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8 June, 23:23

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at the constant conditions of 45∘C and 1500 kPa. A valve in the line is cracked so that air flows slowly into the tank until the pressure equals the line pressure. If the process occurs slowly enough that the temperature in the tank remains at 25∘C, how much heat is lost from the tank? Assume air to be an ideal gas for which CP = (7/2) R and CV = (5/2) R.

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  1. 9 June, 02:24
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    Amount of Energy = 23,467.9278J

    Explanation:

    Given

    Cv = 5/2R

    Cp = 7/2R wjere R = Boltzmann constant = 8.314

    The energy balance in the tank is given as

    ∆U = Q + W

    According to the first law of thermodynamics

    In the question, it can be observed that the volume of the reactor is unaltered

    So, dV = W = 0.

    The Internal energy to keep the tank's constant temperature is given as

    ∆U = Cv ((45°C) - (25°C))

    ∆U = Cv ((45 + 273) - (25 + 273))

    ∆U = Cv (20)

    ∆U = 5/2 * 8.314 * 20

    ∆U = 415.7 J/mol

    Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

    The Initial mole is calculated as

    (P * V) / (R * T)

    Where P = P1 = 101.33kPa = 101330Pa

    V = Volume of Tank = 0.1m³

    R = 8.314J/molK

    T = Initial Temperature = 25 + 273 = 298K

    So, n = (101330 * 0.1) / (8.314*298)

    n = 4.089891232222

    n = 4.089

    Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

    V = Volume of Tank = 0.1m³

    R = 8.314J/molK

    T = Initial Temperature = 25 + 273 = 298K

    n = (1500000 * 0.1) / (8.314*298)

    n = 60.54314465936812

    n = 60.543

    So, tue moles that entered the tank is ∆n

    ∆n = 60.543 - 4.089

    ∆n = 56.454

    Amount of Energy is then calculated as: (∆n) (U)

    Q = 415.7 * 56.454

    Q = 23,467.9278J
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