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 520 cubic centimeters and the pressure is 75 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?

Wouldn't V^1.4 dP/dt+ 1.4V^.4 P dV/dt=0?

solve for dV/dt

To find the rate at which the volume is increasing at a certain instant, we need to use derivatives. Let's define the variables and their rates of change:

V = volume (in cubic centimeters)
P = pressure (in kPa)
t = time (in minutes)

Given:
V = 520 cubic centimeters (volume at a certain instant)
P = 75 kPa (pressure at a certain instant)
dP/dt = -12 kPa/minute (rate at which the pressure is decreasing)

The given relationship between pressure and volume is PV^{1.4} = C, where C is a constant.

To find the rate at which the volume is increasing, we need to find dV/dt at the given instant.

To do this, we will differentiate the equation PV^{1.4} = C with respect to time (t), using the chain rule:

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

Now, let's differentiate each term separately:

For P, we have dP/dt.

For V^{1.4}, we have ((V^{1.4})' * dV/dt), where (V^{1.4})' is the derivative of V^{1.4}.

For C, since it is a constant, dC/dt will be zero.

Differentiating V^{1.4}, we get:

(V^{1.4})' = 1.4V^{0.4} * dV/dt

Now, substituting all the values and derivatives into the equation:

d(PV^{1.4})/dt = dC/dt
d(PV^{1.4})/dt = 0
P * ((V^{1.4})' * dV/dt) + V^{1.4} * dP/dt = 0

Substituting the given values:

75 * (1.4 * (520^{0.4})) * dV/dt + (520^{1.4}) * (-12) = 0

Now, solve for dV/dt:

75 * (1.4 * (520^{0.4})) * dV/dt = (520^{1.4}) * (-12)
dV/dt = ((520^{1.4}) * (-12))/(75 * (1.4 * (520^{0.4})))

Evaluate this expression using a calculator to find the rate at which the volume is increasing at that instant.