦁2−2×10−5, N⦁3+10−5N⦁2}. Let W be the number of microstates associated with this non-equilibrium population set.
(a) What percentage of the molecules are moved in converting the first population set into the second?
(b) How do the energies of these two populations sets differ from one another?
(c) Find Wmp/W. Use Stirling’s approximation and carry as many significant figures as your calculator will allow. You need at least six.
(d) What does this calculation demonstrate?
2. Find the approximate number of energy levels for which ϵ<kt> for a molecule of molecular weight 40 in a box of volume 10−6 m3 at 300 K.
3. The partition function plays a central role in relating the probability of finding a molecule in a particular quantum state to the energy of that state. The energy levels available to a particle in a one-dimensional box are
ϵn=n2h28mℓ2
where m is the mass of the particle and ℓ is the length of the box. For molecular masses and boxes of macroscopic lengths, the factor h2/8mℓ2 is a very small number. Consequently, the energy levels available to a molecule in such a box can be considered to be effectively continuous in the quantum number, n. That is, the partition function sum can be closely approximated by an integral in which the variable of integration, n, runs from 0 to ∞.
(a) Obtain a formula for the partition function of a particle in a one-dimensional box. Integral tables give ∫∞0exp(−an2)dn=√π/4a
(b) The expected value of the energy of a molecule is given by ⟨ϵ⟩=kT2(∂lnz ∂T)V
What is ⟨ϵ⟩ for a particle in a box?
(c) The relationship between the partition function and the per-molecule Helmholtz free energy is A=−kTlnz . For a molecule in a one-dimensional box, we have dA=−SdT−ρℓ, where ρ is the per-molecule “pressure” on the ends of the box and ℓ is the length of the box. (The increment of work associated with changing the length of the box is dw=−ρ dℓ. In this relationship, dℓ is the incremental change in the length of the box and ρ is the one-dimensional “pressure” contribution from each molecule. ρ is, of course, just the force required to push the end of the box outward by a distance dℓ. ρdℓ is the one-dimensional analog of PdV.) For the one-dimensional system, it follows that ρ=−(∂A∂ℓ)T
Use this information to find ρ for a molecule in a one-dimensional box.
(d) We can find ρ for a molecule in a one-dimensional box in another way. The per-molecule contribution to the pressure of a three-dimensional system is related to the energy-level probabilities, Pi, by
Psystemmolecule=−∞∑n=1Pn(∂ϵn∂V)T
By the same argument we use for the three-dimensional case, we find that the per-molecule contribution to the “pressure” inside a one-dimensional box is
ρ=−∞∑n=1Pn(∂ϵn∂ℓ)T
From the equation for the energy levels of a particle in a one dimensional box, find an equation for
(∂ϵn∂ℓ)T
(Hint: We can express this derivative as a simple multiple of ϵn.)
(e) Using your result from part (d), show that the per molecule contribution, ρ, to the “one-dimensional pressure” of N molecules in a one-dimensional box is ρ=2⟨ϵ⟩/ℓ
(f) Use your results from parts (b) and (e) to express ρ as a function of k, T, and ℓ.
(g) Let Π be the pressure of a system of N molecules in a one-dimensional box. From your result in part (c) or part (f), give an equation for Π. Show how this equation is analogous to the ideal gas equation.