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    Volumes and Pressures of Gases in a Thermally Isolated Container

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    Please see the attached file for full description of the problem concerning volume and pressure of gases in a thermally isolated container.

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    Solution Preview

    For monoatomic gas: r (gamma) = 5/3 = 1.67
    and, 1/r = 3/5 = 0.6
    Because, no heat energy is transferred, hence, these processes will be adiabatic processes.
    dQ = dU + dW
    dQ = 0
    dU = -dW

    For gas A:
    PA*VA/TA = PA'*VA'/TA' .....(1)
    PA*VA^r = PA'*VA'^r ....(2)
    where, PA, VA and TA are the pressure, volume and temperature of gas A before unlock and PA', VA', and TA' are at the equilibrium.

    Similarlly for gas B:
    PB*VB/TB = PB'*VB'/TB' .....(3)
    PB*VB^r = PB'*VB'^r ....(4)

    At equilibrium: PA' = PB' ....(5)

    Eqn ...

    Solution Summary

    The solution explains clearly how to find various things about the volume and gas and how they and their equilibria changes in a thermally isolated container according to how the partition in the middle of it moves.