Skip to main content
Chemistry LibreTexts

3.10: Problems

  • Page ID
    206325
  • 1. Consider a system with three non-degenerate quantum states having energies \({\epsilon }_1=0.9\ kT\), \({\epsilon }_2=1.0\ kT\), and \({\epsilon }_3=1.1\ kT\). The system contains \(N=3\times {10}^{10}\) 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 \(\{N^{\textrm{⦁}}_1,N^{\textrm{⦁}}_2,N^{\textrm{⦁}}_3\}\). Let \(W_{mp}\) be the number of microstates associated with the equilibrium population set. Consider the population set when \({10}^{-5}\) of the molecules in \({\epsilon }_2\) are moved to each of \({\epsilon }_1\) and \({\epsilon }_3\). This is the population set \(\{N^{\textrm{⦁}}_1+{10}^{-5}N^{\textrm{⦁}}_2,\ \ \ N^{\textrm{⦁}}_2-2\times {10}^{-5},\ \ \ N^{\textrm{⦁}}_3+{10}^{-5}N^{\textrm{⦁}}_2\}\). 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 \({W_{mp}}/{W}\). 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 \(\epsilon <kt\)> for a molecule of molecular weight \(40\) in a box of volume \({10}^{-6}\ {\mathrm{m}}^3\) 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

    \[{\epsilon }_n=\frac{n^2h^2}{8m{\ell }^2}\]

    where \(m\) is the mass of the particle and \(\ell\) is the length of the box. For molecular masses and boxes of macroscopic lengths, the factor \({h^2}/{8m{\ell }^2}\) 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 \(\infty\).

    (a) Obtain a formula for the partition function of a particle in a one-dimensional box. Integral tables give \[\int^{\infty }_0 \mathrm{exp} \left(-an^2\right) dn=\sqrt{\pi /4a}\]

    (b) The expected value of the energy of a molecule is given by \[\left\langle \epsilon \right\rangle =kT^2{\left(\frac{\partial { \ln z\ }}{\partial T}\right)}_V\]

    What is \(\left\langle \epsilon \right\rangle\) for a particle in a box?

    (c) The relationship between the partition function and the per-molecule Helmholtz free energy is \(A=-kT{ \ln z\ }\). For a molecule in a one-dimensional box, we have \(dA=-SdT-\rho \ell\), where \(\rho\) is the per-molecule “pressure” on the ends of the box and \(\ell\) is the length of the box. (The increment of work associated with changing the length of the box is \(dw=-\rho \ d\ell\). In this relationship, \(d\ell\) is the incremental change in the length of the box and \(\rho\) is the one-dimensional “pressure” contribution from each molecule. \(\rho\) is, of course, just the force required to push the end of the box outward by a distance \(d\ell\). \(\rho d\ell\) is the one-dimensional analog of \(PdV\).) For the one-dimensional system, it follows that \[\rho =-{\left(\frac{\partial A}{\partial \ell }\right)}_T\]

    Use this information to find \(\rho\) for a molecule in a one-dimensional box.

    (d) We can find \(\rho\) 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, \(P_i\), by

    \[P^{\mathrm{system}}_{\mathrm{molecule}}=-\sum^{\infty }_{n=1}{P_n}{\left(\frac{\partial {\epsilon }_n}{\partial V}\right)}_T\]

    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

    \[\rho =-\sum^{\infty }_{n=1}{P_n}{\left(\frac{\partial {\epsilon }_n}{\partial \ell }\right)}_T\]

    From the equation for the energy levels of a particle in a one dimensional box, find an equation for

    \[{\left(\frac{\partial {\epsilon }_n}{\partial \ell }\right)}_T\]

    (Hint: We can express this derivative as a simple multiple of \({\epsilon }_n\).)

    (e) Using your result from part (d), show that the per molecule contribution, \(\rho\), to the “one-dimensional pressure” of \(N\) molecules in a one-dimensional box is \[\rho ={2\left\langle \epsilon \right\rangle }/{\ell }\]

    (f) Use your results from parts (b) and (e) to express \(\rho\) as a function of \(k\), \(T\), and \(\ell\).

    (g) Let \(\mathrm{\Pi }\) 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 \(\mathrm{\Pi }\). Show how this equation is analogous to the ideal gas equation.