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20 March, 10:45

A nitric acid solution flows at a constant rate of 5L/min into a large tank that initially held 200L of a 0.5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 10L/min. If the solution entering the tank is 10% nitric acid, determine the volume of nitric acid in the tank after t minutes.

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  1. 20 March, 11:31
    0
    x (t) = - 39e

    -0.03t + 40.

    Explanation:

    Let V (t) be the volume of solution (water and

    nitric acid) measured in liters after t minutes. Let x (t) be the volume of nitric acid

    measured in liters after t minutes, and let c (t) be the concentration (by volume) of

    nitric acid in solution after t minutes.

    The volume of solution V (t) doesn't change over time since the inflow and outflow

    of solution is equal. Thus V = 200 L. The concentration of nitric acid c (t) is

    c (t) = x (t)

    V (t)

    =

    x (t)

    200

    .

    We model this problem as

    dx

    dt = I (t) - O (t),

    where I (t) is the input rate of nitric acid and O (t) is the output rate of nitric acid,

    both measured in liters of nitric acid per minute. The input rate is

    I (t) = 6 Lsol.

    1 min

    ·

    20 Lnit.

    100 Lsol.

    =

    120 Lnit.

    100 min

    = 1.2 Lnit./min.

    The output rate is

    O (t) = (6 Lsol./min) c (t) = 6 Lsol.

    1 min

    ·

    x (t) Lnit.

    200 Lsol.

    =

    3x (t) Lnit.

    100 min

    = 0.03 x (t) Lnit./min.

    The equation is then

    dx

    dt = 1.2 - 0.03x,

    or

    dx

    dt + 0.03x = 1.2, (1)

    which is a linear equation. The initial condition condition is found in the following

    way:

    c (0) = 0.5% = 5 Lnit.

    1000 Lsol.

    =

    x (0) Lnit.

    200 Lsol.

    .

    Thus x (0) = 1.

    In Eq. (1) we let P (t) = 0.03 and Q (t) = 1.2. The integrating factor for Eq. (1) is

    µ (t) = exp Z

    P (t) dt

    = exp

    0.03 Z

    dt

    = e

    0.03t

    .

    The solution is

    x (t) = 1

    µ (t)

    Z

    µ (t) Q (t) dt + C

    = Ce-0.03t + 1.2e

    -0.03t

    Z

    e

    0.03t

    dt

    = Ce-0.03t +

    1.2

    0.03

    e

    -0.03t

    e

    0.03t

    = Ce-0.03t +

    1.2

    0.03

    = Ce-0.03t + 40.

    The constant is found using x (t) = 1:

    x (0) = Ce-0.03 (0) + 40 = C + 40 = 1.

    Thus C = - 39, and the solution is

    x (t) = - 39e

    -0.03t + 40.
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