Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 21

$
\begin{equation}
\begin{aligned}
\text{a.) } PV &= C\\
\\
V &= \frac{C}{P}\\
\\
\frac{dV}{dP} &= \frac{P \frac{d}{dP}(C) - (C) \cdot \frac{d}{dP} (P) }{P^2}\\
\\
\frac{dV}{dP} &= \frac{P(0) - (C) (1)}{P^2}\\
\\
\frac{dV}{dP} &= \frac{-C}{P^2}
\end{aligned}
\end{equation}
$


b.) When a gas is steadily compressed at constant temperature, the volume and the pressure inversely proportional to each other(as what Boyle's Law stated). From part(a), we know that the change in volume is proportional to $\displaystyle \frac{1}{P^2}$. At the end of 10 minutes, the pressure of the gas must be higher, making $\displaystyle \frac{dV}{dP} = \frac{1}{P^2}$ lower. Therefore, we can conclude that the volume is decreasing more rapidly at the beginning.

c.) Isothermal compressibility is denoted as $\displaystyle \beta = \frac{-1}{V} \frac{dV}{dP}$; but $\displaystyle \frac{dV}{dP} = \frac{-C}{P^2}$ and $\displaystyle V = \frac{C}{P}$ so,


$
\begin{equation}
\begin{aligned}
\beta &= \frac{-1}{\frac{C}{P}} \left( \frac{-C}{P^2}\right)\\
\\
\beta &= -\frac{\cancel{P}}{\cancel{C}} \left( \frac{-\cancel{C}}{P^{\cancel{2}}} \right)\\
\\
\beta &= \frac{1}{P}
\end{aligned}
\end{equation}
$