Answer
(a) $\frac{dV}{dP} = \frac{-C}{P^2}$
(b) At the beginning, the volume is decreasing more rapidly.
Near the end of the 10 minutes, the volume is decreasing less rapidly.
(c) $\beta = \frac{1}{P}$
Work Step by Step
(a) $PV = C$
$V = \frac{C}{P}$
$\frac{dV}{dP} = \frac{-C}{P^2}$
(b) At the beginning, the pressure is low. Then the magnitude of $\frac{dV}{dP}$ is greater since $P^2$ is in the denominator. That is, the volume is decreasing more rapidly.
Near the end of the 10 minutes, the pressure is greater, so the magnitude of $\frac{dV}{dP}$ is less. That is, the volume is decreasing less rapidly.
(c) $\beta = -\frac{1}{V}~\frac{dV}{dP}$
$\beta = -\frac{1}{V}~\frac{-C}{P^2}$
$\beta = \frac{C}{P^2~V}$
$\beta = \frac{C}{(P~V)(P)}$
$\beta = \frac{C}{(C)(P)}$
$\beta = \frac{1}{P}$
