A box of volume V has a movable partition separating it into two compartments. The left compartment contains 3000 particles, the right one c

Question

A box of volume V has a movable partition separating it into two compartments. The left compartment contains 3000 particles, the right one contains 1000 particles, and initially the partition is in the middle so that the compartments are of equal volume. The partition is released and slides horizontally until the system is in a new equilibrium state.
(a) In terms of V, what is the volume of each compartment once the new equilibrium state is reached?
(b) What is the change in the system’s entropy during this process?

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Maris 5 years 2021-09-01T01:32:53+00:00 1 Answers 172 views 0

Answers ( )

  1. quangkhai
    0
    2021-09-01T01:34:00+00:00
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    Answer:

    a) V1 = 4V – V2/3 and V2 = 4V –  3V1

    b) Δe = 4000V – 4000V2 + 9000V1

    Explanation:

    Let V represent volume of the box containing the two compartments

    V1 represents compartment of the left compartment

    V2 represents compartment of the right compartment

    Momentum of the compartments before impact:

    3000V1 + 1000V2

    Momentum of the compartments after impact:

    V(3000 + 1000) = 4000V

    a) To obtain the volume of each compartment, that is, V1 and V2, we say:

    Momentum before impact = Momentum after impact

    3000V1 + 1000V2 = 4000V

    ∴ V1 = 4000V – 1000V2/3000 = 4V – V2/3

    Also, V2 = 4000V – 3000V1/1000 = 4V – 3V1

    b) Change in entropy,Δe = 4000V1 – 1000V2

    By substituting the V1 and V2, we have:

    4000(4V – V2)/3 – 1000(4V – 3V1)

    16000V – 4000V2/3 – 4000V + 3000V1

    16000V –  4000V2 – 12000V + 9000V1

    ∴ Δe = 4000V – 4000V2 + 9000V1

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