System entropy, $S,$ is related to the canonical partition function as $S=\frac{U-A}{T}$

$=-\frac{1}{T}(delln\frac{Q}{d}el{)}_{N,V}+k_{B}lnQ=k_{B}T(delln\frac{Q}{d}$elT${)}_{N,V}+k_{B}lnQ($refer to subchapter $21-6$ on page

$90$ of the pdf material$).$ Answer the following questions. $(10$ points for $($a$)$ and $30$ points for $($b$))$

$($a$)$ When a zero of energy is chosen to be the ground vibrational state of the ground electronic

state, the molecular partition function, $q(V,T),$ can be written as

$q(V,T)=q_{tans}(V,T)q_{\text{r}\text{o}\text{t}}(T)q_{\text{v}\text{i}\text{b}}(T)q_{\text{e}\text{l}\text{e}\text{c}}(T)$

$=\frac{V}{(T{)}^3}{q}_{\text{r}\text{o}\text{t}}(T)pro{d}_j=1^{j_{\text{m}\text{a}\text{x}}}\frac{1}{1-e^{-\frac{{}_{ib,}}{\frac{?}{?}}T}}{g}_{el},$

where the total number, $j_{\text{m}\text{a}\text{x}},$ of vibrational modes for an $n-$atom molecule is given by $3n-5$

for linear molecules and $3n-6$ for nonlinear molecules. For ideal gases, show that

$\frac{S}{N{k}_B}=(\frac{5}{2}+ln(\frac{V}{N{}^3}))+(ln{q}_{\text{r}\text{o}\text{t}}+\frac{f_{\text{r}\text{o}\text{t}}}{2})+{\displaystyle \underset{j=1}{\overset{j_{\text{m}\text{a}\text{x}}}{}}}[\frac{\frac{{}_{rb,j}}{T}}{e^{\frac{{}_{wb,j}}{T}}-1}-ln(1-e^{-\frac{{}_{wb,j}}{T}})]+ln{g}_{e1},$

where the first two terms on the right$-$hand side are originated from $q_{tan}$ and the

indistinguishability correction factor, $N!,$ the second two terms from $q_{\text{r}\text{o}\text{t}},$ the third summation

term from $q_{\text{v}\text{i}\text{b}},$ and the last term from $q_{\text{e}\text{l}\text{e}\text{c}}.f_{\text{r}\text{o}\text{t}}$ indicates the number of rotational degrees

of freedom.

$($b$)$ Using the equation for the system entropy given in $($a$),$ calculate the standard molar

entropies, at $P=P^{**}=1$ bar $\{$:$={10}^5(Pa)={10}^5(N)*m^{-2}),$ of $N_{2}O(g)$ and $N{O}_2(g)$ at $T=$

$25C=298\mathrm{.}15K$ and discuss what plays a major role in making the difference between two

calculated values of $\frac{?}{\text{b}\text{a}\text{r}}(S{)}^n.$ Note that molecular entropy is closely related to the number of

thermally accessible degrees of freedom. Use the ideal gas equation, $PV=N{k}_{B}T,$ to estimate

the volume, $V$ and note that the gas constant, $R,$ is related to $k_B$ as

$R=N_A{k}_B(=8\mathrm{.}3145(J)*K^{-1}*mo{l}^{-1}).$ Values of parameters required to calculate $\frac{?}{\text{b}\text{a}\text{r}}(S^{-1})$ can be

found in Table $18\mathrm{.}4$ on page $59$ of the pdf material and the NIST$-$JANAF thermochemical tables

attached below. It is recommended to compare the calculated values of $\frac{?}{\text{b}\text{a}\text{r}}(S{)}^{**}$ to the experimental

values of $\frac{?}{\text{b}\text{a}\text{r}}(S{)}^{-1}$ presented below. Also note that vibrational frequencies in $c{m}^{-1}$ tabulated below

can be converted into vibrational temperatures in Table $18\mathrm{.}4$ using the relation,

${}_{\text{v}\text{i}\text{b}}=\frac{}{k_B}=h\frac{c}{b}ar\frac{v}{k_B}=\frac{(6\mathrm{.}626{10}^{-34}(J)*s)(3\mathrm{.}0{10}^{10}(cm)*s^{-1})}{1\mathrm{.}380{10}^{-23}(J)*K^{-1}}\frac{}{b}ar(v)(c{m}^{-1})$

with $c$ being the speed of light in vacuum. In fact, vibrational frequencies in $c{m}^{-1}$ tabulated

below correspond to vibrational wavenumbers, tilde$(v).$ In addition, ground state quantum weight

and $$ indicate $g_{e1}$ and the symmetry number, respectively. For $N_{2}O,$ the rotational constant

in $c{m}^{-1}$ can be converted into the rotational temperature using the relation,

${}_{\text{r}\text{o}\text{t}}=\frac{B}{k_B}=\frac{{}^2}{2I{k}_B}=\frac{hc}{k_B}\frac{h}{8{}^{2}cI}=\frac{hc}{k_B}$tilde$(B)=\frac{(6\mathrm{.}626{10}^{-34}(J)*s)(3\mathrm{.}0{10}^{10}(cm)*s^{-1})}{1\mathrm{.}380{10}^{-23}(J)*K^{-1}}$tilde$(B)(c{m}^{-1}),$

where $B$ and tilde$(B)$ denote the rotational constants in units of energy and wavenumber,

respectively.