The energy lost when mixing two boxes of gas

  • #1

Main Question or Discussion Point

Apologize if you feel this question to be so simple and naive.

Suppose a chamber is divided by a insulator into two equal volume (V) chambers, . One chamber contains n moles of hot oxygen at temperature T1, the other chamber contains n moles of cold oxygen at temperature T2. If you move the insulator away and let the hot & cold oxygen molecules to mix well. How much energy is lost during this process? I guess if you know how to compute the enthalpy/free energy/entropy of gas with volume V & amount in moles & temperature, then the answer is simply delta(E) = enthalpy(2n, (T1 + T2)/2, 2V) - enthalpy(n, T1, V) - enthalpy(n, T2, V)? One can replace enthalpy(n, t, v) with free_energy(n, t, v)?

Thanks in advance.
 

Answers and Replies

  • #2
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Apologize if you feel this question to be so simple and naive.

Suppose a chamber is divided by a insulator into two equal volume (V) chambers, . One chamber contains n moles of hot oxygen at temperature T1, the other chamber contains n moles of cold oxygen at temperature T2. If you move the insulator away and let the hot & cold oxygen molecules to mix well. How much energy is lost during this process? I guess if you know how to compute the enthalpy/free energy/entropy of gas with volume V & amount in moles & temperature, then the answer is simply delta(E) = enthalpy(2n, (T1 + T2)/2, 2V) - enthalpy(n, T1, V) - enthalpy(n, T2, V)? One can replace enthalpy(n, t, v) with free_energy(n, t, v)?

Thanks in advance.
Is the chamber itself insulated?
 
  • #3
Is the chamber itself insulated?
Yes, the two chambers and the whole box are insulated.

Thank you!
 
  • #4
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Then the combined gases are not capable of doing work W on the rigid chamber. And, since the chamber is insulated, the amount of heat Q it is capable of exchanging with the surroundings is zero. So from the first law of thermodynamics, ##\Delta E=Q-W##, the combined change in internal energy of the gases is zero. So, no energy is lost during the process you describe. The final temperature will be the average of the hot and cold temperatures, and: $$\Delta E= E(2n, (T1 + T2)/2, 2V)-E(n, T1, V)-E(n, T2, V)=0$$
 
  • #5
Maybe I didn't ask the question the right way. An more appropriate way to ask is: what is the energy (enthalpy?) difference between the hot/cold separated state and the well mixed state. The process of mixing can release workable energy, say, to propel a fan sitting on the insulator.

Thanks.
 
  • #6
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Maybe I didn't ask the question the right way. An more appropriate way to ask is: what is the energy (enthalpy?) difference between the hot/cold separated state and the well mixed state. The process of mixing can release workable energy, say, to propel a fan sitting on the insulator.

Thanks.
Yes. Even in this case, the internal energy is still the appropriate function to quantify the energy change (not the enthalpy). For the case you describe in which work is done to run an external fan (or an internal fan being used as a turbine to generate external electrical energy), work is being done on the surroundings, so that the internal energy of the combination of gases decreases (and the final temperature will not be the average of the initial gas temperatures). The amount of work that is done (and thus the change in internal energy) depends on how the work is done.
 
  • #7
What's the maximum work that can be done with the removal of the insulator?

Thanks.
 
  • #8
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Is your question, "what is the maximum amount of work that can be done by the system if the system is insulated, the initial state is as you described, and the final state is uniform temperature and pressure within the container?"
 
  • #9
You can say that. Or maybe more precisely, given two states: state 1 being the chambers separated by the insulator as described in my previous posts/replies {(n, T1, V), (n, T2, V)}, state 2 being the insulator being removed and gas well mixed {(2n, (T1+T2)/2, 2V)}. What is the difference of the internal energies (or enthalpy/entropy/free energy) of the two states? Or in other words, if the external environment of the box is the same to begin with, what's the difference of work can be down when starting with either of the two states?
 
  • #10
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We're talking about two different situations here. Situation 1:

You can say that. Or maybe more precisely, given two states: state 1 being the chambers separated by the insulator as described in my previous posts/replies {(n, T1, V), (n, T2, V)}, state 2 being the insulator being removed and gas well mixed {(2n, (T1+T2)/2, 2V)}. What is the difference of the internal energies (or enthalpy/entropy/free energy) of the two states?
We already said that the change in internal energy for this situation is zero.

Situation 2:
Or in other words, if the external environment of the box is the same to begin with, what's the difference of work can be down when starting with either of the two states?
In this case, we are interested in determining the maximum amount work that can be done and the maximum decrease in internal energy if the system is subjected to an adiabatic process. For this case, the constraints will be that the final temperatures are identical (not necessarily the average of the initial temperatures), the final pressures are equal, and the final total volume of both gases is 2V. To solve this problem, one would have to devise an adiabatic reversible path between the initial and final states of the system (not involving merely removing the separator from between the two chambers and allowing the system to spontaneously equilibrate).
 

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