# Another Related Rates problem

• Nov 24th 2006, 03:01 PM
Jacobpm64
Another Related Rates problem
A certain quantity of gas occupies a volume of 20cm^3 at a pressure of 1 atmosphere. The gas expans wihtout the addition of heat, ,so, for some constant k, its pressure, P, and volume, V, satisfy the relation

$PV^{1.4} = k$

(a) Find the rate of change of pressure with volume. Give units.
(b) The volume is increasing at 2 cm^3/min when the volume is 30 cm^3. At that moment, is the pressure increasing or decreasing? How fast? Give units.

The thing that confuses me about this problem is that it says the gas occupies a volume of 20cm^3 at a pressure of 1 atmosphere. I didn't use this information when I worked the problem. Could this information not be needed, or did I screw up somewhere?

Anyways, here is my work:

(a)
$PV^{1.4} = k$
$P = \frac{k}{V^{1.4}} = kV^{-1.4}$
$\frac{dP}{dV} = -1.4kV^{-2.4}$
$\frac{dP}{dV} = ( \frac{-1.4k}{V^{2.4}} )$ atm/cm^3

(b)
$P = \frac{k}{V^{1.4}} = kV^{-1.4}$
$\frac{dP}{dt} = -1.4kV^{-2.4} * \frac{dV}{dt}$
$\frac{dP}{dt} = \frac{-1.4k}{V^{2.4}} * \frac{dV}{dt}$
$\frac{dP}{dt} = \frac{-1.4k}{30^{2.4}} * 2$
$\frac{dP}{dt} \approx -0.000798$ atm/min

decreasing at approximately 0.000798 atm/min
• Nov 26th 2006, 04:10 PM
topsquark
Quote:

Originally Posted by Jacobpm64
A certain quantity of gas occupies a volume of 20cm^3 at a pressure of 1 atmosphere. The gas expans wihtout the addition of heat, ,so, for some constant k, its pressure, P, and volume, V, satisfy the relation

$PV^{1.4} = k$

(a) Find the rate of change of pressure with volume. Give units.
(b) The volume is increasing at 2 cm^3/min when the volume is 30 cm^3. At that moment, is the pressure increasing or decreasing? How fast? Give units.

The thing that confuses me about this problem is that it says the gas occupies a volume of 20cm^3 at a pressure of 1 atmosphere. I didn't use this information when I worked the problem. Could this information not be needed, or did I screw up somewhere?

Anyways, here is my work:

(a)
$PV^{1.4} = k$
$P = \frac{k}{V^{1.4}} = kV^{-1.4}$
$\frac{dP}{dV} = -1.4kV^{-2.4}$
$\frac{dP}{dV} = ( \frac{-1.4k}{V^{2.4}} )$ atm/cm^3

(b)
$P = \frac{k}{V^{1.4}} = kV^{-1.4}$
$\frac{dP}{dt} = -1.4kV^{-2.4} * \frac{dV}{dt}$
$\frac{dP}{dt} = \frac{-1.4k}{V^{2.4}} * \frac{dV}{dt}$
$\frac{dP}{dt} = \frac{-1.4k}{30^{2.4}} * 2$
$\frac{dP}{dt} \approx -0.000798$ atm/min

decreasing at approximately 0.000798 atm/min

I didn't check the numbers, but the work is correct. :)

-Dan