Calculus Derivatie app.'s question (Boyle's Law)?

Question

One statement of Boyle's Law is that the pressure of a gas varies inversely as the volume for constant temperature. If a certain gas occupies 650cm^3 when the pressure is 230 kPa and the volume is increasing at the rate of 20.0 cm^3/min., how fast is the pressure changing when the volume is 810 cm^3?

Answer 1

boyle's law -- pV=const = k

So p =k/V

Then dp/dt = kd(1/V)/dt = -k/V^2 * dV/dt

k = p0*V0 = 230 kPa 650cm^5

so dp/dt = -(230kPa)*(650cm^3)/(810 cm^3)^2 *20 cm^3/min

dp/dt = -4.56 kPa/min -note it is decreasing as you would expect

Answer 2

If P and V are the pressure and volume, then here Boyle's Law says that PV = k for some constant k. We're told the volume is 650cm^3 when the pressure is 230 kPa, so k is the product of these. k = 149,500.

We're asked to find how fast the pressure is changing, or dP/ dt (where t is the time in minues). Since P = k/V, taking the derivative gives -kV^(-2) * (dV/dt).

When the volume is increasing at a rate of 20 cm^3/min, this means dV/dt = 20. So when V=810, dP/dt is
dP/dt = -kV^(-2) * (dV/dt)
dP/dt = -149,500(810^-2) * (20)
(you can calculate this out)