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13 May, 16:41

A closed vessel of volume 2.6 L contains a mixture of neon and fluorine gas. The total pressure of the mixture in the vessel is 3.0 atm at 0.0 °C. When the mixture is heated to 14.0 °C, the entropy of the mixture increases by 0.30 J ⋅ K⁻¹. Calculate the number of moles of neon and fluorine in the mixture.

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  1. 13 May, 19:15
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    The number of moles neon in the mixture is 0.148 moles

    The number of fluorine in the mixture is 0.200 moles

    Explanation:

    Step 1: Data given

    Volume = 2.6 L

    Total pressure = 3.0 atm at 0.0 °C

    When the mixture is heated to 14.0 °C, the entropy of the mixture increases by 0.30 J/K

    Step 2: Calculate total moles

    Neon is a monoatomic gas, composed of Ne atoms.

    Fluorine is a diatomic gas, composed of F₂ molecules.

    The heat capacities are different because F₂ has rotational degrees of freedom and for that reason, Cv (F₂) ≅ (5/2) R while Cv (Ne) ≅ (3/2) R.

    n (total) = PV/RT = (3.00 atm) (2.6 L) / (0.08206 L-atm/mol*K) (273.15) = 0.348 mol

    For one mole heated at constant volume,

    Step 3: Define ∆S

    ∆S = Cv*ln (287.15/273.15) = 0.04998*Cv

    So, for 0.348 mol,

    ∆S = (0.348 mol) (0.04998) Cv = 0.30 J/K

    ⇒ Cv = 17.25 J/mol*K for the Ne/F₂ mixture.

    Step 4: Calculate Cv

    For pure Neon (Ne), Cv = (3/2) R = 1.5 * 8.314 J/mol*K = 12.471 J/mol*K

    For pure F₂, Cv = (5/2) R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

    Step 5: Calculate moles of Ne and F2

    if X is the mole fraction of Ne, we can find X by setting the observed ∆S to the weighted average of the ∆S's expected for the two different gases in the mixture:

    17.25 J/mol*K = X * 12.471 J/mol*K + (1 - X) * 20.785 J/mol•K

    17.25 = x12.471 + 20.785 - 20.785x

    -3.535 = - 8.314x

    x = 0.425

    1 - x = 0.575

    moles Ne = (0.425) (0.348 mol) = 0.148 mol

    moles F₂ = (0.575) (0.348 mol) = 0.200 mol

    The number of moles neon in the mixture is 0.148 moles

    The number of fluorine in the mixture is 0.200 moles
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