Question

Help please!

Boyle's law for confined gases states that if the temperature is constant, pv=c where p is pressure,V is volume, and c is a constant. At a certain instant the volume is 75 cubic inches, the pressure is 3 psi, and the pressure is decreasing at the rate of 2 psi every minute. What is the rate of change of the volume at that instant?

Answer 1

This is a calculus my math professor gave the class and I never could get the answer. He worked it out but it made absolutely no sense...Can some one show me how to get the answer?

Answer 2

Oh rate of change :) these are fun.

Answer 3

Do you think you can solve it? Because Im clueless at this point :P

Answer 4

Yah just a sec though, I don't wanna interrupt abbot if he's got this :3

Answer 5

Lol okie dokie :D

Answer 6

It might make it easier to write down all your known's first:

KNOWN:
V = 75 cubic inches
P = 3
\[\frac{ dP }{ dt } = -2\]
negative because it's decreasing.

UNKNOWN:
\[\frac{ dV }{ dt }=?\]

Solve for the above by differentiating w/ respect to time. does that help?

Answer 7

[Note: PV should be differentiated by using the product rule! Also, C = a constant, so it becomes 0]

Answer 8

So basically the prime of P times the second plus the first times the prime of the second?

Answer 9

\[P (\frac{ dV }{ dt })+v(\frac{ dP }{ dt })\]

Answer 10

So 3(dV/dt) + 75(-2)

Answer 11

Then I just do a little algebra and get dV/dt? correct?

Answer 12

Yep. Now, use algebra to rearrange and solve for dV/dt.

Answer 13

So dV/dt equals 50?

Answer 14

Is that the final answer or is there more to this problem?

Answer 15

So dV/dt equals 50?

For these kinds of "physics" problems I would include the units

\[ \frac{dV}{dt}= +50 \text{ in}^3/\text{min} \]
the balloon is getting bigger by that amount every minute, but this rate will change as the volume gets bigger, the pressure goes lower, and the leak slows down.