\documentclass{beamer} % BB Apr 13 2018.

\usetheme{fsuheplat}

%AB: color definitions
\definecolor{abtitlecolor}{rgb}{0.0,0.255,0.494}
\definecolor{absecondarycolor}{rgb}{0.0,0.416,0.804}
\definecolor{abprimarycolor}{rgb}{1.0,0.686,0.0}

%AB: redefinition of beamer colors
\setbeamercolor{title}{fg=abtitlecolor}
\setbeamercolor{frametitle}{fg=abtitlecolor}
\setbeamercolor{palette tertiary}{fg=white,bg=abtitlecolor}
\setbeamercolor{palette secondary}{fg=white,bg=absecondarycolor}
\setbeamercolor{palette primary}{fg=black,bg=abprimarycolor}
\setbeamercolor{structure}{fg=abtitlecolor}

\setbeamerfont{section in toc}{series=\bfseries}
\def\b{\color{blue}} \def\r{\color{red}} \def\s{\color{black}}

%AB: remove navigation icons
\beamertemplatenavigationsymbolsempty

\begin{document}

\frame{\frametitle{\b Ideal Gas - 1} \medskip

{\bf Scuba Diver (1):} Solve 
$$ P_1\,V_1=n\,R\,T_1~~{\rm and}~~P_2\,V_2=n\,R\,T_2\,, $$
where the temperatures are in Kelvin, for $V_2/V_1$ to find $V_2$. To 
calculate the pressures $P_1$ and $P_2$ use $\rho_s=1.025\times 10^3\ 
kg/m^3$ 

for the density of sea water and $1~atm = 101\times 10^3\ N/m^2$.
\bigskip

{\bf Pressure in a Container with Neon Gas (2):} Use
$$ P_0\,V_{\rm mol}=R\,T_0~~{\rm and}~~P\,V = n\,R\,T $$
with $P_0=1~$atm, $V_{\rm mol}=22.4\,$liter, $T_0=273.15\,$K and 
$n=M/m_u$, where $m_u=20.18\,$g is the mol mass of Neon,
to calculate
$$ P = P_0\,n\,\frac{V_{\rm mol}}{V}\,\frac{T}{T_0}\,.$$
\smallskip

{\bf Ideal Gas (3):} The final pressure is $P_f = P_i\,(V_i/V_f)\,
(T_f/T_i)$, where the initial values $P_i$, $V_i$, $T_i$ and 
final values $V_f$, $T_f$ are given.

\vskip 20pt }


\frame{\frametitle{\b Ideal Gas - 2} \medskip

{\bf Escaping Hydrogen (4).} The average kinetic energy is
$$ K_{\rm av} = M_{\rm H_2}\,\frac{v^2}{2} = \frac{3}{2}\,k\,T\,, $$
where $M_{\rm H_2}=2\times 1.673\times 10^{-27}\,$kg is the mass of
one $\rm H_2$ molecule and $k=1.381\times 10^{-23}\,$J/K the Boltzmann 
constant. Use this equation to find $v=v_{\rm rms}$ and calculate
the ratio $v_{\rm rms}/v_{\rm escape}$.
\bigskip

{\bf Kinetic energy of a Gas (5)} ($n$ number of moles):
$$ K = n\,N_{\rm Avogadro}\,\frac{3}{2}\,k_{\rm\ Boltzmann}\,T\,.$$
\bigskip

\vskip 20pt }
\end{document}
