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 330 cubic centimeters and the pressure is 79 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?

Question

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 330 cubic centimeters and the pressure is 79 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?

Answer 1

you have your formula:

pv^1.4 = c, so
you get

v^1.4 dp/dt + 1.4pv^.4 dv/dt = 0
Now just plug in your numbers:

(330^1.4)(-10) + (1.4)(330^.4)(79) dv/dt = 0
-33567.54 + 1125.02 dv/dt = 0
dv/dt = +29.84

Answer 2

To find the rate at which the volume is increasing, we need to differentiate the equation PV^(1.4) = C implicitly with respect to time, t, and then solve for dV/dt.

Differentiating both sides of the equation with respect to t using the product rule, we get:

P * (1.4 * V^(0.4) * dV/dt) + V^(1.4) * dP/dt = 0

Since the problem states that it is an adiabatic process, dP/dt is given as -10 kPa/minute (decreasing at a rate of 10 kPa/minute).

Now, substitute the given values: P = 79 kPa and dP/dt = -10 kPa/minute into the equation:

79 * (1.4 * 330^(0.4) * dV/dt) + 330^(1.4) * (-10) = 0

From here, we can solve for dV/dt, the rate at which the volume is increasing:

dV/dt = [-330^(1.4) * (-10)] / [79 * (1.4 * 330^(0.4))]

Simplifying further, we get:

dV/dt ≈ 0.108 cubic centimeters/minute

Therefore, the volume is increasing at a rate of approximately 0.108 cubic centimeters per minute at this instant.

Answer 3

To find the rate at which the volume is increasing at the given instant, we can differentiate the equation PV^1.4 = C implicitly with respect to time. Using the product rule of differentiation, we get:

d(PV^1.4)/dt = dC/dt

To solve for dV/dt, which represents the rate at which the volume is increasing, we need to isolate it in the equation. Rearranging the equation, we get:

P * dV/dt + 1.4V^(0.4) * dV/dt = 0

Factoring out dV/dt, we have:

dV/dt (P + 1.4V^(0.4)) = 0

Since we are looking for the rate at which the volume is increasing, dV/dt should not be zero. Thus, we can ignore the factor (P + 1.4V^(0.4)) in the equation, and we are left with:

dV/dt = 0

So, at the given instant, the rate at which the volume is increasing is 0 cubic centimeters per minute.