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 380 cubic centimeters and the pressure is 89 kPa and is decreasing at a rate of 10 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
PV^k=C
take the derivative
kPv^(k-1) dV/dt+V^k dP/dt=0
solve for dV/dt
dV/dt= -V/Pk * dP/dt check that.
To find the rate at which the volume is increasing at a given instant, we can use the chain rule of differentiation. Let's start by differentiating the given equation with respect to time (t).
Differentiating both sides of the equation PV^1.4 = C with respect to time, we get:
d(PV^1.4)/dt = 0
Using the chain rule, the left side of the equation can be expanded as follows:
dP/dt * V^1.4 + P * d(V^1.4)/dt = 0
Now we need to find d(V^1.4)/dt, the rate at which the volume is changing with respect to time. To do this, we differentiate V^1.4 separately:
d(V^1.4)/dt = (1.4) * V^(1.4 - 1) * dV/dt
= 1.4 * V^0.4 * dV/dt
Now we can substitute this expression back into the equation:
dP/dt * V^1.4 + P * (1.4 * V^0.4 * dV/dt) = 0
Given information: V = 380 cm^3, P = 89 kPa (which is equal to 89,000 Pa), dP/dt = -10 kPa/min (since the pressure is decreasing)
Substituting these values into the equation:
-10 * (380^1.4) + (89,000) * (1.4 * (380^0.4) * dV/dt) = 0
Simplifying further:
14 * (380^0.4) * dV/dt = 10 * 380^1.4
Dividing both sides by 14 * (380^0.4):
dV/dt = (10 * 380^1.4) / (14 * 380^0.4)
Now we can calculate the value of dV/dt:
dV/dt ≈ 8.579 cm^3/min
Therefore, the volume is increasing at a rate of approximately 8.579 cubic centimeters per minute at that instant.