Fixing typos in chapter 8 and 9.

chapters/ch8.tex

@@ -65,8 +65,9 @@ \section{Vibrational Partition Function}%
 normal coordinates respectively. In solving the above Hamiltonian, the total
 energy becomes,
 \begin{equation*}
-	\varepsilon = \sum_{j=1}^{\alpha}{(n_j + \frac{1}{2})h \nu_j}
-	\quad\text{for}\quad n_j = 0, 1, 2, \ldots
+	\varepsilon = \sum_{j=1}^{\alpha}{\sum_{i=0}^{\infty}{(n_{j,i} + \frac{1}{2})h
+	\nu_{j}}}
+	\quad\text{for}\quad n_{j,i} = 0, 1, 2, \ldots
 \end{equation*}
 where,
 \begin{equation*}
@@ -138,7 +139,13 @@ \subsection{Linear Molecules}
 	q_{rot} = \frac{8 \pi^2 IkT}{\sigma h^2} = \frac{T}{\sigma \Theta_r}.
 \end{equation*}
 Once again $\sigma$ is the symmetry number which represents the number of
-indistinguishable configurations a molecule has.
+indistinguishable configurations a molecule has. The purpose of $\sigma$ is
+important to mention here. Its meaning classically comes from the number of
+different equivalent arrangements of the molecule that are identical over
+specific rotations. In other words, The number of configurations of the atoms
+that make the same structure will have rotations which show them to be
+identical. This prevents their over-counting. In terms of quantum mechanics, the
+particles are indistinguishable so the symmetry number is immediately required.
 
 \subsection{Non-Linear Molecules}
 Non-linear rigid bodies have at least two different diagonal moments of
@@ -172,7 +179,7 @@ \subsubsection{Spherical Tops}
 	v = \frac{4IkT}{\hbar^2} e^{-J^2 \hbar^{2}/2IkT},
 \end{align*}
 Here, I note that the integral is directly analogous to that used to prove the
-law of equipartition derived in Section\ref{sec:eoe}. Then, the final solution
+law of equipartition derived in Section~\ref{sec:eoe}. Then, the final solution
 is
 \begin{equation*}
 	q_{rot} = \frac{1}{\sigma}

chapters/ch9.tex

@@ -26,15 +26,15 @@ \section{Chemical Equilibrium in Terms of Partition Functions}%
 free energy is at a minimum. By the conservation of mass and the definition of
 equilibrium we know
 \begin{equation*}
-	\nu_C C + \nu_D D - \\nu_A A - \nu_B B = 0.
+	\nu_C C + \nu_D D - \nu_A A - \nu_B B = 0.
 \end{equation*}
 Taking the definition of the Helmholtz free energy as,
 \begin{equation*}
 	\d{A} = -S\d{T} - p\d{V} + \sum_{j}{\mu_j \d{N_{j}}},
 \end{equation*}
 and using the fact that we are at constant $T$ and $V$, we get
 \begin{equation*}
-	\d{A} = \sum_{j}{\mu_j \d{N_{j}}} = {\left(\sum_{j}{\nu_j
+	\d{A} = \sum_{j}{\mu_j \d{N_{j}}} = {\left(\sum_{j}{\nu_j \sigma_j
 	\mu_{j}}\right)}\d{\lambda},
 \end{equation*}
 where $\sigma_j \nu_j \d{\lambda} = \d{N_{j}}$ with $\sigma_j$ being $1$
@@ -55,11 +55,12 @@ \section{Chemical Equilibrium in Terms of Partition Functions}%
 Using the previously derived relation,
 \begin{align*}
 	\mu_A &= -kT {\left(\frac{\partial \ln{\Q}}{\partial N_{A}}\right)}_{N_j, V,
-			T} = -kT {\left(\partial \ln{\left(\sum_{i}^{A, B, C, D}
-		{\frac{q_{i}}{N_i !}}\right)} /
-		\partial N_A \right)}
+	T}
+			= -kT {\left(\partial \sum_{i}^{A, B, C, D}
+					{\ln{\left[\frac{q_{i}}{N_i !}}\right]} / \partial N_A
+			\right)}
 			 = -kT {\left(\partial \ln{\left(\frac{q_{A}}{N_A !}\right)} /
-		\partial N_A \right)}\\
+			 \partial N_A \right)}\\
 		  &= -kT \partial(N_{A} \ln{q_{A}} - N_{A}\ln{N_{A}} + N_{A}) / \partial N_A
 		  = -kT {\left( N_A \ln{\left[\frac{q_A}{N_{A}}\right]} + N_A \right)} /
 		  \partial N_A \\
@@ -107,7 +108,7 @@ \section{Chemical Equilibrium in Terms of Partition Functions}%
 equilibrium constant $K_p (T)$,
 \begin{equation*}
 	K_p(T) = \frac{p_{D}^{\nu_D}p_{C}^{\nu_C}}{p_{B}^{\nu_B}p_{A}^{\nu_A}} =
-	(kT)^{\sum_{j}{\nu_{j}}} K_c (T).
+	(kT)^{\sum_{j}{\sigma_j \nu_{j}}} K_c (T).
 \end{equation*}
 
 \section{Thermodynamic Tables}%
@@ -117,7 +118,7 @@ \section{Thermodynamic Tables}%
 \begin{equation*}
 	\mu(T,p) = \mu_0 (T) + kT\ln{p}.
 \end{equation*}
-Using the fact that $\sum_{j}{\nu_j \mu_j} = 0$, we have
+Using the fact that $\sum_{j}{\sigma_j \nu_j \mu_j} = 0$, we have
 \begin{align*}
 	\sum_{j}{\sigma_j \nu_j kT\ln{p_{j}}} + \sum_{j}{\sigma_j \nu_j
 	\mu_{0,j}(T)} &= 0\\