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calc problem.?
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. )
it's to the 1.4th power not 1/4th
2 Answers
- James LLv 51 decade agoFavorite Answer
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.
- dutch_profLv 41 decade ago
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
