Rates & related Rates.

F

ffuh205

Junior Member
Joined
Mar 17, 2010
Messages
51
Ok. So I have the question A certain amount of gas occupies the volume of 20 cm3 at a pressure of 1 atmosphere. The gas expands without the addition of heat, so, for some constant k, its pressure, P, and volume, V, satisfy the relation "
(a)Find the rate of change of pressure with volume. Give units.

This is what I have for part A
First we can solve for k. 1*201.4 = k; k = 66.29; We can then rewrite the equation as P = k/V^1.4 and plugging in k we get P = 66.29/V^1.4 of f(x) = 66.29/V^1.4; To solve the rate of change we find the derivate which gives us f’(x) = -92.8/V^2.4 Units are atmospheres per cubic centimeter;


(b) The volume is increasing at 2 cm3/min when the volume is 30 cm3. At that moment, is the pressure increasing or decreasing? How fast? Give units.

Ok. I am not sure where to start here. It looks like perhaps I should find the inverse function of P = 66.29/V^1.4 which would be V = 1.4th root of 66.29/P But I am not sure. I am even more confused because is added time.
 
Ok. So I have the question A certain amount of gas occupies the volume of 20 cm3 at a pressure of 1 atmosphere. The gas expands without the addition of heat, so, for some constant k, its pressure, P, and volume, V, satisfy the relation "
(a)Find the rate of change of pressure with volume. Give units.

This is what I have for part A
First we can solve for k. 1*201.4 = k; k = 66.29; We can then rewrite the equation as P = k/V^1.4 and plugging in k we get P = 66.29/V^1.4 of f(x) = 66.29/V^1.4; To solve the rate of change we find the derivate which gives us f’(x) = -92.8/V^2.4 Units are atmospheres per cubic centimeter;
Click to expand...

I have no idea where your numbers are coming from: 201.4, 66.29, 1.4, etc. Please explain. In the mean time, I offer this:

Ideal Gas Law:
PV = nRT
P = nRT/V

R is the Ideal Gas Constant, and n and T are held constant in this problem, so

P = k(1/V)
k = (1 atm)(20 cc) = 20 atm*cc

Taking the derivative with respect to time:

dP/dt = (-k/V^2)(dV/dt)
Units are [(atm*cc)/cc^2](cc/(unit time)) = atm/(unit time)
dP/dt = (-20/V^2)(dV/dt) atm/(unit time)

(unit time) means whatever time units you choose to work in: seconds, minutes, hours, etc.

If you choose to convert your pressure or volume to different units, your k value and units will change of course.
 
The given equation is PV^1.4 = k.

A certain quantity of gas occupies a volume of 20 cm3 at a pressure of 1 atmosphere. The gas expands without the addition of heat, so, for some constant k, its pressure, P, and volume, V, satisfy the relation PV^1.4 = k

That is a problem given in the book. I got the other numbers by solving the derivative after rearranging the equation P = k/V^1.4; Plugging in the value of 20cm^3 at 1atm to solve for k I get PV^1.4 = k; Therefore at 1 atm and 20 cm3 Volume I get 1*20^1.4k = 66.29; I then plug 66.29 back into the original equation. Then I find the derivative of P = 66.29*V^1.4
 
The given equation is PV^1.4 = k...

Then I find the derivative of P = 66.29*V^1.4
Click to expand...

Your given equation seems a bit bizarre, as it is at odds with Boyle's Law, which states that PV = k under the given conditions.

However, neglecting that, and going with PV^1.4 = k, you seem to be doing fine as far as you've gone. You must recognize that you need to take the derivative with respect to TIME. Do not write your derivative as "f'(x)"; it has nothing to do with x; it is dP/dt = (something)(dV/dt). Continue.
 
Plugging 30 cm3 into P = 66.29/V1.4 We get P = 66.29/301.4 = .56686 atmospheres. We then plug in 2 cm3/min at get 66.29/281.4 = .62435. The rate of change is therefore .0575 atmospheres per minute.
 
The rate of change is therefore .0575 atmospheres per minute.
Click to expand...

I got about .0529 atm/min. You might want to double-check that. Watch for round-off errors.
 
You must log in or register to reply here.
Top