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16 August, 16:44

Consider a mixture of air and gasoline vapor in a cylinder with a piston and an original volume of 42.5 cm³. The combustion of this mixture releases 925 J of energy. What volume will the gas expand to against a constant external pressure of 1.05 atm if all the energy is converted into work pushing on the piston?

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  1. 16 August, 17:03
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    volume of the gas expands to 8.74 Litres

    Explanation:

    The amount of work done by the gas is given by;

    W = P•ΔV

    Where;

    W = work done by gas

    P = Pressure

    ΔV = change in volume i. e V_f - V_i

    Where V_f is final volume and V_i is initial volume

    Thus;

    W = P (V_f - V_i)

    We are given;

    W = 927J

    V_i = 42.5 cm³ = 42.5 x 10^ (-6) m³

    P = 1.05 atm = 106391.25 N/m²

    Thus;

    925 = 106391.25 (V_f - (42.5 x 10^ (-6))

    925/106391.25 = V_f - (42.5 x 10^ (-6))

    0.00869432402 = V_f - (42.5 x 10^ (-6))

    V_f = 0.00869432402 + (42.5 x 10^ (-6)))

    V_f = 0.00874 m³

    Final volume = 0.00874 m³

    Converting to Litres, gives;

    V_f = 0.00874 x 1000 = 8.74 Litres
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