A container is initially partitioned by a wall into two compartments. The left compartment has volume V1 and the right compartment has volume V2. Both compartments are filled with identical ideal gases at the same temperature T and with an equal number of particles N but at different pressures P1 (left compartment) and P2 (right compartment). The wall is removed and the gases are allowed to mix to an equilibrium state.

Question

A container is initially partitioned by a wall into two compartments. The left compartment has volume V1 and the right compartment has volume V2. Both compartments are filled with identical ideal gases at the same temperature T and with an equal number of particles N but at different pressures P1 (left compartment) and P2 (right compartment). The wall is removed and the gases are allowed to mix to an equilibrium state.

(a) Compute the change in entropy of the system.
(b) Show unambiguously and convincingly that the mixing process is irreversible

Answer 1

To compute the change in entropy of the system, we can use the concept of entropy change in a reversible process. In a reversible process, the change in entropy of the system can be obtained using the equation:

ΔS = ∑(ni * R * ln(Vi₂/Vi₁)) + ∑(ni * R * ln(Pi₂/Pi₁))

where ΔS is the change in entropy, ni is the number of moles of each gas (which are equal in this case), R is the ideal gas constant, Vi₁ and Vi₂ are the initial and final volumes of each compartment, and Pi₁ and Pi₂ are the initial and final pressures of each compartment.

Now, let's begin solving this problem step by step:

(a) Compute the change in entropy:

Given that the volumes V1 and V2 are equal, we can simplify the equation for entropy change to:

ΔS = ∑(ni * R * ln(Pi₂/Pi₁))

Since the number of moles of each gas is the same, ΔS can be expressed as:

ΔS = N * R * ln(P2/P1)

(b) To show unambiguously and convincingly that the mixing process is irreversible, we can examine the change in total entropy of the universe. In a reversible process, the total entropy change of the universe is zero, while in an irreversible process, the total entropy change of the universe is positive.

In this case, when the wall is removed and the gases are allowed to mix, the system tends towards equilibrium. However, due to the difference in pressure between the two compartments, gases will redistribute themselves unevenly. This inhomogeneity leads to a positive change in entropy of the universe.

Therefore, since the change in entropy of the universe is nonzero and positive, we can conclude that the mixing process is irreversible.

It is important to note that irreversibility does not imply that the process is fast or instantaneous. It simply means that the process cannot be undone without affecting other variables or the surroundings.