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25 February, 00:34

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A total of 27.10 KJ of heat is added to a 5.30-L sample of helium at 0.981 atm. The gas is allowed to expand against a fixed external pressure to a volume of 23.70 L.

A) calculate the work done on or by the helium gas units of joules, J.

B) what is the change in the helium a internal energy in units kilojoules, KJ?

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  1. 25 February, 00:45
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    A. Workdone by helium, W = - 1.829KJ

    B. Internal energy, DE = 25.271KJ

    Explanation:

    Workdone can be defined as the force moving through a distance. For a gaseous system, when the volume of the gas expands, the system is losing energy. Therefore,

    W = - P*DV

    Where P is the pressure in pascal

    DV is the change in volume in m3

    DV = Vfinal - Vinitial

    = 23.70 - 5.30

    = 18.4L

    W = - (0.981 * 18.4)

    = - 18.0504L. atm

    Converting L. atm to joule,

    = - 18.0504 * 101.325

    = - 1828.97J

    = - 1.829KJ

    If the system loses heat, Q the rection occurring is Exothermic.

    Heat is the transfer of energy from one system to another.

    Q = mcDT

    Where m is the mass of the system

    C is the specific heat capacity

    Q = 27.20KJ

    Internal energy is the summation of the heat supplied to a system and the workdone by the system

    DE = Q + W

    DE = 27.10 + (-1.829)

    = 25.27KJ
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