How would I solve this problem?
The adiabatic law for expansion of air is $P(V)^{1.4}=C$ when P is pressure V is volume and C is a certain constant.At a given instant the volume is 30 cubic feet and the pressure is 60 psi. At what rate is the pressure changing if the volume is decreasing at a rate of 2 cubic feet per second?
I know that $\frac{dv}{dt}=-2$
$P(1.4)\frac{dv}{dt}+V^{1.4}\frac{dp}{dt}=0$
$60(1.4)(-2)+30^{1.4}\frac{dp}{dt}=0$
$-168+116.9417\frac{dp}{dt}$
$168=116.9417\frac{dp}{dt}$
$\frac{dp}{dt}=1.4366$ but would this be right.
Best Answer
This might help make the answer more sensible: Given $\;PV^{1.04} = C\;$, and differentiating gives us: $$P(1.4)V^{0.4}\frac{dv}{dt}+V^{1.4}\frac{dp}{dt} = 0$$
Now proceed as you did, but keep the equality sign as you substitute values and such: $V = 30, \;\;P = 60,\;\;\frac{dv}{dt} = -2$ to solve for $\;\dfrac{dp}{dt}$.