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Math Help - Another Related Rates problem

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by Jacobpm64 View Post
    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
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