Register to reply

Idea for how to break the second law of thermodynamics

by Bobcent
Tags: thermodynamics
Share this thread:
256bits
#19
Jun27-13, 03:23 AM
P: 1,414
Quote Quote by Bobcent View Post
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?
RE:
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?
But that completely contradicts what this formula says:

K = [A] 2 [B][C]
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 ---
Bobcent
#20
Jun27-13, 05:17 AM
P: 31
Quote Quote by 256bits View Post
For most reactions, the solid or liquid state can be considered pure, and does not vary in concentration.

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.
If the ratio between liquid water and gasous water doesn't matter, how can the reaction be said to be in DYNAMIC equilibrium?


Quote Quote by russ_watters View Post
I've bolded the word "state" a bunch of times here. Do you understand what that word means?
State as in state of matter, right? Liquid or gas in our case.

Quote Quote by russ_watters View Post
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?
It is precisely the increase/decrease of pressure that I use to power the generator.

Quote Quote by russ_watters View Post
Also: 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.
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]^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.

Source:
http://www.youtube.com/watch?v=aNEDU6EL8jc
willem2
#21
Jun27-13, 05:52 AM
P: 1,395
Quote Quote by Bobcent View Post
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]^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.
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.
russ_watters
#22
Jun27-13, 05:55 AM
Mentor
P: 22,237
Quote Quote by Bobcent View Post
State as in state of matter, right? Liquid or gas in our case.
No. The state is all of the unique values of descriptions of the system: Temperature, pressure, specific volume, etc:
http://en.wikipedia.org/wiki/Thermodynamic_state

You need to learn what these things mean.
It is precisely the increase/decrease of pressure that I use to power the generator.
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).
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:
No, it's not. That's a different formula, describing reaction rate, not equilibrium. You've been talking about the equilibrium constant:
https://en.wikipedia.org/wiki/Equilibrium_constant
Read the wiki or other written sources. They contain much more information than you can get from a youtube video.
256bits
#23
Jun27-13, 06:48 AM
P: 1,414
Getting back to the two containers, only under certain conditions will the liquid level become equal in both.

--- continued ---
No need to continue as stated - previous posts provide good explanation.
Bobcent
#24
Jun27-13, 08:44 AM
P: 31
Quote Quote by willem2 View Post
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.
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?


Register to reply

Related Discussions
Thermodynamics- Zemansky Heat and Thermodynamics book question Advanced Physics Homework 0
Physical Chemistry books (Thermodynamics/Statistical Thermodynamics/Kinetics/QM) Science & Math Textbooks 1
Break this. Fun, Photos & Games 5
Thermodynamics: violation of 1st and/or 2nd law of thermodynamics in a heat engine Introductory Physics Homework 7
Prof teaches Statistical thermodynamics in a Classical Thermodynamics class Academic Guidance 10