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31 December, 09:36

2.0801 mol of bromine gas is held at 4.258 atm and 284.92 K. What is the volume of its container in liters?

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Answers (2)
  1. 31 December, 10:02
    0
    11.41L

    Explanation:

    Data obtained from the question include:

    n (number of mole of Br2) = 2.0801 mol

    P (pressure) = 4.258 atm

    T (temperature) = 284.92K

    R (gas constant) = 0.082atm. L/Kmol

    V (volume of the container) = ?

    Applying the ideal gas equation PV = nRT, the volume of the container can be obtained as follow:

    PV = nRT

    4.258 x V = 2.0801x0.082x284.92

    Divide both side by 4.258

    V = (2.0801x0.082x284.92) / 4.258

    V = 11.41L

    Therefore, the volume of the container is 11.41L
  2. 31 December, 11:32
    0
    11.427L

    Explanation:

    From the ideal gas law,

    PV=nRT

    Where;

    P=pressure of the gas

    V=volume of the gas occupied by the container

    n=number of moles of the gas

    R=the ideal gas constant=0.0821atmL/mol/K

    Given

    P=4.258atm

    V=?

    n=2.0801mol

    T=284.92K

    From the equation

    PV=nRT,

    We will make V subject of formula

    V=nRT/P

    Let's now substitute the values

    V = (2.0801*0.0821*284.92) / 4.258

    V=11.427L

    Therefore, the volume of the container is 11.427L
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