Consider a system consisting of two glass bulbs, each with volume V, and each containing n moles of gas. Bulb A contains nA moles of A and bulb B contains nB moles of B (a different gas), nA = nB. The bulbs are connected by a closed stopcock. The stopcock is opened so that the gases are free to move between the bulbs. What is the change in entropy between initial (stopcock closed) and final (stopcock open) states?
Consider the same system as problem above – except that bulb B now contains nA moles of A rather than nB moles of B. The stopcock is opened so that the gases are free to move between the bulbs. What is the change in entropy between initial (stopcock closed) and final (stopcock open) states?
To determine the change in entropy between the initial and final states of the system, we need to consider the different scenarios described.
1. Scenario 1: Bulb A contains nA moles of gas A and Bulb B contains nB moles of gas B, where nA = nB.
In this scenario, when the stopcock is opened, the gases A and B can freely move between the bulbs until they reach an equilibrium. Since both gases have equal amounts in the initial state, they will distribute evenly between the bulbs in the final state.
The change in entropy can be determined by considering the mixing of the gases. Entropy is related to the number of microstates a system can occupy. In the initial state, each gas occupies a separate bulb, resulting in two distinct microstates. In the final state, when the gases have mixed, there is only one microstate because the gases are indistinguishable.
Therefore, the change in entropy, ΔS, is given by:
ΔS = kB * ln(Wf/Wi)
Where kB is the Boltzmann constant, Wf is the number of microstates in the final state, and Wi is the number of microstates in the initial state.
In this case, since there are two gases A and B, the total number of microstates in the initial state is Wi = 2. In the final state, when the gases have mixed, there is only one microstate, Wf = 1.
Therefore, ΔS = kB * ln(1/2) = -kB * ln(2)
2. Scenario 2: Bulb B contains nA moles of gas A instead of nB moles of gas B.
In this scenario, when the stopcock is opened, gas A will move freely between the bulbs until it reaches equilibrium. Since gas A is the same in both bulbs initially, there is no net change or mixing of gases.
In this case, the system remains the same and there is no change in entropy. Thus, ΔS = 0.
Overall, the change in entropy between the initial and final states depends on the scenario. In Scenario 1, there is a change in entropy given by ΔS = -kB * ln(2), while in Scenario 2, there is no change in entropy, ΔS = 0.