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8 April, 14:49

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K. The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K. - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K.

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  1. 8 April, 14:54
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    Complete Question:

    A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K. The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K. - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K.

    a) what is the work for one cycle

    b) what is the thermal efficiency

    Answer:

    a) Work done for 1 cycle = 402.13

    b) Thermal efficiency = 20%

    Explanation:

    Number of moles, n = 0.300 mol

    Initial Volume, V₁ = 1000 cm³

    Temperature, T = 500 K

    Isothermal expansion to 5000 cm³

    Final volume, V₂ = 5000 cm³

    R = 8.314 J / mol. K

    Work done, W = nRT ln (V₂/V₁)

    W = (0.3 * 8.314 * 500) * ln (5000/1000)

    W = 1247.1 * ln5

    W₁ = 2007.13 J

    Isochoric cooling

    In an Isochoric process, volume is constant i. e. V₂ = V₁ = V

    W = nRT ln (V/V)

    But ln (V/V) = ln 1 = 0

    Work done, W₂ = 0 Joules

    Isothermal Compression to 1000 cm³

    V₂ = 1000 cm³

    V₁ = 5000 cm³

    W = nRT ln (V₂/V₁)

    W = 0.3 * 8.314 * 400 ln (1000/5000)

    W₃ = - 1605 J

    Isochoric heating to 500 K

    Since there is no change in volume, no work is done

    W₄ = 0 J

    a) Work done for 1 cycle

    W = W₁ + W₂ + W₃ + W₄

    W = 2007.13 + 0 + 0 - 1605+0

    W = 402.13 Joules

    b) Thermal efficiency

    Thermal efficiency = (Net workdone for 1 cycle) / (Heat absorbed)

    Heat absorbed = Work done due to thermal expansion = 2007.13 J

    Thermal efficiency = 402.13/2007.13

    Thermal efficiency = 0.2

    Thermal efficiency = 0.2 * 100% = 20 %
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