Answer
\begin{aligned}
\left.\frac{\partial P}{\partial T}\right|_{V=50)} &=0.2
\end{aligned}
\begin{aligned}
\left.\frac{\partial V}{\partial P}\right|_{(T=80, V=50)} &=-3.125
\end{aligned}
Work Step by Step
We are given that
\[
\frac{k T}{V}=P
\]
where $\mathrm{k}$ is constant and equal to $10 ~ l b . i n / k$
Thus, we have to find the instantaneous rate of change of pressure for temperature at $T=80$, while Volume remains constantat 50 $in^{3}$
\[
\begin{aligned}
&\frac{k T}{V}
\frac{\partial}{\partial T}=\frac{\partial P}{\partial T} \\
& \frac{\partial}{\partial T}(T) \frac{k}{V}=\frac{\partial P}{\partial T} \\
\frac{\partial P}{\partial T} &=\frac{k}{V} \\
\left.\frac{\partial P}{\partial T}\right|_{(V=50)} &=\frac{10}{50} \\
\left.\frac{\partial P}{\partial T}\right|_{(V=50)} &=0.2
\end{aligned}
\]
$b$) Similarly, we should find the instantaneous rate of change of volume with respect to pressure at $50=V$ $in^{3}$ and $80 \mathrm{\ Kelvin}$. Temperature remains constant.
\[
\begin{aligned}
&\frac{k T}{P} \frac{\partial}{\partial P}=\frac{\partial V}{\partial P} \\
&k T \frac{\partial}{\partial T}\left(\frac{1}{P}\right) =\frac{\partial V}{\partial P} \\
\frac{\partial V}{\partial P} &=-\frac{k T}{P^{2}} \\
\frac{\partial V}{\partial P} &=-\frac{k T}{\left(\frac{k T}{V}\right)^{2}} \\
\frac{\partial V}{\partial P} &=-\frac{V^{2}}{k T} \\
\left.\frac{\partial V}{\partial P}\right|_{(T=80, V=50)} &=-\frac{(50)^{2}}{10 * 80} \\
\left.\frac{\partial V}{\partial P}\right|_{(T=80, V=50)} &=-3.125
\end{aligned}
\]
