(a) The energies and degeneracies of the two lowest electronic levels of atomic iodine are listed below: Calculate the temperature at which 2% of the iodine atoms will be in the excited electronic state. NOTE: 1 cm^-1 = 1.986 × 10^-23 J, kB = 0.69509 cm^-1 K^-1. (b) Two monoatomic ideal gases

Question

(a) The energies and degeneracies of the two lowest electronic levels of atomic iodine are listed below:

[latex]\begin{array}{l|l|l}\hline ext { Level } & ext { Energy }\left( cm ^{-1} ight) & ext { Degeneracy } \\hline 1 & 0.0 & 4 \\hline 2 & 7603.2 & 2 \\hline\end{array}[/latex]

Calculate the temperature at which 2% of the iodine atoms will be in the excited electronic state.

NOTE: [latex]1 cm ^{-1}=1.986 imes 10^{-23} J ext {, and } k_{B}=0.69509 cm ^{-1} K ^{-1} ext {. }[/latex]

(b) Two monoatomic ideal gases consisting of [latex]N_{1} ext { atoms and } N_{2}[/latex] atoms, respectively, are mixed in a vessel of volume, [latex]\underline{V} [/latex], at temperature, T. It is known that the masses of the atoms in each gas are [latex]m_{1} ext { and } m_{2}[/latex], respectively.

(i) Calculate the Canonical partition function of the binary gas mixture.
(ii) Calculate the energy and pressure of the binary gas mixture.

Accepted Answer

(a) The probability that the iodine atoms will be in excited electronic (e) state i is given by:

[latex]p_{e i}=\frac{g_{e i} \exp \left(-\frac{\epsilon_{i}}{k_{B} T} ight)}{q_{e}}[/latex] (1)

where [latex]q_{e}[/latex], the electronic partition function, is given by:

[latex]q_{e}=\sum_{i} g_{e i} \exp \left(-\frac{\epsilon_{i}}{k_{B} T} ight)[/latex] (2)

and [latex]g_{e i}[/latex] is the degeneracy of the electronic energy level i.
Assuming that only the ground (i = 1) and the first (i = 2) excited electronic states contribute significantly to [latex]q_{e}[/latex], that is, assuming that the energy of the second (i = 3) and the higher (i > 3) excited electronic states are extremely high, [latex]q_{e}[/latex]in Eq. (2) is given by:

[latex]q_{e}=g_{e 1} \exp \left(-\frac{\epsilon_{1}}{k_{B} T} ight)+g_{e 2} \exp \left(-\frac{\epsilon_{2}}{k_{B} T} ight)[/latex] (3)

Using the values of [latex]g_{e 1}, g_{e 2}, \epsilon_{1}, ext { and } \epsilon_{2}[/latex] given in the Table in the Problem Statement, Eq. (3) yields:

[latex]q_{e}=4+2 x, ext { where } x=\exp \left(-\frac{7603.2}{0.69509 T} ight)[/latex] (4)

where in Eq. (4), T has units of Kelvin. We are asked to find the temperature at which 2% of the iodine atoms will be in the excited (i = 2) electronic state. From Eqs. (1) and (4), it follows that:

[latex]p_{e 2}=\frac{g_{e 2} \exp \left(-\frac{\epsilon_{2}}{k_{B} T} ight)}{q_{e}}=\frac{2 x}{4+2 x}=0.02[/latex] (5)

Clearly, x can be determined from Eq. (5). Once x is known, T can be calculated using Eq. (4). Rearranging Eq. (5) yields:

[latex]0.08+0.04 x=2 x ightarrow 0.08=1.96 x[/latex] (6)

[latex]x=0.040816[/latex] (7)

Using Eq. (4), T can be calculated from x as follows:

[latex]T=-\frac{10938.44}{\ln (x)}=-\frac{10938.44}{\ln (0.040816)}[/latex] (8)

[latex]T=3419.67 K[/latex] (9)

(bi) Because the atoms in a monoatomic ideal gas are independent and atoms of type 1 and 2 are distinguishable, the Canonical partition function of a mixture of monoatomic ideal gases can be written as the product of the Canonical partition functions of each gas, that is:

[latex]Q\left(N_{1}, N_{2}, \underline{V}, T ight)=Q_{1}\left(N_{1}, \underline{V}, T ight) Q_{2}\left(N_{2}, \underline{V}, T ight)[/latex] (10)

Because in each gas the atoms are independent and indistinguishable, it follows that:

[latex]Q_{1}\left(N_{1}, \underline{V}, T ight)=\frac{\left[q_{1}(\underline{V}, T) ight]^{N_{1}}}{N_{1} !}[/latex] (11)

and

[latex]Q_{2}\left(N_{2}, \underline{V}, T ight)=\frac{\left[q_{2}(\underline{V}, T) ight]^{N_{2}}}{N_{2} !}[/latex] (12)

where [latex]q_{1} ext { and } q_{2}[/latex] are the atomic partition functions of gases 1 and 2, respectively.
For a monoatomic ideal gas, these are given by:

[latex]q_{1}(\underline{V}, T)=\left(\frac{2 \pi m_{1} k_{B} T}{h^{2}} ight)^{3 / 2} \underline{V} g_{e 1,1}[/latex] (13)

where [latex]g_{e 1,1}[/latex] is the degeneracy of the ground electronic state for gas 1. Similarly:

[latex]q_{2}(\underline{V}, T)=\left(\frac{2 \pi m_{2} k_{B} T}{h^{2}} ight)^{3 / 2} \underline{V }g_{e 1,2}[/latex] (14)

where [latex]g_{e 1,2}[/latex] is the degeneracy of the ground electronic state for gas 2.
Substituting Eqs. (11) through (14) in Eq. (10) yields the desired result:

[latex]Q\left(N_{1}, N_{2}, \underline{V}, T ight)=\frac{1}{N_{1} ! N_{2} !}\left[\left(\frac{2 \pi m_{1} k_{B} T}{h^{2}} ight)^{3 / 2} \underline{V} g_{e 1,1} ight]^{N_{1}}\left[\left(\frac{2 \pi m_{2} k_{B} T}{h^{2}} ight)^{3 / 2} \underline{V} g_{e 1,2} ight]^{N_{2}}[/latex] (15)

(bii) To calculate the energy of the gas mixture, we use the expression presented in Part III, that is:

[latex]<\underline{E}>=\underline{U}=k_{B} T^{2}\left(\frac{\partial \ln Q}{\partial T} ight)_{N_{1}, N_{2}, \underline{V}}[/latex] (16)

It is convenient to first write down the expression for lnQ where we isolate the terms that depend explicitly on T. From Eq. (15), it follows that:

[latex]\ln Q=\frac{3 N_{1}}{2} \ln T+\frac{3 N_{2}}{2} \ln T+ ext { Terms that do not depend on } T[/latex] (17)

Differentiating Eq. (17) with respect to T, at constant [latex]N_{1}, N_{2}, ext { and } \underline{V}[/latex], yields:

[latex]\left(\frac{\partial \ln Q}{\partial T} ight)_{N_{1}, N_{2}, \underline{V}}=\frac{3 N_{1}}{2 T}+\frac{3 N_{2}}{2 T}[/latex] (18)

Using Eq. (18) in Eq. (16) yields the desired result:

[latex]<\underline{E}>=\underline{U}=\frac{3}{2}\left(N_{1}+N_{2} ight) k_{B} T[/latex] (19)

To calculate the pressure of the gas mixture, we use the expression presented in Part III, that is:

[latex]P=k_{B} T\left(\frac{\partial \ln Q}{\partial \underline{V}} ight)_{N_{1}, N_{2}, T}[/latex] (20)

Again, it is convenient to isolate the terms that depend explicitly on V in the expression of lnQ before differentiation with respect to V. Specifically:

[latex]\ln Q= N _{1} \ln \underline{V}+ N _{2} \ln \underline{V}+ ext { Terms that do not depend on } \underline{V}[/latex] (21)

Differentiating Eq. (21) with respect to V, at constant [latex]N_{1}, N_{2} ext {, and } T, ext { yields: }[/latex]

[latex]\left(\frac{\partial \ln Q}{\partial \underline{V}} ight)_{N_{1}, N_{2}, T}=\frac{N_{1}}{\underline{V}}+\frac{N_{2}}{\underline{V}}[/latex] (22)

Using Eq. (22) in Eq. (20) yields the desired result:

[latex]P=\frac{k_{B} T\left(N_{1}+N_{2} ight)}{\underline{V}}[/latex] (23)

Equations (19) and (23) indicate that a gas mixture of the two monoatomic ideal gases behaves like an ideal gas consisting of [latex]\left(N_{1}+N_{2} ight)[/latex] atoms.

Question 25.1: (a) The energies and degeneracies of the two lowest electron...

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