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 580 cubic centimeters, and the pressure is 75 kPa and is decreasing at a rate of 9 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
To find the rate at which the volume is increasing, we need to differentiate the given equation with respect to time. This will give us an expression relating the rates of change of pressure and volume.
Differentiating both sides of the equation PV^1.4 = C with respect to time, we get:
d/dt (PV^1.4) = d/dt (C)
Using the product rule for differentiation, the left side of the equation becomes:
(V^1.4) * dP/dt + P * d/dt (V^1.4) = 0
Since the problem states that the expansion is adiabatic and there is no heat transfer, dP/dt = -9 kPa/min (negative because the pressure is decreasing), we can substitute this value into the equation:
(V^1.4) * (-9 kPa/min) + P * d/dt (V^1.4) = 0
Next, we need to find the rate at which the volume is increasing, dV/dt. To do this, we rearrange the equation to solve for d/dt (V^1.4):
d/dt (V^1.4) = (9 kPa/min) * (V^1.4)/P
Notice that the term (V^1.4)/P appears on both sides of the equation, so we can substitute the known values into the equation:
dV/dt = (9 kPa/min) * (580 cm^3)^1.4 / (75 kPa)
Calculating the expression:
dV/dt = (9 kPa/min) * (580^1.4 cm^3) / (75 kPa)
Finally, we calculate the value:
dV/dt ≈ 68.79 cm^3/min
Therefore, the rate at which the volume is increasing at this instant is approximately 68.79 cubic centimeters per minute.