% this tex file was generated by the Physics Derivation Graph 
\documentclass[12pt]{article}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage[dvipdfmx,colorlinks=true,pdfkeywords={physics derivation graph}]{hyperref}
\usepackage{graphicx} % for including PNG files
% inference rules as newcommand for use in the body
\newcommand\LHSofexprequalsLHSofexpr[3]{LHS of Eq.~\ref{eq:#1} is equal to LHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\RHSofexprequalsRHSofexpr[3]{RHS of Eq.~\ref{eq:#1} is equal to RHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\Xcrossbothsidesby[3]{Take cross product of $#1$ and Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\Xdotbothsides[3]{Take inner product of $#1$ with Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\addXtobothsides[3]{Add $#1$ to both sides of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\addexprtoexpr[3]{Add Eq.~\ref{eq:#1} to Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#2}.}
\newcommand\addzerotoLHS[3]{Add zero to LHS of Eq.~\ref{eq:#2}, where $0=#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\addzerotoRHS[3]{Add zero to RHS of Eq.~\ref{eq:#2}, where $0=#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\applydivergence[2]{Apply divergence to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\applyfunctiontobothsidesofexpression[4]{Apply function $#1$ with argument $#2$ to Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}}
\newcommand\applygradienttoscalarfunction[2]{Apply gradient to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\applyoperatortobra[2]{Apply operator in Eq.~\ref{eq:#1} to bra; yields Eq.~\ref{eq:#2}.}
\newcommand\applyoperatortoket[2]{Apply operator in Eq.~\ref{eq:#1} to ket; yields Eq.~\ref{eq:#2}.}
\newcommand\assumeNdimensions[2]{Assume $#1$ dimensions; decompose vector to be Eq.~\ref{eq:#2}.}
\newcommand\bothsidescrossX[3]{Take cross product of Eq.~\ref{eq:#2} and $#1$; yields Eq.~\ref{eq:#3}}
\newcommand\bothsidesdotX[3]{Take inner product of Eq.~\ref{eq:#2} with $#1$; yields Eq.~\ref{eq:#3}}
\newcommand\boundarycondition[2]{Boundary condition: Eq.~\ref{eq:#2} when Eq.~\ref{eq#1}.}
\newcommand\boundaryconditionforexpr[2]{A boundary condition for Eq.~\ref{eq:#1} is Eq.~\ref{eq:#2}}
\newcommand\changefivevariablesinexpr[9]{\def\ArgOne{{#1}}
\def\ArgTwo{{#2}}
\def\ArgThree{{#3}}
\def\ArgFour{{#4}}
\def\ArgFive{{#5}}
\def\ArgSix{{#6}}
\def\ArgSeven{{#7}}
\def\ArgEight{{#8}}
\def\ArgNine{{#9}}
\changefivevariablesinexprRelay
}
\newcommand\changefivevariablesinexprRelay[3]{Change of variable $ArgOne$ to $ArgTwo$ and $ArgThree$ to $ArgFour$ and $ArgFive$ to $ArgSix$ and $ArgSeven$ to $ArgEight$ and $ArgNine$ to $ArgOne0$ in Eq.~\ref{eq:ArgOne1}; yields Eq.~\ref{eq:ArgOne2}.}
\newcommand\changefourvariablesinexpr[9]{\def\ArgOne{{#1}}
\def\ArgTwo{{#2}}
\def\ArgThree{{#3}}
\def\ArgFour{{#4}}
\def\ArgFive{{#5}}
\def\ArgSix{{#6}}
\def\ArgSeven{{#7}}
\def\ArgEight{{#8}}
\def\ArgNine{{#9}}
\changefourvariablesinexprRelay
}
\newcommand\changefourvariablesinexprRelay[1]{Change of variable $ArgOne$ to $ArgTwo$ and $ArgThree$ to $ArgFour$ and $ArgFive$ to $ArgSix$ and $ArgSeven$ to $ArgEight$ in Eq.~\ref{eq:ArgNine}; yields Eq.~\ref{eq:ArgOne0}.}
\newcommand\changesixvariablesinexpr[9]{\def\ArgOne{{#1}}
\def\ArgTwo{{#2}}
\def\ArgThree{{#3}}
\def\ArgFour{{#4}}
\def\ArgFive{{#5}}
\def\ArgSix{{#6}}
\def\ArgSeven{{#7}}
\def\ArgEight{{#8}}
\def\ArgNine{{#9}}
\changesixvariablesinexprRelay
}
\newcommand\changesixvariablesinexprRelay[5]{Change of variable $ArgOne$ to $ArgTwo$ and $ArgThree$ to $ArgFour$ and $ArgFive$ to $ArgSix$ and $ArgSeven$ to $ArgEight$ and $ArgNine$ to $ArgOne0$ and $ArgOne1$ to $ArgOne2$ in Eq.~\ref{eq:ArgOne3}; yields Eq.~\ref{eq:ArgOne4}.}
\newcommand\changethreevariablesinexpr[8]{Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ in Eq.~\ref{eq:#7}; yields Eq.~\ref{eq:#8}.}
\newcommand\changetwovariablesinexpr[6]{Change variable $#1$ to $#2$ and $#3$ to $#4$ in Eq.~\ref{eq:#5}; yields Eq.~\ref{eq:#6}.}
\newcommand\changevariableXtoY[4]{Change variable $#1$ to $#2$ in Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}.}
\newcommand\claimLHSequalsRHS[1]{Thus we see that LHS of Eq.~\ref{eq:#1} is equal to RHS.}
\newcommand\claimexprequalsexpr[2]{Thus we see that Eq.~\ref{eq:#1} is equivalent to Eq.~\ref{eq:#2}.}
\newcommand\combineliketerms[2]{Combine like terms in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\conjugatebothsides[2]{Conjugate both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\conjugatefunctionX[3]{Conjugate $#1$ in Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\conjugatetransposebothsides[2]{Conjugate transpose of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\declareassumption[1]{Eq.~\ref{eq:#1} is an assumption.}
\newcommand\declarefinalexpr[1]{Eq.~\ref{eq:#1} is one of the final equations.}
\newcommand\declareguesssolution[2]{Judicious choice as a guessed solution to Eq.~\ref{eq:#1} is Eq.~\ref{eq:#2},}
\newcommand\declareidentity[1]{Eq.~\ref{eq:#1} is an identity.}
\newcommand\declareinitialexpr[1]{Eq.~\ref{eq:#1} is an initial equation.}
\newcommand\differentiatewithrespectto[3]{Differentiate Eq.~\ref{eq:#2} with respect to $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\distributeconjugatetofactors[2]{Distribute conjugate to factors in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\distributeconjugatetransposetofactors[2]{Distribute conjugate transpose to factors in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\dividebothsidesby[3]{Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\divideexprbyexpr[3]{Divide Eq.~\ref{eq:#1} by Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\dropnondominantterm[3]{Based on the assumption $#1$, drop non-dominant term in Eq.~\ref{#2}; yeilds Eq.~\ref{#3}}
\newcommand\evaluatedefiniteintegral[2]{Evaluate definite integral Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\expandLHS[2]{Expand the LHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\expandRHS[2]{Expand the RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\expandintegrand[2]{Expand integrand of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\expandmagnitudetoconjugate[3]{Expand $#1$ in Eq.~\ref{eq:#2} with conjugate; yields Eq.~\ref{eq:#3}.}
\newcommand\exprisequivalenttoexprunderthecondition[3]{Eq.~\ref{eq:#1} is equivalent to Eq.~\ref{eq:#2} under the condition in Eq.~\ref{eq:#3}.}
\newcommand\expristrueunderconditionexpr[3]{Eq.~\ref{eq:#1} is valid when Eq.~\ref{eq:#2} occurs; yields Eq.~\ref{eq:#3}.}
\newcommand\factoroutX[3]{Factor $#1$ from Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\factoroutXfromLHS[3]{Factor $#1$ from the LHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\factoroutXfromRHS[3]{Factor $#1$ from the RHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\functioniseven[5]{$#1$ is even with respect to $#2$, so replace $#1$ with $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.}
\newcommand\functionisodd[5]{$#1$ is odd with respect to $#2$, so replace $#1$ with $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.}
\newcommand\indefiniteintegralover[3]{Indefinite integral of both sides of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\indefiniteintegrateLHSover[3]{Indefinite integral of LHS of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\indefiniteintegrateRHSover[3]{Indefinite integral of RHS of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\indefiniteintegration[2]{Indefinite integral of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\integrate[2]{Integrate Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}.}
\newcommand\integrateoverfromto[5]{Integrate Eq.~\ref{eq:#4} over $#1$ from lower limit $#2$ to upper limit $#3$; yields Eq.~\ref{eq:#5}.}
\newcommand\makeexprpower[3]{Make Eq.~\ref{eq:#2} the power of $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\maximumofexpr[3]{The maximum of Eq.~\ref{eq:#2} with respect to $#1$ is Eq.~\ref{eq:#3}}
\newcommand\multiplyLHSbyunity[3]{Multiply LHS of Eq.~\ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~\ref{eq:#3}}
\newcommand\multiplyRHSbyunity[3]{Multiply RHS of Eq.~\ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~\ref{eq:#3}}
\newcommand\multiplybothsidesby[3]{Multiply both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\multiplyexprbyexpr[3]{Multiply Eq.~\ref{eq:#1} by Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\normalizationcondition[1]{Normalization condition is Eq.~\ref{eq:#1}.}
\newcommand\partiallydifferentiatewithrespectto[3]{Partially differentiate Eq.~\ref{eq:#2} with respect to $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\raisebothsidestopower[3]{Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\replaceconstantwithvalue[5]{Replace constant $#1$ with value $#2$ and units $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}}
\newcommand\replacecurlwithLeviCevitasummationcontravariant[2]{Replace curl in Eq.~\ref{eq:#1} with Levi-Cevita contravariant; yields Eq.~\ref{eq:#2}.}
\newcommand\replacescalarwithvector[2]{Replace scalar variables in Eq.~\ref{eq:#1} with equivalent vector variables; yields Eq.~\ref{eq:#2}.}
\newcommand\replacesummationnotationwithvectornotation[2]{Replace summation notation in Eq.~\ref{eq:#1} with vector notation; yields Eq.~\ref{eq:#2}.}
\newcommand\selectimaginaryparts[2]{Select imaginary parts of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\selectrealparts[2]{Select real parts of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\separatethreevectorcomponents[4]{Separate three vector components in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3} and Eq.~\ref{eq:#4}.}
\newcommand\separatetwovectorcomponents[3]{Separate two vector components in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3}}
\newcommand\separatevectorintotwotrigonometricratios[4]{Separate vector in Eq.~\ref{eq:#2} into components related by angle $#1$; yields Eq.~\ref{eq:#3} and Eq.~\ref{eq:#4}.}
\newcommand\simplify[2]{Simplify Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\solveforX[3]{Solve Eq.~\ref{eq:#2} for $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\squarerootbothsides[3]{Take the square root of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3}.}
\newcommand\substituteLHSofexprintoexpr[3]{Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\substituteLHSoffiveexpressionsintoexpr[7]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} and LHS of Eq.~\ref{eq:#5} into Eq.~\ref{eq:#6}; yields Eq.~\ref{eq:#7}.}
\newcommand\substituteLHSoffourexpressionsintoexpr[6]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} into Eq.~\ref{eq:#5}; yields Eq.~\ref{eq:#6}.}
\newcommand\substituteLHSofsixexpressionsintoexpr[8]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} and LHS of Eq.~\ref{eq:#5} and LHS of Eq.~\ref{eq:#6} into Eq.~\ref{eq:#7}; yields Eq.~\ref{eq:#8}.}
\newcommand\substituteLHSofthreeexpressionsintoexpr[5]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} into Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.}
\newcommand\substituteLHSoftwoexpressionsintoexpr[4]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} into Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}.}
\newcommand\substituteRHSofexprintoexpr[3]{Substitute RHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\subtractXfrombothsides[3]{Subtract $#1$ from both sides of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\subtractexprfromexpr[3]{Subtract Eq.~\ref{eq:#1} from Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#2}.}
\newcommand\sumexponents[2]{Sum exponents on LHS and RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\sumexponentsLHS[2]{Sum exponents on LHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\sumexponentsRHS[2]{Sum exponents on RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\swapLHSwithRHS[2]{Swap LHS of Eq.~\ref{eq:#1} with RHS; yields Eq.~\ref{eq:#2}.}
\newcommand\takecurlofbothsides[2]{Apply curl to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\title{coefficient of isothermal compressibility using the equation of state for an ideal gas}
\date{\today}
\author{none}
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{9in}
\setlength{\oddsidemargin}{0in}
\setlength{\textwidth}{6.5in}
\begin{document}
\maketitle
\begin{abstract}
Generated by the \href{http://derivationmap.net/}{Physics Derivation Graph}.
https://notendur.hi.is/hj/EE2/HD1lausn.pdf
\end{abstract}
% step ID = 3831533
\declareinitialexpr{4239912}
\begin{equation}
\kappa_T = \frac{-1}{V} \left( \frac{ \partial V}{\partial P} \right)_T
\label{eq:4239912}
\end{equation}
% step ID = 9872032
\declareinitialexpr{4454896}
\begin{equation}
P V = n R T
\label{eq:4454896}
\end{equation}
% step ID = 3862516
\dividebothsidesby{P}{4454896}{5840241}
\begin{equation}
V = \frac{n R T}{P}
\label{eq:5840241}
\end{equation}
% step ID = 9225696
\substituteLHSofexprintoexpr{5840241}{4239912}{5196207}
\begin{equation}
\kappa_T = \frac{-1}{V} \left( \frac{ \partial }{\partial P}\left(\frac{nRT}{P}\right) \right)_T
\label{eq:5196207}
\end{equation}
% step ID = 6914144
\simplify{5196207}{3915956}
\begin{equation}
\kappa_T = \frac{-nRT}{V} \left( \frac{ \partial }{\partial P}\left(\frac{1}{P}\right) \right)_T
\label{eq:3915956}
\end{equation}
% step ID = 1873674
\simplify{3915956}{6275836}
\begin{equation}
\kappa_T = \frac{-nRT}{V} \left( \frac{-1}{P^2}\right)
\label{eq:6275836}
\end{equation}
% step ID = 5391858
\substituteLHSofexprintoexpr{4454896}{6275836}{1003658}
\begin{equation}
\kappa_T = \frac{-PV}{V} \left( \frac{-1}{P^2}\right)
\label{eq:1003658}
\end{equation}
% step ID = 8648371
\simplify{1003658}{2206759}
\begin{equation}
\kappa_T = \frac{1}{P}
\label{eq:2206759}
\end{equation}
% step ID = 1059017
\declarefinalexpr{2206759}
\bibliographystyle{plain}
\bibliography{pdg.bib}
\end{document}
% EOF
