Answer
(a)
$\frac{dV}{dP}$ = $-\frac{C}{P^{2}}$
(b)
From the formula for $\frac{dV}{dP}$ in part (a), we see that as $P$ increases, the absolute value of $\frac{dV}{dP}$ decreases.
Thus, the volume is decreasing more rapidly at the beginning.
(c)
$β$ = $-\frac{1}{V}{\frac{dV}{dP}}$ = $-\frac{1}{V}(-\frac{C}{P^{2}})$ = $\frac{C}{(PV)P}$ = $\frac{C}{CP}$ = $\frac{1}{P}$
Work Step by Step
(a)
$PV$ = $C$
$V$ = $\frac{C}{P}$
$\frac{dV}{dP}$ = $-\frac{C}{P^{2}}$
(b)
From the formula for $\frac{dV}{dP}$ in part (a), we see that as $P$ increases, the absolute value of $\frac{dV}{dP}$ decreases.
Thus, the volume is decreasing more rapidly at the beginning.
(c)
$β$ = $-\frac{1}{V}{\frac{dV}{dP}}$ = $-\frac{1}{V}(-\frac{C}{P^{2}})$ = $\frac{C}{(PV)P}$ = $\frac{C}{CP}$ = $\frac{1}{P}$
