Consider a system of N identical, but distinguishable, particles, each having two energy levels with energies 0 and ε 0, respectively. The upper level is g-fold degenerate and the lower level is nondegenerate. The total energy of the system is E.(a) Use the Micro-Canonical ensemble to calculate the entropy of the system.

Question

Consider a system of N identical, but distinguishable, particles, each having two energy levels with energies 0 and ε > 0, respectively. The upper level is g-fold degenerate and the lower level is nondegenerate. The total energy of the system is E.

(a) Use the Micro-Canonical ensemble to calculate the entropy of the system. Express your result in terms of g, N, and the occupation numbers of the upper and lower levels, [latex]n_{+} ext {and } n_{0}[/latex], respectively, where [latex]n_{+}+n_{0}=N ext { and } \underline{E}=n_{+} \varepsilon[/latex]. Note that your result corresponds to the entropy fundamental equation, S = S (E, N), where in the present case, there is no explicit dependence on the system volume, V.

(b) Use the result in Part (a) to derive an expression for the temperature of the system.

(c) Use the result in Part (b) to derive expressions for the occupation numbers [latex]n_{0}[/latex] and [latex]n_{+}[/latex]. Show that the same expressions can be derived much more readily in the context of the Canonical ensemble.

Accepted Answer

In this problem, we are asked to find the entropy, S, of a system of N identical, but distinguishable, particles using the Micro-Canonical ensemble, where the independent variables are E, V, and N. Knowing S as a function of E, V, and N, T can be evaluated using concepts from classical thermodynamics discussed in Part I. Lastly, we are asked to compare the results obtained using the Micro-Canonical ensemble to the results obtained using the Canonical ensemble, where the independent variables are T, V, and N.

Part (a): Calculating W and S

We begin with Boltzmann’s celebrated expression for the entropy in the context of the Micro-Canonical ensemble. Specifically:

[latex]\underline{ S }=k_{B} \ln ( W )[/latex] (1)

In Eq. (1), [latex]k_{B}[/latex] is the Boltzmann constant, and W is the number of distinct states of the system for a given E, V, and N, that is, the degeneracy corresponding to E, V, and N. Note that W is not the degeneracy, g, of an energy level. Recall that according to the Problem Statement, we have:

[latex]\underline{ E }=n_{+} \varepsilon[/latex] (2)

Therefore, we can also regard W as the number of distinct ways in which we can arrange N particles such that the total energy is always E. An examination of Eq. (2) shows that only the number of particles in the upper energy level, [latex]n_{+}[/latex], contributes to the total energy. Calculation of W, therefore, involves counting the distinct number of ways in which [latex]n_{+}[/latex] distinguishable particles out of the available N particles can occupy the upper energy level, ε, including recognizing that the [latex]n_{+}[/latex] particles can be assigned to any one of the g available degenerate states. Accordingly, W can be expressed as follows:

[latex]W=\left(\frac{N !}{n_{+} !\left(N-n_{+} ight) !} ight)\left(g^{n_{+}} ight)[/latex] (3)

In Eq. (3), the first quantity in parentheses represents the number of distinct ways in which [latex]n_{+}[/latex] distinguishable particles can occupy the upper energy level, and the second quantity in parentheses represents the number of distinct ways of assigning the [latex]n_{+}[/latex] particles to g degenerate states.
Using Eq. (3) in Eq. (1), we obtain:

[latex]\underline{ S }=k_{B} \ln \left[\left(\frac{N !}{n_{+} !\left(N-n_{+} ight) !} ight)\left(g^{n_{+}} ight) ight][/latex] (4)

Rearranging Eq. (4) yields:

[latex]\underline{ S }=k_{B}\left\{\ln N !-\ln n_{+} !-\ln \left(N-n_{+} ight) !+n_{+} \ln g ight\}[/latex] (5)

Because N >>1, we can use Stirling’s approximation, that is:

[latex]\ln y !=y \ln y-y ext {, for } y>>1[/latex] (6)

Using Eq. (6) in Eq. (5), we obtain:

[latex]\underline{ S }=k_{B}\left\{N \ln N-N-n_{+} \ln n_{+}+n_{+}-\left(N-n_{+} ight) \ln \left(N-n_{+} ight)+N-n_{+}+n_{+} \ln g ight\}[/latex]

Rearranging and simplifying the last expression yields:

[latex]\underline{ S }=-k_{B}\left\{n_{+} \ln \frac{n_{+}}{N}+\left(N-n_{+} ight) \ln \frac{\left(N-n_{+} ight)}{N}-n_{+} \ln g ight\}[/latex] (7)

For convenience, let us define the fraction of molecules in the upper level as follows:

[latex]x=\frac{n_{+}}{N}=\frac{\underline{ E } / \varepsilon}{N}=\frac{\underline{ E }}{\varepsilon N}[/latex] (8)

Recall that [latex]\underline{ E }=n_{+} \varepsilon[/latex] (see the Problem Statement). Substituting Eq. (8) in Eq. (7) yields:

[latex]\underline{ S }=-k_{B} N\{x \ln x+(1-x) \ln (1-x)-x \ln g\}[/latex] (9)

Equation (9) can also be expressed as follows.

[latex]\underline{ S }=k_{B} N\left\{\ln \frac{1}{1-x}+x \ln \frac{1-x}{x}+x \ln g ight\}[/latex] (10)

Part (b): Obtaining T

Note that because [latex]x=\underline{ E } / N \varepsilon[/latex], Eq. (9), or Eq. (10), expresses S as a function of E and N (the volume V does not appear explicitly in Eq. (9), although it could affect the value of ε which is assumed to be a constant). Therefore, Eq. (9), or Eq. (10), is essentially the entropy fundamental equation corresponding to this system. As discussed in Part I, we can calculate the temperature recalling that S and E are related as follows:

[latex]\left(\frac{\partial \underline{S}}{\partial \underline{E}} ight)_{N}=\frac{1}{T}[/latex] (11)

Because [latex]\underline{E}=N \varepsilon x[/latex], Eq. (11) can be expressed as follows:

[latex]\left(\frac{\partial \underline{S}}{\partial \underline{E}} ight)_{N}=\frac{1}{N \varepsilon}\left(\frac{\partial \underline{S}}{\partial x} ight)_{N}[/latex] (12)

Differentiating Eq. (9) (or Eq. (10)) with respect to x, at constant N, yields:

[latex]\left(\frac{\partial \underline{S}}{\partial x} ight)_{N}=-k N\left\{\ln x+\frac{x}{x}-\ln (1-x)-\frac{1-x}{1-x}-\ln g ight\}[/latex]

[latex]\left(\frac{\partial \underline{S}}{\partial x} ight)_{N}=k N\{\ln (1-x)+\ln g-\ln x\}=k N \ln \frac{g(1-x)}{x}[/latex] (13)

Using Eq. (3) in Eq. (12) yields:

[latex]\left(\frac{\partial \underline{S}}{\partial \underline{E}} ight)_{N}=\frac{1}{N \varepsilon} k N \ln \frac{g(1-x)}{x}=\frac{k}{\varepsilon} \ln \frac{g(1-x)}{x}[/latex] (14)

Using Eqs. (11) and (14), we can now calculate the temperature which is given by:

[latex]T=\frac{\varepsilon}{k}\left[\ln \frac{g(1-x)}{x} ight]^{-1}, x=\frac{n_{+}}{N}=\frac{\underline{E}}{N \varepsilon}[/latex] (15)

Part (c): Determining [latex]n_{+} ext {and } n_{o}[/latex]

For large values of N, [latex]\frac{n_{+}}{N}[/latex] represents the probablility of finding a particle in the upper energy level [latex]\left(p_{+} ight)[/latex]. In that case, rearranging Eq. (15) for x yields:

[latex]x=\frac{n_{+}}{N}=p_{+}=\frac{\operatorname{gexp}\left(\frac{-\varepsilon}{k T} ight)}{1+\operatorname{gexp}\left(\frac{-\varepsilon}{k T} ight)}=\frac{\operatorname{gexp} (-\varepsilon \beta)}{1+\operatorname{gexp}(-\varepsilon \beta)}[/latex] (16)

Equation (16) can be readily derived using the molecular partition function in the context of the Canonical ensemble, where q is given by:

[latex]q=\exp \left(-\beta \varepsilon_{o} ight)+g \exp (-\beta \varepsilon)=1+g \exp (-\beta \varepsilon)[/latex]

recalling that [latex]\varepsilon_{o}=0[/latex]. The fraction of particles in the lower-energy level, or [latex]p _{o}[/latex], is given by:

[latex]p_{o}=\frac{1}{1+\operatorname{gexp}(-\beta \varepsilon)}[/latex] (17)

Similarly, [latex]p_{+}[/latex] is given by:

[latex]p_{+}=\frac{\operatorname{gexp}(-\varepsilon \beta)}{1+\operatorname{gexp}(-\varepsilon \beta)}[/latex] (18)

Upon comparison, Eq. (18) (obtained using the Canonical ensemble) is identical to Eq. (16) (obtained using the Micro-Canonical ensemble).
We can next calculate [latex]n_{o} ext { and } n_{+}[/latex] using Eqs. (17) and (18), respectively.
Specifically:

[latex]n_{o}=\frac{N}{1+\operatorname{gexp}(-\beta \varepsilon)}[/latex] (19)

[latex]n_{+}=\frac{\operatorname{Ngexp}(-\varepsilon \beta)}{1+\operatorname{gexp}(-\varepsilon \beta)}[/latex] (20)

Question 23.1: Consider a system of N identical, but distinguishable, parti...

Related Answered Questions

(a) Calculate the vibrational partition function of a diatomic molecule treating the vibrational mode classically. (b) Calculate the average vibrational energy of a diatomic molecule treating the vibrational mode classically.(c) Compare your results in Parts (a) and (b) with the corresponding ...

Verified Answer:

Verified Answer:

For a binary liquid mixture of components 1 and 2, if the partial molar enthalpy H1 is known as a function of T, P, and x1, calculate H2 and H. The data H1 = f (x1) is available at constant T and P. ...

Verified Answer: