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What happens to entropy when doubling the volume?

  1. Dec 13, 2015 #1
    1. The problem statement, all variables and given/known data
    A container of volume 2V is divided into two compartments of equal volume by an impenetrable wall. One of the compartments is filled with an ideal gas with N particles. The gas is in equilibrium and has a temperature T. How does the total energy, the entropy, the temperature, and the pressure change when the separating wall is suddenly removed? (There is no interaction with the surroundings).

    2. Relevant equations
    I'm guessing there are several, but perhaps dE=TdS-PdV+μdN, dS/dV=P/T, and PV=NkT are the most important ones.

    3. The attempt at a solution
    Here's what I think is the answer, but I want to be sure:
    Since there is no energy added to the system, and there is no reason energy would be removed from the system, the total energy should remain the same.

    Since no thermal energy has been added or removed, the temperature should also remain unchanged.

    Since the temperature is unchanged, no particles have been added or removed, and the volume has been doubled, we can use PV=NkT to deduce that the pressure has to be halved.

    The problem is that I haven't found a good way of showing how the entropy would increase. It's obvious that it should increase, but I don't know by how much. Help????
     
  2. jcsd
  3. Dec 13, 2015 #2

    TSny

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    Your answers for the energy, temperature, and pressure are correct.

    As a hint for getting the change in entropy, can you think of a reversible way to get from the initial to the final state?
     
  4. Dec 13, 2015 #3
    Moving the wall to one side of the container through a quasistatic process (i.e. "infinitely slowly"). But this still doesn't help me find how much the entropy increases.

    EDIT: Wait, since entropy does not increase in a reversible process, is the entropy the same? But hasn't the number of microstates increased?
     
  5. Dec 13, 2015 #4

    TSny

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    The entropy of a system may or may not change during a particular reversible process. It depends on the process.

    You said that you would move the wall quasi-statically. That's good. But that doesn't fully describe the process that you have in mind. You want to end up in a particular final state. You can allow the system to interact with the environment during the reversible process in such a way that you get to the desired final state. And you can give the environment whatever property you wish.
     
  6. Dec 13, 2015 #5
    But there is no interaction with the surroundings as stated in the original problem. I don't think comparing this to a reversible process is of much help. What I need help with is finding a relation I can use to explain how much the entropy increases.

    I've tried using dS/dN=-μ/T , dS/dV=P/T , and dE=TdS-PdV+μdN, but they don't really tell me anything (or I'm missing something).
     
  7. Dec 13, 2015 #6

    TSny

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    Entropy is a state variable. No matter what particular process you use to get the system to go from the given initial state to the given final state, the change in a state variable of the system is independent of the particular process. This is true whether the process is reversible or irreversible and whether the system interacts with the environment or not.
     
  8. Dec 13, 2015 #7
    I think I figured it out. Since the process is isothermal, we know that ΔQ=ΔW. We also know that W=NkT⋅ln(V2/V1).

    Since ΔS=ΔQ/T, it follows that ΔS=Nk⋅ln(V2/V1)), which in out case yields ΔS=Nk⋅ln2

    Right?
     
  9. Dec 13, 2015 #8

    TSny

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    Which process are you thinking about here?
     
  10. Dec 13, 2015 #9
    The process of the original problem.
     
  11. Dec 13, 2015 #10

    TSny

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    What is the value of ΔQ for the original process where there is no interaction with the surroundings?
    What is the value of ΔW for the original process?

    Even though the final state has the same temperature as the initial state, the original process is not isothermal. The process is irreversible. During the process, the system is not in thermal equililbrium. So, the temperature is not defined during the process.
     
  12. Dec 13, 2015 #11
    To add to what TSny said, once you have established the initial and final equilibrium states of the system as a result of the actual irreversible process path, the next step is to completely forget about the actual process path. To get the entropy change, you need to devise (dream up) a reversible path that takes you between the same initial and final equilibrium states, and calculate the integral of dq/T for that path. Any reversible path will do (there are an infinite number of reversible paths, and they all give the same value for the integral). Just choose one that is easy to calculate the integral for.

    Chet
     
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