(a) Use the Grand-Canonical ensemble to calculate the standard-state chemical potential, μ0(T), of Ar (g) at 298 K. It is known that (i) the standard-state pressure, P0, equals 1 bar, (ii) the mass of argon is 0.03995 kg/mol, and (iii) the first electronic state of argon is nondegenerate.How does

Question

(a) Use the Grand-Canonical ensemble to calculate the standard-state chemical potential, [latex]\mu_{o}(T)[/latex], of Ar (g) at 298 K. It is known that (i) the standard-state pressure, [latex]P_{0}[/latex], equals 1 bar, (ii) the mass of argon is 0.03995 kg/mol, and (iii) the first electronic state of argon is nondegenerate. How does your result compare with the experimentally measured value of 39.97 kJ/mol.

(b) It has been suggested that the triangular potential:

[latex]\Phi(r)=\left\{\begin{array}{rr}\infty, & ext { for } r \leq \sigma_{0} \-\epsilon\left(\frac{r-\sigma_{1}}{\sigma_{0}-\sigma_{1}} ight), & ext { for } \sigma_{0}\sigma_{1}\end{array} ight.[/latex]

may provide adequate second viral coefficients, [latex]B_{2}(T)[/latex]. Use the model for Φ(r) above to derive an expression for [latex]B_{2}(T)[/latex] in the high-temperature limit (to linear order in [latex]\epsilon \beta[/latex]) in terms of the model parameters [latex]\sigma_{0}, \sigma_{1}, ext { and } \epsilon[/latex].

Accepted Answer

(a) We are asked to calculate the chemical potential, μ, using the Grand-Canonical ensemble. The Grand-Canonical ensemble partition function is given by:

[latex]\Xi=\sum_{N=0}^{\infty} Q(N, \underline{V}, T) \exp \left(N \mu / k_{B} T ight)[/latex] (43)

where Q (N, V, T) is the Canonical partition function. We anticipate that at [latex]P_{0}=1[/latex] bar and T = 298 K, argon behaves like an ideal gas. In that case, the Canonical partition function is given by:

[latex]Q(N, \underline{V}, T)=\frac{[q(\underline{V}, T)]^{N}}{N !}[/latex] (44)

where the atomic partition function, q, is given by:

[latex]q (\underline{V}, T)=\left(\frac{2 \pi m k_{B} T}{h^{2}} ight)^{3 / 2} \underline{V }g_{e 1}[/latex] (45)

We are asked to calculate μ when [latex]P_{0}=1 ext { bar and } T=298 K[/latex]. Therefore, if we could express [latex]\Xi ext { in terms of } P_{0}[/latex], we could use Eq. (43) to calculate μ as a function of [latex]P_{0}[/latex]. Recall that [latex]\Xi[/latex] and P are related by:

[latex]\Xi=\exp \left(P \underline{V} / k_{B} T ight)[/latex] (46)

Using Eq. (46) in Eq. (43) and substituting Eqs. (44) and (45) yields:

[latex]\exp \left(P \underline{V} / k_{B} T ight)=\sum_{N=0}^{\infty} \frac{[q(\underline{V}, T)]^{N}}{N !} \lambda^{N}, ext { where } \quad \lambda=\exp \left(\mu / k_{B} T ight)[/latex] (47)

Recall that:

[latex]e^{x}=\sum_{N=0}^{\infty} \frac{x^{N}}{N !}[/latex] (48)

In view of Eq. (48), Eq. (47) can be written as follows:

[latex]\exp \left(P \underline{V} / k_{B} T ight)=\exp (q(\underline{V}, T) \lambda)[/latex] (49)

Substituting Eq. (45) for q(V, T) in Eq. (49), and subsequently taking the natural logarithm of both sides of the resulting equation, yields:

[latex]\frac{P \underline{V}}{k_{B} T}=\left(\frac{2 \pi m k_{B} T}{h^{2}} ight)^{3 / 2} \underline{V }g_{e 1} \lambda[/latex] (50)

Rearranging Eq. (50) yields:

[latex]\lambda=\frac{P}{g_{e 1}}\left(k_{B} T ight)^{-5 / 2}\left(\frac{h^{2}}{2 \pi m} ight)^{3 / 2}[/latex] (51)

Recalling that [latex]\lambda=\exp \left(\mu / k_{B} T ight)[/latex], we obtain the desired expression:

[latex]\mu=k_{B} T \ln \left[\frac{P}{g_{e 1}}\left(k_{B} T ight)^{-5 / 2}\left(\frac{h^{2}}{2 \pi m} ight)^{3 / 2} ight][/latex] (52)

Note that the expression for μ in Eq. (52) is the same as the expression obtained in Part III using the Canonical ensemble approach.
Equation (52) enables calculation of the chemical potential at a given P and T.
Note that the standard-state chemical potential is simply the chemical potential at the reference pressure [latex]P=P_{0}[/latex], that is:

[latex]\mu_{0}\left(T, P_{0} ight)=k_{B} T \ln \left[\frac{P_{0}}{g_{e 1}}\left(k_{B} T ight)^{-5 / 2}\left(\frac{h^{2}}{2 \pi m} ight)^{3 / 2} ight][/latex] (53)

Next, let us calculate the chemical potential of argon at [latex]P=P_{0}=1[/latex] bar and T = 298 K. The values of the various terms that appear in Eq. (53) are as follows:

[latex]\begin{gathered}g_{e 1}=1 ext { (nondegenerate first electronic ground state) } \h=6.626 imes 10^{-34} Js \m=0.03995 kg / mol =6.633 imes 10^{-26} kg / molecule \k_{B}=1.381 imes 10^{-23} J / K \T=298 K\end{gathered}[/latex]

[latex]P_{0}=1 ext { bar }=1.0 imes 10^{5} N / m ^{2}[/latex] (54)

Substituting the values in Eq. (54) in Eq. (53) yields:

[latex]\mu_{0}(298 K , 1 bar )=-6.636 imes 10^{-23} kJ / ext { molecule }[/latex] (55)

We next need to convert the value in Eq. (55) from a per molecule basis to a per mole basis and then compare the resulting predicted value with the experimental value given in the Problem Statement. To this end, we simply multiply the result in Eq. (55) by Avogadro’s number. This yields:

[latex]\mu_{0}(298 K , 1 bar )=-39.97 kJ / mol[/latex] (56)

It turns out that [latex]\mu_{0}^{ ext {experiment }}(1 bar , 298 K )=-39.97 kJ / mol[/latex]. It then follows that the statistical mechanical prediction is in remarkable agreement with the experimental result. It is important to note that the calculated [latex]\mu(1 bar , 298 K )[/latex] is based on the assumption that the ground electronic state of the argon atom is zero. If this is not the case, additional contributions may appear in the expression for [latex]\mu_{0}(1 ext { bar, } 298 K )[/latex].

(b) This problem involves calculating the second virial coefficient from a given interaction potential.

We begin with the expression for [latex]B_{2}(T)[/latex] in terms of u(r) presented in Part III:

[latex]B_{2}(T)=-2 \pi \int_{0}^{\infty}\left[e^{-\beta u(r)}-1 ight] r^{2} d r[/latex] (57)

The given interaction potential, u(r), is plotted in Fig. 1 below:
Because u(r) shows distinct behaviors in three regions (see Fig. 1), we break the integration range in Eq. (57) into three regions as follows:

[latex]B_{2}(T)=-2 \pi\left[\int_{0}^{\sigma_{0}}\left[e^{-\beta u(r)}-1 ight] r^{2} d r+\int_{\sigma_{0}}^{\sigma_{1}}\left[e^{-\beta u(r)}-1 ight] r^{2} d r+\int_{\sigma_{1}}^{\infty}\left[e^{-\beta u(r)}-1 ight] r^{2} d r ight][/latex]

[latex]B_2(T)=-2\pi [I_1+I_2+I_3][/latex] (58)

We will next calculate [latex]I_{1}, I_{2}, ext { and } I_{3}[/latex] as follows:

[latex]I_{1}=\int_{0}^{\sigma_{0}}\left[e^{-\infty}-1 ight] r^{2} d r=\int_{0}^{\sigma_{0}}[-1] r^{2} d r=-\frac{\sigma_{0}^{3}}{3}[/latex] (59)

[latex]I_{2}=\int_{\sigma_{0}}^{\sigma_{1}}\left[e^{\beta \varepsilon \frac{r-\sigma_{1}}{\sigma_{0}-\sigma_{1}}}-1 ight] r^{2} d r[/latex] (60)

[latex]I_{3}=\int_{\sigma_{1}}^{\infty}\left[e^{-\beta(0)}-1 ight] r^{2} d r=\int_{0}^{\sigma_{0}}[0] r^{2} d r=0[/latex] (61)

Using Eqs. (59), (60), and (61) in Eq. (58) yields:

[latex]B_{2}(T)=\frac{2 \pi \sigma_{0}^{3}}{3}-2 \pi \int_{\sigma_{0}}^{\sigma_{1}}\left[e^{\beta \varepsilon \frac{r-\sigma_{1}}{\sigma_{0}-\sigma_{1}}}-1 ight] r^{2} d r[/latex] (62)

At high temperatures, [latex]\beta \varepsilon \ll 1[/latex], and we can expand the exponential term in Eq. (62) as follows:

[latex]e^{\beta \frac{r-\sigma_{1}}{\sigma_{0}-\sigma_{1}}}-1=1+\beta \varepsilon \frac{r-\sigma_{1}}{\sigma_{0}-\sigma_{1}}-1+O\left(\left(r-\sigma_{0} ight)^{2} ight) \approx \frac{\beta \varepsilon r}{\sigma_{0}-\sigma_{1}}-\frac{\beta \varepsilon \sigma_{1}}{\sigma_{0}-\sigma_{1}}[/latex] (63)

Using Eq. (63) in Eq. (60) yields:

[latex]I_{2}=\int_{\sigma_{0}}^{\sigma_{1}} \frac{\beta \varepsilon r}{\sigma_{0}-\sigma_{1}} r^{2} d r-\int_{\sigma_{0}}^{\sigma_{1}} \frac{\beta \varepsilon \sigma_{1}}{\sigma_{0}-\sigma_{1}} r^{2} d r[/latex] (64)

[latex]I_{2}=\frac{\beta \varepsilon}{\sigma_{0}-\sigma_{1}}\left(\frac{\sigma_{1}^{4}-\sigma_{0}^{4}}{4} ight)-\frac{\beta \varepsilon \sigma_{1}}{\sigma_{0}-\sigma_{1}}\left(\frac{\sigma_{1}^{3}-\sigma_{0}^{3}}{3} ight)[/latex] (65)

[latex]\begin{aligned}I_{2}=& \frac{\beta \varepsilon}{\sigma_{0}-\sigma_{1}}\left(\frac{\left(\sigma_{1}^{2}-\sigma_{0}^{2} ight)\left(\sigma_{1}^{2}+\sigma_{0}^{2} ight)}{4} ight)-\frac{\beta \varepsilon \sigma_{1}}{\sigma_{0}-\sigma_{1}} \& imes\left(\frac{\left(\sigma_{1}^{2}+\sigma_{0} \sigma_{1}+\sigma_{0}^{2} ight)\left(\sigma_{1}-\sigma_{0} ight)}{3} ight)\end{aligned}[/latex] (66)

or

[latex]I_{2}=-\beta \varepsilon\left(\frac{\left(\sigma_{1}+\sigma_{0} ight)\left(\sigma_{1}^{2}+\sigma_{0}^{2} ight)}{4} ight)+\beta \varepsilon \sigma_{1}\left(\frac{\left(\sigma_{1}^{2}+\sigma_{0} \sigma_{1}+\sigma_{0}^{2} ight)}{3} ight)[/latex] (67)

Combining the two terms in Eq. (67) yields:

[latex]I_{2}=\beta \varepsilon\left(\frac{\left(\sigma_{1}^{3}+\sigma_{0}^{2} \sigma_{1}+\sigma_{1}^{2} \sigma_{0}-3 \sigma_{0}^{3} ight)}{12} ight)[/latex] (68)

Equation (68) can be further simplified as follows:

[latex]I_{2}=\beta \varepsilon\left(\frac{\left(\sigma_{1}^{2}+2 \sigma_{0} \sigma_{1}+3 \sigma_{0}^{2} ight)\left(\sigma_{1}-\sigma_{0} ight)}{12} ight)[/latex] (69)

Using Eqs. (59), (61), and (69) in Eq. (58) yields the desired result:

[latex]B_{2}(T)=\frac{2 \pi \sigma_{0}^{3}}{3}-\pi \beta \varepsilon\left(\frac{\left(\sigma_{1}^{2}+2 \sigma_{0} \sigma_{1}+3 \sigma_{0}^{2} ight)\left(\sigma_{1}-\sigma_{0} ight)}{6} ight)[/latex] (70)

The first term in Eq. (70) is due to the hard-sphere repulsive part of the interaction potential, u(r), and, therefore, has a positive contribution to [latex]B _{2}[/latex]. As Fig. 1 shows, ɛ > 0 corresponds to an attractive interaction and, consequently, has a negative contribution to [latex]B _{2}[/latex] On the other hand, ɛ < 0 corresponds to a repulsive interaction and, consequently, has a negative contribution to [latex]B _{2}[/latex].

Question 25.3: (a) Use the Grand-Canonical ensemble to calculate the standa...

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