Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV =

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Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant?

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  1. Answer:

    The volume is decreasing at 160 cm³/min

    Explanation:

    Given;

    Boyle’s law,  PV = C

    where;

    P is pressure of the gas

    V is volume of the gas

    C is constant

    Differentiate this equation using product rule:

    [tex]V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}[/tex]

    Given;

    [tex]\frac{dP}{dt}[/tex] (increasing pressure rate of the gas) = 40 kPa/min

    V (volume of the gas) =  600 cm³

    P (pressure of the gas) = 150 kPa

    Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( [tex]\frac{dv}{dt}[/tex]);

    (600 x 40) + (150 x [tex]\frac{dv}{dt}[/tex]) = 0

    [tex]\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min[/tex]

    Therefore, the volume is decreasing at 160 cm³/min

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