1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Idea for how to break the second law of thermodynamics

  1. Jun 25, 2013 #1

    I have an idea for how to break the second law of thermodynamics, and I don't see where it cracks. I would really appreciate if someone could explain this to me.

    Kelvin-Planck statement of the second law:

    "No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of this heat into work."

    Clausius statement of the second law:

    "No process is possible whose sole result is the transfer of heat from a cooler to a hotter body."

    This is my idea:

    It involves vapor-liquid equilibrium, which is a state where the rate of evaporation equals the rate of condensation on a molecular level such that there is no net vapor–liquid interconversion. Imagine a chemical substance in this state, so that for example 50 % of the substance is in liquid state, and 50 % of the substance is in gas state for a certain temperature at atmospheric pressure, to make it simple. The substance is in a completely isolated (hypothetically) container, and since the liquid has a higher density than the gas, they will naturally separate from each other.

    Once equilibrium is achieved, the liquid is separated from the gas by a wall (see attached drawing). To introduce this wall shouldn't have to cost any energy. This will result in that the substance is not in equilibrium anymore. Since the substance will strive to achieve equilibrium again, some of the gas in the gas part of the container will try to condense, which will reduce the pressure in this part of the container. In the liquid part of the container, some of the liquid will try to evaporate, which will increase the pressure in this part of the container.

    So now, can't you just let the two pressures power a generator? Let the gas power a generator through contraction, and let the liquid power a generator through expansion. Once equilibrium is achieved, and the pressure has equalized, you just remove the wall, and do the same thing over and over again. The energy that powers the generator would come from the thermal energy of the substance, so after each cycle you would have to heat up the substance a bit. Theoretically this could have an efficiency of 100%. But this clearly violates the second law. According to the second law, heat can't purely be converted into work (see the first statement above).

    So why doesn't this work?

    Grateful for response!

    Attached Files:

    Last edited: Jun 25, 2013
  2. jcsd
  3. Jun 25, 2013 #2


    User Avatar

    Staff: Mentor

    That's not how that sort of equilibrium works. In a closed, insulated container, an equilibrium is about the states themselves being stable and balanced. If a vapor can exist at a certain temperature and pressure, it doesn't matter if all of the walls of the container are solid or if one "wall" is the liquid of the same substance. That does not create an instability or imbalance.

    My suggestion would be to get a steam table and pick a mass, volume, temperature and pressure and calculate what the equilibrium looks like. Then see what changes (nothing) when you divide the two sections.
  4. Jun 25, 2013 #3
    Thanks for your reply!

    I thought vapor-liquid equilibrium was just like chemical equilibrium, but for states of matter instead of a chemical reaction with products and reactants.

    I know that for a chemical reaction, if you remove a product, or a reactant, you will disrupt the equilibrium, and the reaction will keep going until equilibrium is reached again.

    This is exactly what I'm doing through putting a wall between the gas and the liquid, I'm separating the "reactant" from the "product", to make the reaction keep going.

    Are you sure a vapor-liquid equilibrium doesn't work the same way? If it doesn't work the same way, my idea works for any chemical equilibrium, as long as its easy to separate products from reactants, and as long as the chemical reaction creates a pressure. Ideally would be if reactants are in liquid state, and product(s) are in gas state, or vice versa.
  5. Jun 25, 2013 #4


    User Avatar

    Staff: Mentor

    Completely sure. This is pretty basic thermodynamics.
  6. Jun 25, 2013 #5


    User Avatar
    Gold Member

    I guess what your scenario boils down to is the situation of two similar containers connected by a tube, with one container holding liquid water and the other gaseous water. After time, does the level of water in both containers become even? And why?
  7. Jun 26, 2013 #6
    I still don't understand. Once I separated the gas from the liquid (whether its a vapor-liquid equilibrium, or a chemical equilibrium for an ordinary chemical reaction), why wouldn't 50 % of the gas in the gas container condense, and why wouldn't 50 % of the liquid in the liquid container evaporate? Thats the ratio for the equilibrium!

    Are you completely sure that this doesn't work for a chemical equilibrium for an ordinary chemical reaction with reactants and a product aswell?
    Last edited: Jun 26, 2013
  8. Jun 26, 2013 #7
    This discussion seems to be about saturated vapours where a liquid and its vapour are in equilibrium. As far as I remember that doesn't require 50 percent of each state for them to be in equilibrium at any particular temperature and pressure.As long as there's a mixture of states,in any ratio,its a saturated vapour.
    Another point is ,how will you introduce a wall without the need for energy?
  9. Jun 26, 2013 #8


    User Avatar

    Staff: Mentor

    The liquid completely fills its container, so there is no room for gas. If some of the gas in the other condenses, the pressure drops and it moves away from the stable state it was just in.

    The part that you are misunderstanding is that you think equilibrium is about the ratio. It isn't, it is about the states. When you divide the container, you don't change the states. There can be any ratio at all of water and gas for a given state that is in the stable range for both. Heck, consider your house: why doesn't some of the water vapor in the air spontaneously condense and fall to the floor when you close your front door? There's no water in your house for it to be in equilibrium with and you just separated it from the ocean by closing the door. In addition, how does vapor in the middle of the container know if there is water at the bottom or a solid divider at the bottom?

    Last night I suggested looking at the actual properties of water and steam in a steam table. Have you done it yet? Here's one: http://www.efunda.com/materials/water/steamtable_sat.cfm

    One sample problem: Say you have 1kg of H20 in a 1 cubic meter container, at 110 C. How much liquid and how much gas is there and what is the pressure?

    After you answer that, we'll go into what happens when you divide the two sides of the container.
    Last edited: Jun 26, 2013
  10. Jun 26, 2013 #9
    I appreciate your help russ_watters, but I don't see how a steam table can answer this question.

    Forget about a vapor-liquid equilibrium, it is a chemical equilibrium instead. I thought I was just simplifying the problem, but apparently not.

    Below is the formula for chemical equilibrium:

    K = [itex]\frac{[A]^2}{\times[C]}[/itex]


    K = The equilibrium constant for the specific chemical reaction for a certain temperature

    [A] = The concentration of substance A (product)

    = The concentration of substance B (reactant)

    [C] = The concentration of substance C (reactant)

    K varies with temperature.

    So for a certain chemical reaction, at a certain temperature, the concentration between reactants and product can be 50 %. Say that the product is in gas state, and the reactants in liquid state, to make it easy to separate the two. Such a chemical reaction should exist.

    According to this formula, if you lower the concentration of the product, or in this case, if you seperate the product from the reactants by a wall, you will disrupt the equilibrium, and the chemical reaction will keep going in order to compensate, until equilibrium is reached again.

    In response to Dadface, sure, introducing the wall will cost a litte energy, but imagine the contaner to be huge and thin, so that the wall is very small in comparison. The pressure required to insert the wall should then be negligable.

    From wikipedia below:
    In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present at concentrations which have no further tendency to change with time. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but equal. Thus, there are no net changes in the concentrations of the reactant(s) and product(s).
    Last edited: Jun 26, 2013
  11. Jun 26, 2013 #10


    User Avatar

    Staff: Mentor

    OK, so you are saying you recognize it doesn't work for a liquid-vapor mix? If that's the case, then yes, we don't need a steam table and to walk through a problem.

    For a chemical reaction, the issue works pretty much the same way though. If you mix an acid and a base, you produce CO2. The reaction stops when the pressure in the container is high enough to stop it. Sectioning-off the liquid and vapor doesn't change that, only reducing the pressure does.

    Again, equilibrium between phases is not about the ratio of quantities, it is about the state properties.
    Last edited: Jun 26, 2013
  12. Jun 26, 2013 #11
    You mean mixing an acid and a base produces H2O, not CO2.

    But that completely contradicts what this formula says:

    K = [itex]\frac{[A]^2}{\times[C]}[/itex]
    Last edited: Jun 26, 2013
  13. Jun 26, 2013 #12


    User Avatar

    Staff: Mentor

    I think maybe the problem here is about mixtures and concentrations. A "mixture" of a liquid and vapor is not a mixture in the chemical sense, it is two separate volumes of two different substances, with a boundary between them. The relevant mixture for the reaction rate analysis is the CO2 dissolved in the solution.
  14. Jun 26, 2013 #13


    User Avatar

    Staff: Mentor

  15. Jun 26, 2013 #14


    User Avatar

    Staff: Mentor

  16. Jun 26, 2013 #15
    That would mean that it is impossible for a chemical reaction involving different states of matter to exist in dynamic equilibrium. I don't think thats true. See this link:


    I read your links, but nothing there explains why my idea doesn't work?
  17. Jun 26, 2013 #16


    User Avatar

    Staff: Mentor

  18. Jun 26, 2013 #17
    What else would russ_watters mean by this?:

  19. Jun 26, 2013 #18


    User Avatar

    Staff: Mentor

    No, it doesn't. Chemical reactions can happen at the boundary, but they are not limited by an equilibrium of concentrations, they are limited by the states of the two substances.

    I've bolded the word "state" a bunch of times here. Do you understand what that word means?
    Ok.... did you get from those links that pressure changes the equilibrium point of a reaction? So then, if you have a container that is filled solid with a liquid, how can any of it become a gas without an enormous increase in pressure?

    Where did you get that formula from and what exactly have you read about it? Because links I see describe exactly what I'm describing. In fact, there are two different equilibrium constants, one based on pressure, like I've said, and one based on concentration, like you said. One thing you are missing here is that your suggestion involves a massive change in pressure.
    Last edited: Jun 26, 2013
  20. Jun 27, 2013 #19


    User Avatar
    Gold Member


    The chemical reaction for water-vapour is
    H20(l) [itex]\Leftrightarrow[/itex] H20(g)
    The K is
    Kc = [H20(g)] / [H20(l)]
    For most reactions, the solid or liquid state can be considered pure, and does not vary in concentration.
    So K will reduce to
    Kc = [ H20(g) ]

    For gases, we can use the partial pressures
    Kp = P[itex]_{H20(g)}[/itex]

    That is the state that Russ is mentioning.

    The liquid water is not included in the formula and it does not matter how much there is whether 1 litre of 1000 litres or a lake full, or none. The gaseous water will have values of temperature and pressure. Removing the reactant, ie liquid water, does not change the state of the gaseous water.

    What matters is the pressure on the gas, and by another formula PV = nRT, other properties of the state of the gas can be determined.

    Getting back to the two containers, only under certain conditions will the liquid level become equal in both.

    --- continued ---
    Last edited: Jun 27, 2013
  21. Jun 27, 2013 #20
    If the ratio between liquid water and gasous water doesn't matter, how can the reaction be said to be in DYNAMIC equilibrium?

    State as in state of matter, right? Liquid or gas in our case.

    It is precisely the increase/decrease of pressure that I use to power the generator.

    The formula I've been talking about is called "The Law of Mass Action". I noticed that I've written it wrong, here is how its supposed to be:

    For the chemical reaction:

    [itex]aA + bB ⇔ cC + dD [/itex]

    AT EQUILIBRIUM, the equilibrium constant can be expressed as:

    [itex]K = \frac{[A]^a \times ^b}{[C]^c \times [D]^d}[/itex]

    The small letters a,b,c, and d represent the number of moles or molecules for the reaction to be balanced.

    Last edited by a moderator: Sep 25, 2014
  22. Jun 27, 2013 #21

    you have to modify this for a liquid vapour transition.

    There's only one molecule reactant and one molecule of reaction product, and the concentration of the liquid is constant, while the concentration of the vapour does depend on the pressure.

    so you just get

    [tex] K = \frac{[Water vapour] } {Constant} [/tex]

    This means the concentration of water vapour is a constant and does not depend on how much liquid is around, but only on the temperature.
  23. Jun 27, 2013 #22


    User Avatar

    Staff: Mentor

    No. The state is all of the unique values of descriptions of the system: Temperature, pressure, specific volume, etc:

    You need to learn what these things mean.
    That's not what I asked/not what this is about. I asked how it is possible for the liquid to turn to gas in a fixed volume without an increase in pressure. Then, if the pressure changes, if you recognize that moves the equilibrium point.

    If you have a pressurized anything you can of course use it to power a turbine. But only once. As soon as you start using the energy of the system to drive a turbine, the energy - manifested as pressure and temperature - starts to drop. You're thinking - incorrectly - that you can get it to rise back up again to where it started.

    By the way, even if the liquid had enough initial energy and you were willing to lower the pressure enough to vaporize all of it, you could still only do this once. On the other side, the vapor just stays a vapor. None of it will condense without an external application of pressure or removal of temperature (heat energy).
    No, it's not. That's a different formula, describing reaction rate, not equilibrium. You've been talking about the equilibrium constant:
    Read the wiki or other written sources. They contain much more information than you can get from a youtube video.
    Last edited by a moderator: Sep 25, 2014
  24. Jun 27, 2013 #23


    User Avatar
    Gold Member

    No need to continue as stated - previous posts provide good explanation.
  25. Jun 27, 2013 #24
    Ok, I understand. That clearly makes my idea impossible with vapor-liquid equilibrium. Thanks!

    But in order for the second law to hold, it would also have to be that the equilibrium constant of any chemical reaction that can change pressure, or indirectly perform work, can never be dependent on both the concentration of reactants and products, if it doesn't cost any work to separate reactants from products.

    Is this true? Sorry for complicated sentence.

    What about water freezing? Pressure will increase. Is the equilibrium constant not dependent on both reactant (liquid water) or product (ice) there either?
    Last edited: Jun 27, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook