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 570 cubic centimeters and the pressure is 91 kPa and is decreasing at a rate of 13 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at
this instant?
(Pa stands for Pascal – it is equivalent to one Newton/(meter
squared); kPa is a kiloPascal or 1000 Pascals. )
2006-10-13 08:09:10 · 2 answers · asked by M45-S355 l_l532 2 in Science & Mathematics ➔ Mathematics
it's to the 1.4th power not 1/4th
2006-10-13 08:24:43 · update #1
Note: I assume you meant 1.4 in your equation, while the previous answerer assumed 1/4. Either way, the approach is the same.
Differentiate the equation
PV^(1.4) = C
with respect to t. You get
dP/dt V^(1.4) + (1.4)PV^(0.4) dV/dt = 0
since C is constant.
Plug in V=570, P=91, and dP/dt=-13. Then you can solve for dV/dt.
2006-10-13 08:20:34 · answer #1 · answered by James L 5 · 1⤊ 1⤋
p v^(1/4) = C
differentiate using product rule and chain rule:
p' v^(1/4) + 1/4 p v^(-3/4) v' = 0
v' / v= -4 p' / p
In your case, p' / p = 13 / 91 = 1/7; v = 570; so
v' / 570 = - 4 / 7
v' = 325.7 cc/min
2006-10-13 08:19:33 · answer #2 · answered by dutch_prof 4 · 0⤊ 2⤋