A Tricky Problem Involving Entropy

[BucketOfFish]

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?

 

[mfb]

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).

 

[BucketOfFish]

Okay, that makes sense. I see now from my thermo references that the relation is only true for $(\frac{∂S}{∂V})_{E,N}$. 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?

 

[mfb]

The entropy of a subsystem does not have to reach a maximum.
Just consider a refrigerator.

The equilibrium position for the wall requires p_left=p_right.

 

[sardar]

This is indeed a tricky problem involving entropy. At first glance, it may seem counterintuitive that the wall would not slide in this scenario. However, we must remember that entropy is not just a measure of energy, but also of disorder or randomness. In this case, the gas on the hotter side has a higher temperature and therefore a greater amount of thermal energy. This results in a higher degree of molecular motion and therefore a greater degree of randomness or disorder. On the other hand, the gas on the colder side has a lower temperature and less thermal energy, leading to a lower degree of randomness.

When the wall is allowed to slide, the gas particles on the hotter side will move into the colder side in an attempt to equalize the temperatures. However, this movement also leads to a decrease in the randomness or disorder of the system. This decrease in entropy must be taken into account when considering the overall change in entropy of the system. The decrease in entropy due to the movement of gas particles will offset the increase in entropy due to the equalization of pressure.

Therefore, in this scenario, the wall will not slide because the decrease in entropy from the movement of gas particles is greater than the increase in entropy from equalizing pressure. This highlights the importance of considering all factors, not just energy, when analyzing thermodynamic systems.