k = P V^1.4
k = 40 * 14^1.4
k = 1609
so
P = 1609 V^-1.4
dP/dt = 1609 (-1.4)(V^-2.4) dV/dt
= -2253 (14)^-2.4 * -1
= 4 psi/s
Air expands adiabatically in accordance with the law PV^1.4=Const. If at a given time the volume is 14 cubic feet and the pressure is 40 pounds per square inch, at what rate is the pressure changing when the volume is decreasing 1 cubic foot per second?
1 answer