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1 month ago

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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?

1 answer:

Softa [3K]1 month ago

6 0

Answer:

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

b) Δe = 4000V - 4000V2 + 9000V1

Explanation:

Let V be the volume of the box with two compartments

V1 is for the left compartment

V2 is for the right compartment

Momentum for the compartments before impact:

3000V1 + 1000V2

Momentum after the impact:

V(3000 + 1000) = 4000V

a) To find the volume of each compartment, V1 and V2, we note:

Momentum prior to impact = Momentum post-impact

3000V1 + 1000V2 = 4000V

Thus, V1 = (4000V - 1000V2)/3000 = 4V - V2/3

And V2 = (4000V - 3000V1)/1000 = 4V - 3V1

b) The change in entropy, Δe = 4000V1 - 1000V2

Substituting for V1 and V2 gives:

4000(4V - V2)/3 - 1000(4V - 3V1)

Results in 16000V - 4000V2/3 - 4000V + 3000V1

Finally yielding Δe = 4000V - 4000V2 + 9000V1

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