When air expands adiabatically (without gaining or losing heat), its pressure and volume are related by the equation PV^1,4=C where C is a constant. Suppose that a a certain instant the volume is 520 cubic centimeters and the pressure is 99kPa and is decreasing at a rate of 11 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

PV^1.4=C

V^1.4 dp/dt+ 1.4pV^.4 dv/dt=0
So you are given, V, P, dp/dt, solve for dv/dt

noob

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To find the rate at which the volume is increasing, we can differentiate the given equation with respect to time.

Given:
PV^1.4 = C
where P is the pressure, V is the volume, and C is a constant.

Differentiating both sides of the equation with respect to time (t), we get:
(P)(1.4V^0.4)(dV/dt) = 0
Simplifying this equation, we have:
1.4PV^0.4(dV/dt) = 0
(dV/dt) = 0 / (1.4PV^0.4)
(dV/dt) = 0

Since the rate of pressure change (dP/dt) is given as -11 kPa/minute, we substitute the given values into the equation to find the rate of volume change (dV/dt) at that instant.

(dV/dt) = 0
(dV/dt) = 0 / (1.4 × 99 × (520)^0.4) [Substituting the given values]
(dV/dt) = 0

Therefore, at that instant, the rate at which the volume is increasing is 0 cubic centimeters per minute.