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 =

Question

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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Linh Đan 3 months 2021-07-19T22:36:52+00:00 1 Answers 178 views 0

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    2021-07-19T22:38:44+00:00

    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:

    V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}

    Given;

    \frac{dP}{dt} (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 ( \frac{dv}{dt});

    (600 x 40) + (150 x \frac{dv}{dt}) = 0

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

    Therefore, the volume is decreasing at 160 cm³/min

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