Suppose a bottle is filled with a hypothetical liquid that cannot absorb air/gas. The cap is tightened enough to prevent any loss or gain of mass. Now, this bottle is vigorously shaken and then allowed to rest. Will the pressure build-up within the bottle last for an indefinite period of time ? If no, how do you explain the transfer of entropy developed due to shaking, to the surroundings ? Will the temperature differential created as a result of shaking, alone, be sufficient enough to transfer the entropy to the surroundings (considering that the system of bottle+fluid is closed and hence, there is no entropy transfer due to mass) ?
Assume the liquid in the bottle is initially at the same temperature of the surrounding air, and that the bottle can transfer heat (glass, metal, etc.). Shaking a liquid results in viscous dissipation (friction) within the liquid. This, in turn, increases the temperature of the liquid. Heat transfer to the surroundings occurs, transferring entropy to the surroundings.
Hope this helps.
You are saying the shaking has warmed the fluid and the air above it, and this has caused a pressure increase. If it was in equilibrium before, it was the same temperature as the bottle, and the bottle was the same temperature as the surroundings.
Molecules in one small element of the fluid bounce off molecules in the next. The hotter one loses energy, and the cooler gains energy. Heat flows until they become the same temperature. The same thing happens at the interfaces between fluid and air, fluid and bottle, within the wall of the bottle, and the interface between bottle and surroundings.
It might be that the bottle is suspended by a thread in a vacuum chamber. In that case, the bottle will stay warm for a longer time. But both the bottle and the chamber are warm. At room temperature, they glow with infrared blackbody radiation. The warmer one is brighter and on the average gives off a higher frequency. The warmer will lose more energy through radiation than it absorbs. The bottle and chamber will slowly come to the same temperature.
It may be that the bottle is a good insulator. That means heat flows slowly through it.
It may be that the fluid has two parts, like salad oil and water. In this case, the denser water will slowly sink toward the bottom. It loses a little potential energy as it does so. Also water and oil are more strongly attracted to each other than to the other type. These intermolecular forces also contribute to a potential energy loss and kinetic energy gain when like is near like. It is the same as fluid being at the bottom and air at the top.
The kinetic energy is quickly randomized by collisions with other molecules. The fluid will become slightly warmer. At equilibrium, the water will mostly be on the bottom and oil mostly on top. There will be a little mixing because the molecules are moving. There will be less than there would be if mixing was not energetically unfavorable. So oil and water will come to a lower entropy equilibrium than a mixable fluid like alcohol and water.
Fluid at the bottom and air at the top is another example of the same thing. When you say that the fluid cannot absorb air, it is because fluid is saturated. The air and fluid have mixed as much as is consistent with equilibrium.
I believe I get your point. But how about a little more intricate scenario ? What I am saying is, let's assume that the liquid is hypothetically non-absorbent (it does not absorb any gas). Also, the bottle is completely filled with the liquid to the brim and then closed. So, under these situations, if the bottle is completely insulated (theoretically) with zero possibility of heat transfer, then what would happen to the the fluid inside the bottle ? The extra amount of energy cannot escape. So, will the molecules be in a state of randomness for an indefinite period of time ? Can this apparatus (hypothetical) be used as a reservoir of heat, assuming that the shaking is too violent ? (Of course, the wall of the material able to withstand the high stresses)