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Given a bottle of water, a closed system, some of the water molecules in the liquid phase will have enough energy to escape that phase, forming water vapour, contributing to vapour pressure.

Now consider the simple setup below:

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Water is used as an example, does not have to be water. The work for pumping water is converted to gravitational potential energy of water. But the system seems to be capable of doing extra work on the mass while retaining that gravitational potential energy gained. Where does this capability come from? Doesn't the existence of this capability already violate First Law?

P/S: I know the pressure increment due to formation of vapour will be small, but just because it is small and so negligible does not mean that the 'minor violation' of First Law can be overlooked, so I believe something must have been wrong with the argument above.

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    $\begingroup$ Anywhere you see a bias in movement you can attempt to exploit it to do work, even if it may not be worth your while. $\endgroup$
    – Buck Thorn
    Commented Feb 6, 2021 at 9:40
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    $\begingroup$ The increase of pressure is of the order of 3% at room temperature. So this phenomena may produce a negligible amount of work $\endgroup$
    – Maurice
    Commented Feb 6, 2021 at 10:46
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    $\begingroup$ But at boiling water temperatures, this force is sufficient to drive locomotives and engines all over the world. $\endgroup$
    – Maurice
    Commented Feb 6, 2021 at 11:22
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    $\begingroup$ If you partially-filled a water bottle at 25 °C, and then replaced the remaining air with completely dry air at 1 atm and sealed it, the total pressure inside would increase to 1.03 atm, because the vapor pressure of water at 25 °C is 0.03 atm. However, the air in water bottles is probably already partially saturated with water when the bottles are sealed. If it's, say, 80% relative humidity, then the increase in pressure is only going to be 0.006 atm (20% of 0.03 atm). That's why sealed water bottles don't seem to be pressurized (assuming you're at the same $T$ at which they were sealed). $\endgroup$
    – theorist
    Commented Feb 6, 2021 at 18:43
  • $\begingroup$ Yeah, the pressure increment is small, and could produce small amount of work, doesn't that violate First Law? Or are you saying 'violation of First Law by a little' does not matter? $\endgroup$
    – TheLearner
    Commented Feb 6, 2021 at 23:23

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I believe the vapour produced in the step 3 of you experiment is due to the particles with high order speed in the Maxwell distribution of speeds curve that escape because of their higher kinetic energies. When escaping, they will reduce the average kinetic energy of particles in the water sample, which will then reduce the net temperature of sample.

This will then prompt the system (water sample) to take energy, in the form of heat, from the surroundings to reestablish the thermal equilibrium. Hence taking back the work done by system on surrounding via expansion of the bottle through heat to maintain its temperature. Although, as you can see in the Maxwell distribution of speed, the particles with enough kinetic energy to escape from the water surface are very less at low temperatures.

So, if you take a system with just enough temperature and manage to isolate it from the surroundings, I think you should see a visible drop in the temperature of water proportional to the work done in expanding the water bottle.

Mathematical description of the above Explanation:

Let's say the total energy change taking place after the particles escape water surface is $\Delta U$ out of which $W$ is used to do work on the surroundings. The remaining energy ($\Delta U - W$) is used to establish an equilibrium temperature inside the bottle and hence is given back to the air-water system inside the bottle.

This step may be happening simultaneously with the next steps I am writing but I am assuming that the air- water system inside the bottle will reach an equilibrium first before any work is done. This is purely to make the math easier and because thermodynamic is the study of equilibriums rather than processes so it won't affect the results as long as this equilibrium is established even when we are studying the final change taken place.

So the net change in temperature inside the bottle is $$\frac{-W}{C_{system}}$$ where $C_{system}$ is the heat capacity of the system.

Net temperature difference between system and surrounding $$\Delta T=\frac{W}{C_{system}}$$

So the system will absorb energy in the form of heat to bring the temperature back to equilibrium $$Q=C_{system}\Delta T=>Q=C_{system}\biggr[\frac{W}{C_{system}}\biggr]=>Q=W$$

So we can see mathematically that the system will take back all the energy it gave to the surroundings by doing work in the form of heat.

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  • $\begingroup$ Using equations will make your answer clearer .(Equilibrium and Thermodynamic Expressions) $\endgroup$
    – user102687
    Commented Feb 14, 2021 at 7:37

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