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 900 cm3, the pressure is 100 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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Ximena 2 hours 2021-09-15T20:17:40+00:00 1 Answer 0

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    2021-09-15T20:19:09+00:00

    Answer:

    The value of rate of decrease of volume = – 3600 \frac{k pa}{min}

    Step-by-step explanation:

    According to Boyle’s law P V = C ——- (1)

    Pressure ( P ) = 100 k pa = 10 \frac{N}{cm^{2} }

    Volume ( V ) = 900 cm^{3}

    Put these values in equation ( 1 ) we get,

    ⇒ C = 10 × 900 = 9000 N-cm = 90 N-m

    Differentiate Equation ( 1 ) with respect to time we get,

    ⇒ V \frac{dP}{dt} + P \frac{dV}{dt} = 0

    ⇒ V \frac{dP}{dt} = – P \frac{dV}{dt}

    \frac{dV}{dt} = – \frac{V}{P} \frac{dP}{dt} ———- (2)

    This equation gives the rate of decrease of volume.

    Given that Rate of increase of pressure = \frac{dP}{dt} = 40 \frac{k pa}{min}

    C = 90 N-m

    P =  10^{5} pa = 10 \frac{N}{cm^{2} }

    V = 900 cm^{3}

    Put all the above values in equation 2 we get,

    ⇒  \frac{dV}{dt} = – \frac{900}{10} × 40

    ⇒  \frac{dV}{dt} = – 3600 \frac{k pa}{min}

    This is the value of rate of decrease of volume.

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