A Tricky Problem Involving Entropy

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Discussion Overview

The discussion revolves around the behavior of entropy in a thermodynamic system with two partitions and a movable wall, particularly focusing on how the wall's position is influenced by pressure and temperature differences across the partitions. The scope includes theoretical considerations of entropy maximization, equilibrium conditions, and the effects of thermal isolation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that the derivative of entropy with respect to volume is given by the relation ∂S/∂V = P/T, and discusses the conditions under which the wall will slide to maximize entropy.
  • Another participant notes that if the gas is thermally isolated, changing the volume will also affect the temperature, leading to additional entropy changes.
  • A different participant clarifies that the relation holds only under specific conditions (E and N constant) and questions the scenario where energy is maintained constant by heating/cooling elements while keeping a temperature difference.
  • One participant argues that the entropy of a subsystem does not necessarily need to reach a maximum, using the example of a refrigerator to illustrate their point.
  • It is mentioned that the equilibrium position for the wall requires equal pressures on both sides, regardless of temperature differences.

Areas of Agreement / Disagreement

Participants express differing views on the implications of thermal isolation and the conditions under which entropy maximization occurs. There is no consensus on how the wall behaves under varying temperature and pressure conditions, indicating ongoing debate.

Contextual Notes

The discussion highlights limitations regarding assumptions about thermal isolation and the conditions necessary for the entropy relations to hold. The impact of external heating/cooling elements on the system's entropy is also noted as an area of complexity.

BucketOfFish
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So we learn in basic thermo that for any system, the derivative of entropy with respect to volume is pressure over temperature:

∂S/∂V＝P/T.

Suppose we have a box with two partitions, and a moveable wall in between the partitions. The box and wall are both insulated to prevent heat transfer. The wall will slide until the entropy of the entire system is maximized. This occurs when the change in entropy with volume is equal for the two sides:

(∂S/∂V)1＝(∂S/∂V)2.
(P/T)1＝(P/T)2.

When both sides are at the same temperature, this result is obvious. The wall simply stops sliding when the pressure is the same on both sides. However, what happens when the temperatures are very different? If you fill one side of the box with a gas at 100K and 1atm, and the other side with a gas at 200K and 2atm, will the wall really not slide, despite there being a pressure difference of 1atm between the two sides?
 
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If your gas is thermally isolated, changing the volume will also change the temperature, which leads to an additional entropy change.
If the gas is not thermally isolated, the equilibrium has the same temperature on both sides (=obvious).
 
Okay, that makes sense. I see now from my thermo references that the relation is only true for [itex](\frac{∂S}{∂V})_{E,N}[/itex]. The wall must be heat-permeable for E to remain constant as volume is shifted.

However, what if the energy in each compartment is kept constant by other means, such as through heating/cooling elements, while keeping the temperature difference between compartments intact? I realize that now the entire system, including the heating components, may not be tending towards an entropic maximum, but can we still think about the entropy of the subsystem containing only the box? If so, would the input of the heating/cooling elements cause the wall to remain in equilibrium despite a pressure difference between the two sides?
 
The entropy of a subsystem does not have to reach a maximum.
Just consider a refrigerator.

The equilibrium position for the wall requires pleft=pright.
 

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