Please provide a clear and step-by-step solution for a scientific problem in the categories of Chemistry, Physics, or Mathematics. The problem will specify the unit of measurement, which should not be included in the answer. Express the final answer as a decimal number with three digits after the decimal point. Conclude the answer by stating 'Therefore, the answer is \boxed[ANSWER].

Promblem 1: For an ensemble consisting of 1.00 moles of particles having two energy levels separated by $h v=1.00 \times 10^{-20} \mathrm{~J}$, at what temperature will the internal energy of this system equal $1.00 \mathrm{~kJ}$ ?
Explanation for Problem 1: Using the expression for total energy and recognizing that $N=n N_A$,
$$
U=-\left(\frac{\partial \ln Q}{\partial \beta}\right)_V=-n N_A\left(\frac{\partial \ln q}{\partial \beta}\right)_V
$$
Evaluating the preceding expression and paying particular attention to units, we get
$$
\begin{aligned}
& U=-n N_A\left(\frac{\partial}{\partial \beta} \ln q\right)_V=-\frac{n N_A}{q}\left(\frac{\partial q}{\partial \beta}\right)_V \\
& \frac{U}{n N_A}=\frac{-1}{\left(1+e^{-\beta h \nu}\right)}\left(\frac{\partial}{\partial \beta}\left(1+e^{-\beta h \nu}\right)\right)_V \\
&=\frac{h \nu e^{-\beta h \nu}}{1+e^{-\beta h \nu}}=\frac{h \nu}{e^{\beta h \nu}+1} \\
& \frac{n N_A h \nu}{U}-1=e^{\beta h \nu} \\
& \ln \left(\frac{n N_A h \nu}{U}-1\right)=\beta h \nu=\frac{h \nu}{k T}
\end{aligned}
$$
$$
\begin{aligned}
T & =\frac{h \nu}{k \ln \left(\frac{n N_A h \nu}{U}-1\right)} \\
= & \frac{1.00 \times 10^{-20} \mathrm{~J}}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right) \ln \left(\frac{(1.00 \mathrm{~mol})\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)\left(1.00 \times 10^{-20} \mathrm{~J}\right)}{\left(1.00 \times 10^3 \mathrm{~J}\right)}-1\right)} \\
& =449 \mathrm{~K}
\end{aligned}
$$
Therefore, the answer is \boxed{449}.

Promblem 2: At $298.15 \mathrm{~K}, \Delta G_f^{\circ}(\mathrm{C}$, graphite $)=0$, and $\Delta G_f^{\circ}(\mathrm{C}$, diamond $)=2.90 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Therefore, graphite is the more stable solid phase at this temperature at $P=P^{\circ}=1$ bar. Given that the densities of graphite and diamond are 2.25 and $3.52 \mathrm{~kg} / \mathrm{L}$, respectively, at what pressure will graphite and diamond be in equilibrium at $298.15 \mathrm{~K}$ ?
Explanation for Problem 2: At equilibrium $\Delta G=G(\mathrm{C}$, graphite $)-G(\mathrm{C}$, diamond $)=0$. Using the pressure dependence of $G,\left(\partial G_m / \partial P\right)_T=V_m$, we establish the condition for equilibrium:
$$
\begin{gathered}
\Delta G=\Delta G_f^{\circ}(\mathrm{C}, \text { graphite })-\Delta G_f^{\circ}(\mathrm{C}, \text { diamond }) \\
+\left(V_m^{\text {graphite }}-V_m^{\text {diamond }}\right)(\Delta P)=0 \\
0=0-2.90 \times 10^3+\left(V_m^{\text {graphite }}-V_m^{\text {diamond }}\right)(P-1 \mathrm{bar}) \\
P=1 \mathrm{bar}+\frac{2.90 \times 10^3}{M_C\left(\frac{1}{\rho_{\text {graphite }}}-\frac{1}{\rho_{\text {diamond }}}\right)} \\
=1 \mathrm{bar}+\frac{2.90 \times 10^3}{12.00 \times 10^{-3} \mathrm{~kg} \mathrm{~mol}^{-1} \times\left(\frac{1}{2.25 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}-\frac{1}{3.52 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}}\right)}\\
=10^5 \mathrm{~Pa}+1.51 \times 10^9 \mathrm{~Pa}=1.51 \times 10^4 \mathrm{bar}
\end{gathered}
$$
Fortunately for all those with diamond rings, although the conversion of diamond to graphite at $1 \mathrm{bar}$ and $298 \mathrm{~K}$ is spontaneous, the rate of conversion is vanishingly small.

Therefore, the answer is \boxed{1.51}.

Promblem 3: The vibrational frequency of $I_2$ is $208 \mathrm{~cm}^{-1}$. What is the probability of $I_2$ populating the $n=2$ vibrational level if the molecular temperature is $298 \mathrm{~K}$ ?
Explanation for Problem 3: Molecular vibrational energy levels can be modeled as harmonic oscillators; therefore, this problem can be solved by employing a strategy identical to the one just presented. To evaluate the partition function $q$, the "trick" used earlier was to write the partition function as a series and use the equivalent series expression:
$$
\begin{aligned}
q & =\sum_n e^{-\beta \varepsilon_n}=1+e^{-\beta h c \widetilde{\nu}}+e^{-2 \beta h c \tilde{\nu}}+e^{-3 \beta h c \widetilde{\nu}}+\ldots \\
& =\frac{1}{1-e^{-\beta h c \widetilde{\nu}}}
\end{aligned}
$$
Since $\tilde{\nu}=208 \mathrm{~cm}^{-1}$ and $T=298 \mathrm{~K}$, the partition function is
$$
\begin{aligned}
q & =\frac{1}{1-e^{-\beta h c \widetilde{\nu}}} \\
& =\frac{1}{1-e^{-h c \widetilde{\nu} / k T}} \\
& =\frac{1}{1-\exp \left[-\left(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(208 \mathrm{~cm}^{-1}\right)}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)(298 \mathrm{~K})}\right)\right]} \\
& =\frac{1}{1-e^{-1}}=1.58
\end{aligned}
$$
This result is then used to evaluate the probability of occupying the second vibrational state $(n=2)$ as follows:
$$
\begin{aligned}
p_2 & =\frac{e^{-2 \beta h c \tilde{\nu}}}{q} \\
& =\frac{\exp \left[-2\left(\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}^{-1}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(208 \mathrm{~cm}^{-1}\right)}{\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)(298 \mathrm{~K})}\right)\right]}{1.58} \\
& =0.086
\end{aligned}
$$
Therefore, the answer is \boxed{0.086}.

Promblem 4: In a rotational spectrum of $\operatorname{HBr}\left(B=8.46 \mathrm{~cm}^{-1}\right)$, the maximum intensity transition in the R-branch corresponds to the $J=4$ to 5 transition. At what temperature was the spectrum obtained?
Explanation for Problem 4: The information provided for this problem dictates that the $J=4$ rotational energy level was the most populated at the temperature at which the spectrum was taken. To determine the temperature, we first determine the change in occupation number for the rotational energy level, $a_J$, versus $J$ as follows:
$$
\begin{aligned}
a_J & =\frac{N(2 J+1) e^{-\beta h c B J(J+1)}}{q_R}=\frac{N(2 J+1) e^{-\beta h c B J(J+1)}}{\left(\frac{1}{\beta h c B}\right)} \\
& =N \beta h c B(2 J+1) e^{-\beta h c B J(J+1)}
\end{aligned}
$$
Next, we take the derivative of $a_J$ with respect to $J$ and set the derivative equal to zero to find the maximum of the function:
$$
\begin{aligned}
\frac{d a_J}{d J} & =0=\frac{d}{d J} N \beta h c B(2 J+1) e^{-\beta h c B J(J+1)} \\
0 & =\frac{d}{d J}(2 J+1) e^{-\beta h c B J(J+1)} \\
0 & =2 e^{-\beta h c B J(J+1)}-\beta h c B(2 J+1)^2 e^{-\beta h c B J(J+1)} \\
0 & =2-\beta h c B(2 J+1)^2 \\
2 & =\beta h c B(2 J+1)^2=\frac{h c B}{k T}(2 J+1)^2 \\
T & =\frac{(2 J+1)^2 h c B}{2 k}
\end{aligned}
$$
Substitution of $J=4$ into the preceding expression results in the following temperature at which the spectrum was obtained:
$$
\begin{aligned}
T & =\frac{(2 J+1)^2 h c B}{2 k} \\
& =\frac{(2(4)+1)^2\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3.00 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}\right)\left(8.46 \mathrm{~cm}^{-1}\right)}{2\left(1.38 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\right)} \\
& =4943 \mathrm{~K}
\end{aligned}
$$
Therefore, the answer is \boxed{4943}.

Promblem 5: Determine the diffusion coefficient for Ar at $298 \mathrm{~K}$ and a pressure of $1.00 \mathrm{~atm}$.
Explanation for Problem 5: ￼￼Using Equation (17.10) and the collisional cross section for Ar provided in Table 17.1,
 $$
\begin{aligned}
D_{Ar} &= \frac{1}{3} \nu_{ave, Ar} \lambda_{Ar} \\
&= \frac{1}{3} \left(\frac{8RT}{\pi M_{Ar}}\right)^{\frac{1}{2}} \left(\frac{RT}{PN_A\sqrt{2}\sigma_{Ar}}\right) \\
&= \frac{1}{3} \left(\frac{8(8.314~\mathrm{J~mol^{-1}~K^{-1}}) \times 298~\mathrm{K}}{\pi(0.040~\mathrm{kg~mol^{-1}})}\right)^{\frac{1}{2}} \\
&\quad \times \left(\frac{(8.314~\mathrm{J~mol^{-1}~K^{-1}}) \times 298~\mathrm{K}}{(101,325~\mathrm{Pa}) \times (6.022 \times 10^{23}~\mathrm{mol^{-1}})} \times \frac{1}{\sqrt{2}(3.6 \times 10^{-19}~\mathrm{m^2})}\right) \\
&= \frac{1}{3} \times (397~\mathrm{m~s^{-1}}) \times (7.98 \times 10^{-8}~\mathrm{m}) \\
&= 1.1 \times 10^{-5}~\mathrm{m^2~s^{-1}}
\end{aligned}
$$
Therefore, the answer is \boxed{1.1}.

