From 63620fcbe5ac4f9749f300a65abc381efabd0105 Mon Sep 17 00:00:00 2001 From: Deepak Date: Wed, 16 Mar 2022 21:54:55 -0500 Subject: [PATCH] fix: typo fixes --- tex/3.11.tex | 4 ++-- tex/3.8.tex | 2 +- 2 files changed, 3 insertions(+), 3 deletions(-) diff --git a/tex/3.11.tex b/tex/3.11.tex index 3d6349d..b637639 100644 --- a/tex/3.11.tex +++ b/tex/3.11.tex @@ -114,7 +114,7 @@ \subsection{(a)} Want to show - \item \begin{equation} + \begin{equation} \left. \pdv{T}{V} \right|_{S, N} = - \left. \pdv{P}{S} \right|_{V, N} \end{equation} @@ -133,7 +133,7 @@ \pdv{T}{V_{S, N}} &= - \pdv{P}{S_{V, N}} \end{align} - \subection{(b)} + \subsection{(b)} We want: \begin{equation} diff --git a/tex/3.8.tex b/tex/3.8.tex index ca687ce..fd66d81 100644 --- a/tex/3.8.tex +++ b/tex/3.8.tex @@ -107,7 +107,7 @@ The equipartition theorem tells us that each degree of freedom in the kinetic energy should satisfy $\frac12 N \kb T$. So, using the three degrees of freedom as described above, we can just write that $K = \left< E_1 \right> = \frac32 N \kb T$. - Our specific heat per particle satisfies $\N c_v^{(1)}) \left.\left( \pdv{E_1}{T} \right)\right|_{V, N}$, which is clearly $c_v^{(1)} = \frac32 \kb$, as Sethna wants us to find. + Our specific heat per particle satisfies $N c_v^{(1)} = \left.\left( \pdv{E_1}{T} \right)\right|_{V, N}$, which is clearly $c_v^{(1)} = \frac32 \kb$, as Sethna wants us to find. Now lets use \cref{eq:3} and solve for $c_v^{(2)}$. \begin{align}