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23 August, 21:33

A welder using a tank of volume 7.20 x 10^-2 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol) at a gauge pressure of 3.20 x 10^5 Pa and temperature of 39.0°C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 20.9 °C, the gauge pressure of the oxygen in the tank is 1.85 x 10^5 Pa.

Part A) Find the initial mass of oxygen.

Part B) Find the mass of oxygen that has leaked out.

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  1. 23 August, 21:42
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    a) the initial mass of O₂ is 373.92 gr

    b) the mass leaked of O₂ is 104.26 gr

    Explanation:

    we can assume ideal gas behaviour of oxygen, then we can calculate the mass using the ideal gas equation

    P*V = n*R*T,

    where P = absolute pressure, V = volume occupied by the gas, n = number of moles of gas, R = ideal gas constant = 8.314 J/mol K, T = absolute temperature

    Initially

    P = Pg + Pa (1 atm) = 3.20 * 10⁵ Pa + 101325 Pa = 4.21*10⁵ Pa

    where Pg = gauge pressure, Pa=atmospheric pressure

    T = 39 °C = 312 K

    V = 7.20 * 10⁻² m³

    therefore

    P*V = n*R*T → n = P*V / (R*T)

    replacing values

    n = P*V / (R*T) = 4.21*10⁵ Pa*7.20 * 10⁻² m³ / (8.314 J/mol K*312 K) = 11.685 mol

    since

    m = n*M, where m = mass, n = number of moles, M = molecular weight of oxygen

    then

    m = n*M = 11.685 mol * 32.0 g/mol = 373.92 gr of O₂

    therefore the initial mass of O₂ is 373.92 gr

    for the part B)

    P₂ = Pg₂ + Pa (1 atm) = 1.85*10⁵ Pa + 101325 Pa = 2.86*10⁵ Pa

    T₂ = 20.9 °C = 293.9 K

    V = 7.20 * 10⁻² m³

    therefore

    n₂ = P₂*V / (R*T₂) = 2.86*10⁵ Pa*7.20 * 10⁻² m³ / (8.314 J/mol K*293.9 K) = 8.427 mol

    m₂ = n*M = 8.427 mol*32.0 g/mol = 269.66 gr of O₂

    thus the mass leaked of oxygen is

    m leaked = m - m₂ = 373.92 gr - 269.66 gr = 104.26 gr
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