Asked by claudia

When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^{1.4}=C where C is a constant. Suppose that at a certain instant the volume is 340 cubic centimeters and the pressure is 87 kPa and is decreasing at a rate of 14 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Answered by bobpursley
take the differential..

PV^1.4=0
1.4 P v^.4 dv/dt + V^1.4 dp/dt=0

Solve for dv/dt.
Answered by bobpursley
oopss. STart off with PV^1.4=C
typo.
Answered by claudia
what do i do with the 14kpa/min?
Answered by Collin
Plug in dp/dt =-14 kpa/min because it is decreasing at a rate. So solving for dv/dt= (1-(V^(1/4)dP/dt))/((1/4)PV^(-3/4)). Your answer so be
dv/dt= (1-((340)^(1/4)(-14))/((1/4)(87)(340)^(-3/4)).