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Calc rate problem

  1. Mar 12, 2009 #1
    hen air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV^{1.4}=C where C is a constant. Suppose that at a certain instant the volume is 630 cubic centimeters and the pressure is 97 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?

    (Pa stands for Pascal -- it is equivalent to one Newton/(meter squared); kPa is a kiloPascal or 1000 Pascals. )

    I think i start by doing (97)(630)^1.4 = 805097.5471 what is my next step? How do i solve this problem
  2. jcsd
  3. Mar 12, 2009 #2


    Staff: Mentor

    You can't start by putting numbers in--P and V are functions of time. Find a relationship between the rates of change (i.e., time derivatives) of P and V, and then solve for dV/dt.

    Then you can put your numbers in.
  4. Mar 12, 2009 #3
    I don't get it. So i differentiate pv^1.4 ?

    so the answer would be 1.4pv' + v^1.4 ... would that be correct?
  5. Mar 12, 2009 #4


    Staff: Mentor

    No. Both p and v are functions of t, so you need to differentiate with respect to t. I would recommend using d/dt (Leibniz) notation rather than ' (Newton) notation.

    Start with your equation, PV1.4 = C, and differentiate both sides with respect to t. You should get another equation with P, dP/dt, V, and dV/dt. Solve that equation for dV/dt. Then you can find the value of dV/dt at the particular time in question.
  6. Mar 12, 2009 #5
    Is this correct ???

    dP/dt = -[P 1.4*V^.4 dV/dt]/[V^1.4]
  7. Mar 12, 2009 #6
    That can be simplified but, yes, I think it is correct. I have trouble reading the formula like that so...

    [tex]\frac{dP}{dt} = \frac{-1.4P\frac{dV}{dt}}{V}[/tex]

    Now that you have an equation that relates the rates, the problem is figuring out what these values are from the problem. You solved for dP/dt and I'm not sure why, but it stills works. The problem asks for the rate at which volume is changing, and it gives you a volume, pressure, and a rate at which pressure is changing. Your new equation has 4 variables, Pressure, Volume, the rate at which Pressure changes, and the rate at which Volume changes. Plug in your known variables and solve the problem.
  8. Mar 13, 2009 #7
    Thank you i got the answer . :smile:
  9. Mar 13, 2009 #8
    The answer is 32.47422681
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