A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically from 500 cm3

Question

A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically from 500 cm3 to 2.10 dm3. What is the work done by the gas in the units of J? Enter your answer with the sign ( + or -) to 3 significant figures. DO NOT include the units.

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Eirian 1 year 2021-09-01T12:07:30+00:00 1 Answers 18 views 0

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    2021-09-01T12:08:36+00:00

    Answer:

    182 to 3 s.f

    Explanation:

    Workdone for an adiabatic process is given as

    W = K(V₂¹⁻ʸ – V₁¹⁻ʸ)/(1 – γ)

    where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

    For an adiabatic process

    P₁V₁ʸ = P₂V₂ʸ = K

    K = P₁V₁ʸ

    We need to calculate the P₁ using ideal gas equation

    P₁V₁ = mRT₁

    P₁ = (mRT₁/V₁)

    m = 2.80 g = 0.0028 kg

    R = 188.92 J/kg.K

    T₁ = 27°C = 300 K

    V₁ = 500 cm³ = 0.0005 m³

    P₁ = (0.0028)(188.92)(300)/0.0005

    P₁ = 317385.6 Pa

    K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

    W = K(V₂¹⁻ʸ – V₁¹⁻ʸ)/(1 – γ)

    V₁ = 0.0005 m³

    V₂ = 2.10 dm³ = 0.002 m³

    1 – γ = 1 – 1.28 = – 0.28

    W =

    18.89 [(0.002)⁻⁰•²⁸ – (0.0005)⁻⁰•²⁸]/(-0.28)

    W = -67.47 (5.698 – 8.4)

    W = 182.3 = 182 to 3 s.f

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