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Calc AB Related Rates Problem

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    A container of constant volume contains a quantity of gas under pressure. At t=0, the pressure is 4 psi and the temperature is 15°C per minute. What is the rate of change of the pressure inside the container at time t=0?

    2. Relevant equations

    [itex]\frac{P}{T}[/itex] = constant

    3. The attempt at a solution

    Taking the derivative of both sides and then simplifying, I ended up with:

    T[itex]\frac{dP}{dT}[/itex] - P[itex]\frac{dT}{dT}[/itex] = 0

    So solving for dP/dt:

    [itex]\frac{dP}{dt}[/itex] = [itex]\frac{P}{T}[/itex][itex]\frac{dT}{dt}[/itex]

    This is where I get stuck. I know P = 4 psi and dT/dt = 15°C per minute, but I can't figure out how to determine T. Without knowing the actual value of the constant in the original formula, I don't see how it is possible. The wording of the temperature value in the problem sounds odd to me. The only thing I can think of is that the temperature should be assumed to be 0°C at t=0 and so dP/dt does not exist. Is this it, or am I missing some way to find T?
     
  2. jcsd
  3. Feb 7, 2012 #2

    Mark44

    Staff: Mentor

    15°C per minute is not a temperature. It's the time rate of change of temperature. What is the exact wording of this problem?
     
  4. Feb 7, 2012 #3
    That is the exact wording of the problem. That's why I said the wording was odd.
     
  5. Feb 7, 2012 #4

    Mark44

    Staff: Mentor

    Whoever wrote the problem made a mistake. I would get some clarification from the instructor as to what the real problem is.
     
  6. Feb 7, 2012 #5
    That's what I expected, thank you!
     
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