Answer
$\dfrac{dP}{dt}=-1.92 \dfrac{kPa}{min}$
Work Step by Step
Firstly differentiate $PV^b=C$ with respect to time $t$ using product rule and chain rule.
We get, $\dfrac{dP}{dt}V^b+bV^{b-1}P\dfrac{dV}{dt}=0$
$\implies \dfrac{dP}{dt}V+bP\dfrac{dV}{dt}=0$
Substitute $b = 1.2$, $ P = 8 $kPa, $ V = 100 cm^3$, and $\dfrac{dV}{dt} =
20 \dfrac{cm^3}{min}$ in $\dfrac{dP}{dt}V+bP\dfrac{dV}{dt}=0$.
$\dfrac{dP}{dt}100+1.2\cdot8\cdot20=0$
$\implies\dfrac{dP}{dt}100+192=0$
$\implies \dfrac{dP}{dt}=\dfrac{-192}{100}=-1.92 \dfrac{kPa}{min}$
