Question on Thermodynamics Problem

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Hello PF people! This is my first topic on the forum. I basically have a question on a problem that was on a thermodynamics exam. I have the solution that was posted by the professor after the exam but I have a question on it. I asked the professor but I wasn't completely satisfied. So:

Homework Statement



Within a container of constant volume V0, there are initially 100 moles of N2 at pressure P1 = 400 kPa. Using a heat exchanger, the temperature inside tha container is held constant at equal to the environment's temperature, T0 = 300 K. The container's wall is made of a ceramic that allows O2 to pass through, but not N2. N2 and O2 are considered ideal gases and the air in the environment contains 79% N2 and 21% O2. We expose the container to the environment and wait for equilibrium to be reached. Calculate:
a) The final pressure and air composition inside the container.
b) The change in the entropy of the universe during the transition to the final state.

Homework Equations



We basically need the state equation of the ideal gases, PV = nRT
We also need the equations for entropy change during mixing/unmixing transfering heat and compressing ideal gases.

The Attempt at a Solution



The first one (a) is easy. The chamical potential of O2 will be equal inside and outside of the container, which means, since the gases are ideal, that the partial pressure of O2 will be equal inside and outside. So we have (Dalton's law for partial pressures):

P1*xO2,inside = PO2,outside => 400*nO2,inside/(100+nO2,inside) = 21kPa => nO2,inside = 5.54 moles and the final pressure inside the container will be 421 kPa.

My question is in (b): If we consider our system to be the gas inside the container and the 5.54 moles of oxygen that enter the container during the transition, is any wirk done on the system?? I would say no but the professor disagrees! Sure, the nitrogen inside the container and the moles that enter the container are compressed, but this is not done via a piston that would give energy to the molecules. The molecules merely pass through the ceramic wall... The professor, in his solution, after calculating the entripy change of the system (unmixing the oxygen from the environment, mixing it with the nitrogen inside the container and then compressing it), calculates the work for isothermal compression of the nitrogen and oxygen to the final pressure. Then this work returns to the environment as heat (since the compression is isothermal), increasing its entropy. But I cannot see how is work done on the system. Any insights? The professor told me that the work is done by the wall but i didn't get it... :smile:

P.S.Sorry for the lengthy post but I wanted to be clear.
 
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Answers and Replies

  • #2
Mapes
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No work is done in the actual process. But to calculate the entropy change, it is convenient to replace the original (irreversible) process with a substitute sequence of reversible processes that starts at the same state and ends at the same state. In this conceptual process, work is done, possibly by a moving wall or a piston (it doesn't matter, as long as the process is reversible). I can see why your professor's answer was confusing; I would have emphasized that the work is done by a fictional piston in a fictional process. Does this help clarify things?
 
  • #3
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Ahh I understand now... Your answer was very helpful. Thanks a lot! :smile:
 

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