1. ## practice work help

It's a word problem.

When air expands without gaining or losing heat, it's pressure P and volume V are related by the equation PV^1.4=C (c is constant). Instant volume is 400 cm^3 and pressure is 80kPa and is decreasing at a rate of 10kPa/min. At what rate is the volume increasing at this instant?

Should I solve for V and the take the derivative? What is the answer worked out?

Thankies, first time doing these types of problems.

2. Originally Posted by mathman66
It's a word problem.

When air expands without gaining or losing heat, it's pressure P and volume V are related by the equation PV^1.4=C (c is constant). Instant volume is 400 cm^3 and pressure is 80kPa and is decreasing at a rate of 10kPa/min. At what rate is the volume increasing at this instant?

Should I solve for V and the take the derivative? What is the answer worked out?

Thankies, first time doing these types of problems.
PV^{1.4} = C
implies

dP/dt * V^{1.4} + P * 1.4*V^{0.4}*dV/dt = 0

So:
dV/dt = - [dP/dt * V^{1.4}]/[1.4*P*V^{0.4}*dV/dt] = -V/1.4 * [dP/dt]/P

dV/dt = -(400 cm^3)/1.4 * [-10 kPa/min]/(80 kPa) = 35.714285714 cm^3/min

So the volume is increasing at a rate of about 35.7 cm^3/min.

-Dan

PS Yes, you CAN solve for V first, though I think in this case it's a bit harder:
PV^{1.4} = C

V^{1.4} = C/P

V = (C/P)^{1/1.4} = (C/P)^{5/7}

dV/dt = (5/7)(C/P)^{-2/7}*(-C/P^2)*dP/dt

Now C = PV^{1.4} = PV^{7/5}:

dV/dt = -(5/7)([PV^{7/5}]/P)^{-2/7}*([PV^{7/5}]/P^2)*dP/dt

dV/dt = -(5/7)V*(1/P)*dP/dt

which is the same as I had above.

-Dan