\documentclass[11pt,fleqn]{article}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site 
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% Page setup
\topmargin -1.5cm      
\oddsidemargin -0.04cm   
\evensidemargin -0.04cm  
\textwidth 16.59cm
\textheight 24cm 
\setlength{\parindent}{0cm} 
\setlength{\parskip}{0cm} 


% Fonts
\usepackage{latexsym}
\usepackage{amsmath} 
\usepackage{amssymb} 
\usepackage{bm}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{subfigure}
\usepackage{epsfig}


% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}

% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}

% Mass symbol
\newcommand{\mass}{\mathsf{m}} 
\newcommand{\Mass}{\mathsf{M}} 

% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}} 
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}} 
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}

% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}

% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}

% heading
\newcommand{\heading}[1]{\begin{center} \Large {#1} \end{center}}
\newcommand{\auname}[1]{\begin{center} \bf Submitted by: #1 \end{center}}


\begin{document} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\heading{5955: Change of boiling point with altitude} 
\auname{Boris Shif}


{\bf The problem:}


\Dn
Consider an atmosphere as an ideal gas whose average mass is $m_A=30$ gr/mole, atmospheric pressure at sea level ($h=0$) equals $P_0$ and is at uniform temperature of $T_A={27 \,{^\circ}C}$.\\
We take liquid whose latent heat $Q$ is ${1000}$ cal/mole and we find that its boiling point is ${95 \,{^\circ}C}$ at the top of a mountain and ${105 \,{^\circ}C}$ at the bottom (sea level).\\
Asume that the gas phase of the liquid is an ideal gas with density much lower than that of the liquid.
\Dn

(1) Calculate the atmospheric pressure $P_A$ as a function of height $h$. \\

(2) Calculate the liquid pressure as a function of its temerature.\\

(3) From above deduce the height of the mountain.\\
\Dn

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}
\Dn

(1)
\\The probability to find a particle at height $h$ is 
\[
p(h)\propto e^{-\beta mgh}
\]
the number of particles at height $h$ is
\[
N_A(H)=N_A(0)e^{-\beta mgh}
\]
The gas is an ideal gas at uniform temperature and therefore
\[
P_A(h)=P_A(0)e^{-\beta mgh}
\]
%%%%%%%%%%%%%%%
\Dn

(2) 
\\We know that $F_H$ is an analitical function and $dF_H=-SdT-PdV$.
From the analitical property of the $F_H$ function
\[
\frac{dP}{dT}=\frac{\partial^2F_H}{\partial T\partial V}=\frac{\partial^2F_H}{\partial V\partial T}=\frac{dS}{dV}=\frac{TdS}{TdV}\approx\frac{Q}{TV_{gas}}=\frac{Q}{k_bT^2}P
\]
\\therefore
\[
\frac{dP}{P}=-\frac{Q}{k_bT^2}dT
\]
\[
\Rightarrow Ln(P(T_h))-Ln(P(T_{h=0}))=\frac{Q}{k_b}(\frac{1}{T_{h=0}}-\frac{1}{T_h})\Leftrightarrow P_g(T)=P_g(T_{h=0})e^{\frac{Q}{k_b}(\frac{1}{T_{h=0}}-\frac{1}{T_h})}
\]%%%%%%%%%%%%%%%
\Dn
\Dn

(3) 
\\We define $T_{h=0}=T_0$ 
\\At steady state the pressure of the atmosphere and the gas must be equal
\[
P_g(T_h)=P_A(h)\Leftrightarrow P_0e^{\frac{Q}{k_b}(\frac{1}{T_0}-\frac{1}{T_h})}=P_0e^{-\frac{m_g}{k_bT_A}z}\Leftrightarrow Q(\frac{1}{T_0}-\frac{1}{T_h})=-\frac{m_Ag}{T_A}h
\]
\\We know that $1 cal/mol=4.184 J/mol$ and therefore:
\[
h=-4184*(\frac{1}{378}-\frac{1}{368})*300/0.3=300.8m
\]
\\Remark:
\\In case of water the latent heat is $\approx40 KJ/mole$ and we get the same way that
\[
h\approx-40000*(\frac{1}{378}-\frac{1}{368})*300/0.3=2875m
\]
This fit with the fact that the boiling point of water decreases by $1^0$C for increase of 286 meters in height.
%%%%%%%%%%%%%%
\end{document} 