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

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

p_{e i}=\frac{g_{e i} \exp \left(-\frac{\epsilon_{i}}{k_{B} T}\right)}{q_{e}}                          (1)

where q_{e}, the electronic partition function, is given by:

q_{e}=\sum_{i} g_{e i} \exp \left(-\frac{\epsilon_{i}}{k_{B} T}\right)                            (2)

and g_{e i} 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 q_{e}, that is, assuming that the energy of the second (i = 3) and the higher (i > 3) excited electronic states are extremely high, q_{e}in Eq. (2) is given by:

q_{e}=g_{e 1} \exp \left(-\frac{\epsilon_{1}}{k_{B} T}\right)+g_{e 2} \exp \left(-\frac{\epsilon_{2}}{k_{B} T}\right)                           (3)

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

q_{e}=4+2 x, \text { where } x=\exp \left(-\frac{7603.2}{0.69509 T}\right)                            (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:

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

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

0.08+0.04 x=2 x \rightarrow 0.08=1.96 x                          (6)

x=0.040816                        (7)

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

T=-\frac{10938.44}{\ln (x)}=-\frac{10938.44}{\ln (0.040816)}                          (8)

T=3419.67  K                           (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:

Q\left(N_{1}, N_{2}, \underline{V}, T\right)=Q_{1}\left(N_{1}, \underline{V}, T\right) Q_{2}\left(N_{2}, \underline{V}, T\right)                            (10)

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

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

and

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

where q_{1} \text { and } q_{2} are the atomic partition functions of gases 1 and 2, respectively.
For a monoatomic ideal gas, these are given by:

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

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

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

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

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

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

<\underline{E}>=\underline{U}=k_{B} T^{2}\left(\frac{\partial \ln Q}{\partial T}\right)_{N_{1}, N_{2}, \underline{V}}                       (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:

\ln Q=\frac{3 N_{1}}{2} \ln T+\frac{3 N_{2}}{2} \ln T+\text { Terms that do not depend on } T                            (17)

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

\left(\frac{\partial \ln Q}{\partial T}\right)_{N_{1}, N_{2}, \underline{V}}=\frac{3 N_{1}}{2 T}+\frac{3 N_{2}}{2 T}                           (18)

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

<\underline{E}>=\underline{U}=\frac{3}{2}\left(N_{1}+N_{2}\right) k_{B} T                            (19)

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

P=k_{B} T\left(\frac{\partial \ln Q}{\partial \underline{V}}\right)_{N_{1}, N_{2}, T}                          (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:

\ln Q= N _{1} \ln \underline{V}+ N _{2} \ln \underline{V}+\text { Terms that do not depend on } \underline{V}                            (21)

Differentiating Eq. (21) with respect to V, at constant N_{1}, N_{2} \text {, and } T, \text { yields: }

\left(\frac{\partial \ln Q}{\partial \underline{V}}\right)_{N_{1}, N_{2}, T}=\frac{N_{1}}{\underline{V}}+\frac{N_{2}}{\underline{V}}                          (22)

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

P=\frac{k_{B} T\left(N_{1}+N_{2}\right)}{\underline{V}}                         (23)

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