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17 July, 06:02

At constant temperature and pressure, 2.05 g of oxygen gas O2 is added to a 1.0 L balloon containing 1.00 g of O2. What is the new volume of the balloon?

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  1. 17 July, 06:26
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    1.50 L.

    Explanation:

    From the general gas law:

    PV = nRT,

    Where, P is the pressure if the gas,

    V is the volume if the gas container,

    n is the no. of gas moles,

    R is the general gas constant,

    T is the temperature of the gas.

    At constant P and T:

    n₁V₂ = n₂V₁.

    V₁ = 1.0 L, V₂ = ? L.

    n₁ = mass/molar mass = (2.05 g) / (32.0 g/mol) = 0.064 mol.

    n₂ is the no. of moles of the total gas (2.05 g + 1.0 g).

    n₂ = n₁ + (1.00 g) / (32.0 g/mol) = 0.0953 mol.

    ∴ V₂ = n₂V₁/n₁ = (0.0953 mol) (1.0 L) / (0.064 mol) = 1.489 L ≅ 1.50 L.
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