Boyles Law?

According to Boyle's Law, when a sample of gas is compressed at a constant temperature, the pressure and volumn satisfy the equation P V = C, where C is a constant. Assume that, at a certain instant, a sample has a volume of 950 cm3, is a pressure of 120 kPa, and the pressure is increasing at a rate of 19 kPa/min. At what rate is the volume decreasing at this instant?

3 Answers

Relevance
  • Como
    Lv 7
    1 decade ago
    Favorite Answer

    PV = C

    C = 950 x 120

    C = 114000

    V = C/P = CP^(-1)

    dV/dP = - C / P ²

    dV/dt = (dV/dP) (dP/dt)

    dv/dt = (- C / P ²) (19)

    dv/dt = (- 114000) (19) / (120 ²)

    dv/dt = - 150.4 cm ³ / min

    Volume is decreasing at a rate of 150.4 cm ³ / min

  • TFV
    Lv 5
    1 decade ago

    Use implicit differentiation with respect to some "time" variable t.

    (dP/dt)V + P(dV/dt) = dC/dt

    (19)(950) + (120)(dV/dt) = 0

    120(dV/dt) = -(19)(950) = -18050

    dV/dt = -18050/120 = -1805/12

    That tell us that the volume is decreasing at a rate of (-1805/12), where the units are ((cm^3)/min).

  • 1 decade ago

    PV=c

    The trick is to differentiate this with respect to time using the product rule because both P and V are functions of time.

    P dV/dt + VdP/dt = 0

    We know the values of P, V and dP/dt

    plug in those values to get dV/dt.

    the value is obviously negative because the volume is decreasing with time.

Still have questions? Get your answers by asking now.