When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 690 cubic centimeters and the pressure is 97 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?
pv^1.4 = c
using the product rule,
dp/dt * v^1.4 + 1.4pv^0.4 dv/dt = 0
No wjust plug in your numbers and solve for dv/dt
To find the rate at which the volume is increasing at a specific instant, we need to differentiate the given equation with respect to time. The equation is PV^1.4 = C.
Differentiating both sides of the equation with respect to time (t), we get:
(1.4 * P * V^0.4 * dV/dt) + (V^1.4 * dP/dt) = 0
We are given the following information:
- Volume (V) = 690 cc
- Pressure (P) = 97 kPa
- Pressure rate of change (dP/dt) = -9 kPa/minute
- We need to find the rate of change of volume (dV/dt) in cc/minute.
Substituting the given values into the equation:
(1.4 * 97 * (690)^0.4 * dV/dt) + (690^1.4 * -9) = 0
Now we can solve for dV/dt:
(1.4 * 97 * (690)^0.4 * dV/dt) = - (690^1.4 * -9)
Simplifying the equation further:
dV/dt = - (690^1.4 * -9) / (1.4 * 97 * (690)^0.4)
Calculating the expression on the right side of the equation will give us the rate at which the volume is increasing at the specified instant.