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 330 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 12 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
not sure of your equation,
is it (PV)^1.4 = C
or
P(V^(1.4)) = C
or
1.4PV = C , (usually we don't write a multiplier after the variable like in yours)
Just noticed that all of the 10 "Related Questions" below are your problem.
Steve assumed P(V^(1/4) = C
http://www.jiskha.com/display.cgi?id=1414421171
To find the rate at which the volume is increasing at a given instant, we need to calculate the derivative of the volume with respect to time. In this case, we are given the rate of change of pressure with respect to time, so we need to use the chain rule to relate the two.
Let's start by differentiating the given equation PV^1.4 = C implicitly with respect to time (t):
d/dt (PV^1.4) = d/dt (C)
Using the product rule of differentiation, the derivative of PV^1.4 with respect to t can be expressed as:
V^1.4 * dP/dt + P * (1.4V^0.4 * dV/dt) = 0
Now, we can substitute the given values into the equation. At the given instant, the volume V is 330 cubic centimeters and the pressure P is 99 kPa, with a rate of change of pressure dP/dt of -12 kPa/minute.
V = 330 cm^3
P = 99 kPa
dP/dt = -12 kPa/min
Plugging these values into the equation, we get:
(330^1.4) * (-12) + (99) * (1.4 * 330^0.4 * dV/dt) = 0
Simplifying the equation:
1980 * (-12) + 1386 * dV/dt = 0
Now, we can solve for dV/dt, the rate at which the volume is increasing:
1386 * dV/dt = 1980 * 12
dV/dt = (1980 * 12) / 1386
Therefore, the rate at which the volume is increasing at this instant is (1980 * 12) / 1386 cubic centimeters per minute.