Adapted from Molecular Driving Forces - Statistical Thermodynamics in Chemistry and Biology by Ken A. Dill and Sarina Bromberg, Garland Science, Taylor & Francis Group, New York and London (2003). Hereafter, we will abbreviate the names of the authors as D&B.(a) Given the intermolecular potential,

Question

Adapted from Molecular Driving Forces - Statistical Thermodynamics in Chemistry and Biology by Ken A. Dill and Sarina Bromberg, Garland Science, Taylor & Francis Group, New York and London (2003). Hereafter, we will abbreviate the names of the authors as D&B.

(a) Given the intermolecular potential, [latex]\varphi_{2}(r)[/latex], as a function of intermolecular separation, r, shown below, derive an expression for the second virial coefficient, [latex]B_{2}(T)[/latex]. Express your result solely in terms of [latex]\varepsilon, \sigma, R \sigma, ext { and } k_{B} T[/latex].

(b) Consider an ideal gas of molecules which possess permanent dipole moments, [latex]\vec{\mu}[/latex], in an external electric field, [latex]\vec{\varepsilon}[/latex]. It is known that the potential energy of a single dipolar molecule in an external electric field is [latex]u=-\mu \varepsilon \cos heta, ext { where } heta[/latex] is the angle between the vectors [latex]\vec{\mu} ext { and } \vec{\varepsilon}[/latex].

The Hamiltonian for a single molecule possessing a permanent dipole moment and interacting with an external electric field can be expressed as the sum of translational, rotational, and potential energy contributions. The contribution to the Hamiltonian related to rotational energy can be modeled using a rigid rotor.

1. Write the Hamiltonian expression for a single molecule possessing a permanent dipole moment in an external electric field.
2. Derive the classical partition function for the single gas molecule in Part (a).
3. Calculate the additional contribution to the ideal gas internal energy resulting from the dipole-electric field interactions.

Accepted Answer

(a) By inspection of the figure in the Problem Statement, it follows that:

[latex]\varphi(r)=\left\{\begin{array}{lll}\infty, & ext { for } r<\sigma & (I) \-\varepsilon, & ext { for } R_{\sigma}>r \geq \sigma & (I I) \\varepsilon, & ext { for } 2 R_{\sigma}>r \geq R_{\sigma} & (I I I) \0, & ext { for } r \geq 2 R_{\sigma} & (I V)\end{array} ight.[/latex] (27)

As discussed in Part III, the second virial coefficient is related to the intermolecular potential as follows:

[latex]B_{2}(T)=-2 \pi \int_{0}^{\infty}\left[e^{-\beta \varphi_{2}(r)}-1 ight] r^{2} d r[/latex] (28)

Using [latex]B_{2}(T)=B_{2}^{I}(T)+B_{2}^{I I}(T)+B_{2}^{I I I}(T)+B_{2}^{I V}(T)[/latex], we find:

[latex]B_{2}^{I}(T)=-2 \pi \int_{0}^{\sigma}\left[e^{-\infty}-1 ight] r^{2} d r=2 \pi \int_{0}^{\sigma} r^{2} d r=\frac{2}{3} \pi \sigma^{3}[/latex] (29)

[latex]B_{2}^{I I}(T)=-2 \pi \int_{\sigma}^{R_{\sigma}}\left[e^{-\beta(-\varepsilon)}-1 ight] r^{2} d r=\frac{2}{3} \pi\left(1-e^{\beta \varepsilon} ight)\left(\left(R_{\sigma} ight)^{3}-\sigma^{3} ight)[/latex] (30)

[latex]\begin{aligned}B_{2}^{I I I}(T) &=-2 \pi \int_{R_{\sigma}}^{2 R_{\sigma}}\left[e^{-\beta \varepsilon}-1 ight] r^{2} d r=\frac{2}{3} \pi\left(R_{\sigma} ight)^{3}\left(1-e^{-\beta \varepsilon} ight)(8-1) \&=\frac{14}{3} \pi\left(R_{\sigma} ight)^{3}\left(1-e^{-\beta \varepsilon} ight)\end{aligned}[/latex] (31)

[latex]B_{2}^{I V}(T)=-2 \pi \int_{2 R_{\sigma}}^{\infty}\left[e^{0}-1 ight] r^{2} d r=-2 \pi \int_{2 R_{\sigma}}^{\infty} 0 r^{2} d r=0[/latex] (32)

Adding up Eqs. (29)–(32) yields the desired result:

[latex]B_{2}(T)=\frac{2}{3} \pi \sigma^{3}\left\{1+\left(1-e^{\beta \varepsilon} ight)\left(R_{\sigma}^{3}-1 ight)+7 R_{\sigma}^{3}\left(1-e^{-\beta \varepsilon} ight) ight\}[/latex] (33)

(b) The solution to this problem involves (i) deriving an expression for the Hamiltonian of a dipolar molecule interacting with an external electric field, (ii) calculating the partition function of the dipolar molecule, and (iii) calculating the additional contribution to the ideal gas internal energy resulting from the dipole-electric field interactions.

In this problem, the kinetic energy is not affected by the presence of the dipoles in the molecules. In addition, while the dipolar molecules do not interact with each other by assumption (ideal gas), they interact with the external electric field, [latex]\vec{\varepsilon}[/latex], according to the interaction given in the Problem Statement [latex](u=-\mu \varepsilon \cos heta)[/latex] (see Fig. 2).

Figure 1 shows only the in-plane projection, where the out-of-plane angle, ϕ, is not shown.
The five degrees of freedom of one dipolar molecule include:

• Translational: [latex]\left(x, y, z, p_{x}, p_{y}, p_{z} ight)-3[/latex] translations. The limits for x, y, and z are [latex](-\infty[/latex] to [latex]+\infty )[/latex].

• Rotational: [latex] heta, \phi, P_{ heta}, P_{\phi}-2[/latex] rotations. The limits for ϴ and ϕ are (0 to π) and (0 to 2π), respectively.

• Because the dipole is fixed, no vibrations are possible.

The Classical Hamiltonian for the Rigid Rotor is given by:

[latex]H=\left(\frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2 m} ight)+\frac{1}{2 I}\left(p_{ heta}^{2}+\frac{p_{\phi}^{2}}{\sin ^{2} \Theta} ight)-\mu \varepsilon \cos \Theta[/latex] (34)

where the first two terms in Eq. (34) were discussed in Part III and the third term is new and corresponds to the potential energy of the dipole in an external electric field.
The classical partition function of the dipolar molecule possessing five degrees of freedom (see above) is then given by:

[latex]q=\frac{\underline{V}}{h^{5}} \int_{-\infty}^{\infty} d p_{x} \int_{-\infty}^{\infty} d p_{y} \int_{-\infty}^{\infty} d p_{z} \int_{-\infty}^{\infty} d p_{\Theta} \int_{-\infty}^{\infty} d p_{\phi} \int_{0}^{2 \pi} d \phi \int_{0}^{\pi} d \Theta e^{-H / k_{B} T}[/latex] (35)

[latex]q=\frac{\underline{V}}{\Lambda^{3}} \frac{2 \pi}{h^{2}} \int_{-\infty}^{\infty} e^{-P_{\Theta}^{2} / 2 I k_{B} T} d p_{\Theta} \int_{-\infty}^{\infty} e^{-P_{\phi}^{2} / 2 I k_{B} T \sin ^{2} \Theta} d p_{\phi} \int_{0}^{\pi} e^{\mu \varepsilon \cos \Theta / k_{B} T} d \Theta[/latex] (36)

The integrals over [latex]p_{\Theta} ext { and } p_{\phi}[/latex] in Eq. (36) are both Gaussian integrals that can be evaluated using the expression presented in Part III, which is repeated below for completeness:

[latex]\int_{-\infty}^{\infty} e^{-a^{2} x^{2}} d x=\frac{\sqrt{\pi}}{a}[/latex] (37)

Carrying out the two Gaussian integrals in Eq. (36) using Eq. (37) yields:

[latex]q =\left(\frac{\underline{ V }}{\Lambda^{3}} ight)\left(\frac{8 \pi^{2} Ik _{ B } T }{ h ^{2}} ight) \int_{0}^{\pi} \frac{1}{2} e ^{\mu \varepsilon \cos \Theta / k _{ B } T } \sin \Theta d \Theta[/latex] (38)

As shown in Part III, in Eq. (38), the first two terms in parentheses correspond to the translational and the rotational contributions to the molecular partition function of a rigid rotor, respectively. In other words, we can rewrite Eq. (38) as follows:

[latex]q=q_{ ext {tran }} q_{r o t} \int_{0}^{\pi} \frac{1}{2} e^{\mu \varepsilon \cos \Theta / k_{B} T} \sin \Theta d \Theta[/latex] (39)

An examination of Eq. (39) reveals that the additional contribution of the μ-ɛ interaction to the molecular partition function, that is, of [latex]q_{ ext {dipole }-\varepsilon}[/latex] in Eq. (39), corresponds to:

[latex]q_{ ext {dipole }-\varepsilon}=\int_{0}^{\pi} \frac{1}{2} e^{\mu \varepsilon \cos \Theta / k_{B} T} \sin \Theta d \Theta[/latex] (40)

To calculate the integral in Eq. (40), it is convenient to change variables as follows:

[latex]x =\cos \Theta, d x =-\sin \Theta d \Theta[/latex] (41)

Using Eq. (41) in Eq. (40) yields:

[latex]q_{ ext {dipole }-\varepsilon}=\int_{1}^{-1} \frac{1}{2} e^{\frac{\mu e }{k_{B} T}}(-d x)= -\frac{1}{2} \int_{1}^{-1} e^{\frac{\mu \varepsilon x }{k_{B} T}} d x[/latex] (42)

[latex]q_{ ext {dipole }-\varepsilon}=\frac{1}{2} \int_{-1}^{1} e^{\frac{\mu \varepsilon x }{k_{B} T}} d x[/latex] (43)

[latex]q_{ ext {dipole }-\varepsilon}=\frac{1}{2} \frac{k_{B} T}{\mu \varepsilon}\left(e^{\frac{\mu \varepsilon x }{k_{B} T}} ight)_{-1}^{1}= \frac{k_{B} T}{\mu \varepsilon}\left(\frac{e^{\frac{\mu \varepsilon}{k_{B} T}}-e^{-\frac{\mu \varepsilon}{k_{B} T}}}{2} ight)[/latex] (44)

[latex]q_{ ext {dipole }-\varepsilon}= \frac{k_{B} T}{\mu \varepsilon} \sinh \left(\frac{\mu \varepsilon}{k_{B} T} ight)[/latex] (45)

Accordingly, the expression for the total partition function of a dipolar molecule interacting with an external electric field is given by:

[latex]q=\left(\frac{\underline{V}}{\Lambda^{3}} ight)\left(\frac{8 \pi^{2} I k_{B} T}{h^{2}} ight) \frac{k_{B} T}{\mu \varepsilon} \sinh \left(\frac{\mu \varepsilon}{k_{B} T} ight)[/latex] (46)

(c) The additional contribution (i.e., excluding the translations and the rotations) to the total internal energy of an ideal gas of N dipolar molecules, each interacting with an external electric field, is given by:

[latex]\underline{U}_{ ext {dipole }-\varepsilon}(N, \underline{V}, T, \varepsilon) = N k_{B} T^{2}\left[\frac{\partial \ln q_{ ext {dipole }-\varepsilon}}{\partial T} ight]_{N, \underline{ V }, \varepsilon}[/latex] (47)

Using Eq. (46) in Eq. (47), including taking the temperature partial derivative, we obtain the desired result:

[latex]\underline{U}_{ ext {dipole }-\varepsilon}(N, \underline{V}, T, \varepsilon)=N k_{B} T^{2}\left[\frac{\partial\left(\ln \frac{k_{B} T}{\mu \varepsilon} ight)}{\partial T}+\frac{\partial \ln \left(\sinh \left(\frac{\mu \varepsilon}{k_{B} T} ight) ight)}{\partial T} ight]_{N, \underline{ V }, \varepsilon}[/latex] (48)

[latex]\underline{U}_{ ext {dipole- } \varepsilon}(N, \underline{V}, T, \varepsilon) = N k_{B} T^{2}\left[\frac{k_{B} / \mu \varepsilon}{k_{B} T / \mu \varepsilon}-\frac{\mu \varepsilon}{k_{B} T^{2}} \frac{\cosh \left(\frac{\mu \varepsilon}{k_{B} T} ight)}{\sinh \left(\frac{\mu \varepsilon}{k_{B} T} ight)} ight][/latex] (49)

[latex]\underline{U}_{ ext {dipole }-\varepsilon}(N, \underline{V}, T, \varepsilon)= N k_{B} T\left[1-\frac{\mu \varepsilon}{k_{B} T} \cosh \left(\frac{\mu \varepsilon}{k_{B} T} ight) ight][/latex] (50)

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