A rigid tank contains 0.5 kmol of Ar and 2 kmol of N2 at 250 kPa and 280 K. The mixture is now heated to 400 K. Determine the volume of the

Question

A rigid tank contains 0.5 kmol of Ar and 2 kmol of N2 at 250 kPa and 280 K. The mixture is now heated to 400 K. Determine the volume of the tank and the final pressure of the mixture.

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Thái Dương 3 years 2021-08-02T14:51:09+00:00 2 Answers 145 views 0

Answers ( )

  1. thanhcong
    0
    2021-08-02T14:52:17+00:00
    Reply

    Answer:

    The volume of the tank is 23.28 m^3.

    The final pressure of the mixture is 357.14 kPa.

    Explanation:

    Total moles of gas mixture (n) = 0.5 kmol Ar + 2 kmol N2 = 2.5 kmol

    From the ideal gas equation, V = nRT/P

    V is volume of the tank

    n is the total moles of gas mixture = 2.5 kmol

    R is gas constant = 8314.34 J/kmol.K

    T is temperature = 280 K

    P is pressure = 250 kPa = 250×1000 = 250,000 Pa

    V = 2.5×8314.34×280/250,000 = 23.28 m^3

    From Pressure law:

    P1/T1 = P2/T2

    P2 = P1T2/T1

    P1 (initial pressure) = 250 kPa

    T1 (initial temperature) = 280 K

    T2 (final temperature) = 400 K

    P2 (final pressure) = 250×400/280 = 357.14 kPa

  2. Latifah
    0
    2021-08-02T14:52:24+00:00
    Reply

    Answer:

    Explanation:

    Given:

    Moles of Argon, na = 0.5 kmol

    Moles of N2, nn = 2 kmol

    Pressure, P = 250 kPa

    = 2.5 × 10^5 Pa

    Temperature, T1 = 280 K

    Final temperature, T2 = 400 K

    Total number of moles in the system, n = na + nn

    = 2 + 0.5

    = 2.5 kmol

    = 2500 mol.

    Using ideal gas equation,

    PV = nRT

    Volume, V = (2500 × 8.314 × 280)/2.5 × 10^5

    = 23.28 m^3

    B.

    Final pressure, P2 using pressure law,

    P1/T1 = P2/T2

    P2 = (P1 × T2)/T1

    = (2.5 × 10^5 × 400)/280

    = 3.57 × 10^5 Pa

    = 357 kPa

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