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\begin{document}

\frame{\frametitle{\b 1st Law of Thermodynamics - 1} \medskip

{\bf Monatomic Gas Isobar (1):} 1.~The number of moles follows from 
$P\,V_i=n\,R\,T_i$, where $P$ and the initial values $V_i$, $T_i$ are 
given, and $R=8.31\,J/(K\cdot mol)$ is the gas constant.

2.~Final temperature from $P\,V_f=n\,R\,T_f$, where $P$ and $V_f$
   are given.

3.~Work $W=P\,(V_f-V_i)$.
\bigskip

{\bf $P\,V$ Diagram Work (2):} 1.~Work on path A: $W=P_1\,(V_2-V_1)$.
\smallskip

2.~Work on path B: $W=\int_{V_1}^{V_2} P(V)\,dV$, where we have to
determine $P(V)$ first, which is a straight line. From the figure the 
slope $a$ of $P(V)$ is seen to be
$$ a=\frac{P_2-P_1}{V_2-V_1}~~\Rightarrow~~P(V)=P_1+a\,(V-V_1) $$
and the integration can be performed: $W=\int_{V_1}^{V_2}P(V)\,dV=
(V_2-V_1)\,P_1+a\,(V_2^2-V_1^2)/2-(P_2-P_1)\,V_1$.
\smallskip

3.~Work on path C: $W=P_2\,(V_2-V_1)$.

\vskip 20pt }


\frame{\frametitle{\b 1st Law of Thermodynamics - 2} \medskip

{\bf Isothermal Expansion Work (4):} $P\,V=n\,R\,T$ with $T$ constant
implies
$$ W = \int_{V_1}^{V_2}P(V)\,dV = n\,R\,T\int_{V_1}^{V_2}\frac{dV}{V} 
     = n\,R\,T\,\ln\left(\frac{V_2}{V_1}\right)\,. $$
Note: $n\,R\,T=P_1\,V_1$ and one liter$=(0.1)^3\,m^3$.
\bigskip

{\bf Thermodynamic system (5):} $U$, $W$, $Q$. Find the sign of
$\triangle U$ from $U=c_V\,T$ and $P\,V=n\,R\,T$, from $\triangle 
W=\int_{V_1}^{V_2} P(V)\,dV$ the sign of $\triangle W$. Then, if 
the sum of $\triangle U$ and $\triangle W$ is positive (negative) 
$\triangle Q$ is positive (negative) because of energy conservation.
\bigskip

\vskip 20pt }
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