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28 February, 04:00

A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b) the molar - and mass-based specific volumes, in m³/kmol and m³/kg respectively. Let g=9.81m/s².

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  1. 28 February, 04:21
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    a) m=336.18N

    b) Vn=16.67m/kmol

    Vm=0.1459m^3/kg

    Explanation:

    To calculate the mass of the octane (m):

    Number of mole of octane (n) = 0.3kmol (given)

    Molarmass of octane (M) = 114.23kg/kmol

    m=n*M

    m = (0.3kmol) * (114.23kg/kmol)

    m=34.269kg

    To calculate for the weight of octane (W):

    W=g*m

    W = (9.81m/s^2) * (34.269kg)

    W=336.18N

    b) For specific volumes of Vn and Vm:

    Given volume of octane (V) = 5m^3

    Vm=V/m

    Vm=5m^3/34.269kg

    Vm=0.1459m^3/kg

    And Vn will be:

    Vn=V/m=5m^3/0.3kmol

    Vn=16.67m/Kmol

    Therefore, the answers are:

    a) m=336.18N

    b) Vn=16.67m/kmol

    Vm=0.1459m^3/kg
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