1. Sep 18, 2010

### Cosmossos

we know that when the energy drops the entropy increases. but when the energy of a gas drops then it will become eventualy solid which has less degrees of freedem and therfore the entropy should decrease and not increase.
Moreover high temp. has more entropy (gas for ex.) and as long as the temp increases -> energy increases and so is the entropy . but the entropy should DECREASE AS LONG AS THE ENERGY INCREASES.
where is my mistake?

2. Sep 18, 2010

### Staff: Mentor

You have that backwards. dS = dQ/T. When energy is removed from a system, the entropy decreases.

3. Sep 19, 2010

### phywjc

entropy never decreases！

4. Sep 19, 2010

### Staff: Mentor

Total entropy including everything never decreases. But the entropy of some particular system can certainly decrease. (If one system loses entropy, some other system must gain at least that much.)

5. Sep 19, 2010

### Cosmossos

I don't understand. It is know that if I have some gas and its energy drops then the entropy should increase It's written in every textbook.
But in that case, the gas will eventually become solid which has less degrees of freedom and therefore less entropy...

you can look here http://en.wikipedia.org/wiki/Entropy
under classical thermo.

6. Sep 19, 2010

### Staff: Mentor

Give one example.

Where does it say anywhere that removing energy from a gas causes its entropy to increase?

7. Sep 19, 2010

### Cosmossos

It says that if the energy drops/decrease the entropy increase.
So gas is an example

8. Sep 19, 2010

### Staff: Mentor

Where does it say this? Cut and paste an exact quote.

9. Sep 19, 2010

### Drakkith

Staff Emeritus
From wikipedia on entropy: "An air conditioner, for example, may cool the air in a room, thus reducing the entropy of the air of that system. The heat expelled from the room (the system), involved in the operation of the air conditioner, will always make a bigger contribution to the entropy of the environment than will the decrease of the entropy of the air of that system. Thus, the total of entropy of the room plus the entropy of the environment increases, in agreement with the second law of thermodynamics."

If you decrease the entropy of a system (in this case the air in the room) then you MUST increase it in another system (In this case, the air outside).

10. Sep 20, 2010

### Cosmossos

Ok thanks
and I have another question.
Entropy is non-conserved state function and in isolated systems it can't decrease and therefore ds>=0
In case ds=0 it does conserved because it doesn't change. how can it be a non conserved state function and also may have a case which ds=0?

11. Sep 21, 2010

### AlexChandler

Here is a paper written by an old physics professor of mine. He really does a great job of explaining all of this, but at the same time, it is more in the scope of a freshman physics course than wikipedia. Read the paper and take notes! After that you will feel much better about entropy.
If you like this paper, then you should check out a few others. He goes through Force, Work/Energy, Rotational Dynamics, Collections of Particles.. and does a paper on Thermodynamics in Chemistry.
http://orchard.wccnet.org/~gkapp/

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12. Sep 21, 2010

### Andrew Mason

Entropy is a state function. This means it depends only upon the thermodynamic state not on the path taken between thermodynamic states.

Entropy is non-conserved. If the sum of all entropy changes was always 0, entropy would be a conserved quantity. But it isn't. The sum of all changes in entropy of a system and its surroundings can theoretically be as close to 0 as you wish to make it (provided you have enough time to carry out a process). But that special case does not mean it is always conserved. In real life, entropy of the universe (system + surroundings) increases during any process.

AM

13. Sep 22, 2010

### brainstorm

I am confused. Wiki says that entropy increases as substances "disgregate." What is disgregating in the a/c example? Clouds of relatively dense cold air?

14. Sep 22, 2010

### lionelwang

From The Second Law of Thermodynamics，there is no contradiction for a spontaneous process in which energy drops while entropy decreases.

15. Sep 22, 2010

### brainstorm

Don't you have to specify some empirical/contextual parameters to make this statement testable?

16. Sep 22, 2010

### Drakkith

Staff Emeritus
The heat is transferred from the air in the room and ends up outside. The outside air disgregates, as it is heated up by the transfer of heat from the inside of the room.

17. Sep 22, 2010

### brainstorm

Isn't the a/c itself reducing entropy by compressing air and removing heat from it in that way? Then, logically, both the hot output of the evaporator and the cold output of the condenser would be introducing aggregated pockets of air into the surrounding air. At that point, the surrounding air goes to work increasing system entropy by mixing and exchanging energy with the hot/cold pocket of air. So there are several different systems going on with different forms of entropy/disagregation, no?

18. Sep 23, 2010

### Drakkith

Staff Emeritus
I think i can say yes to this. The AC used work to remove the heat from the air inside the room and transfers it to the outside air. Entropy inside the room decreases, and entropy outside increases. Yes, there are several systems going on, such as the transfer of heat from the outside to the walls of the room and then into the room, gradually heating the room and increasing its entropy at the same time that the AC is cooling the room and reducing it.

19. Sep 23, 2010

### brainstorm

It sounds like you are talking about a decrease in entropy as the heat is aggregated outside. Disgregation would take place if the a/c was turned off and the heat was allowed to seep in from outside. When temperature equilibrium is reached between outside and inside, that's when you've reached maximum entropy, correct? But that only considers inside & outside together as a single system. If you were just looking at the inside system along, I think you'd be dealing with disgregation of heat from the air to the condenser or something like that.

20. Sep 23, 2010

### Andrew Mason

Just a little helpful advice: Disgregation is an archaic term and is no longer used in thermodynamics. So disregard disgregation. Stick to entropy and its thermodynamic meaning:

$$\Delta S_{a->b} = \Delta S_{surr:a->b} + \Delta S_{sys: a->b} = \int_a^b \frac{dQ_{surr}}{T_{surr}} + \int_a^b \frac{dQ_{sys}}{T_{sys}}$$

where the integral is measured along the reversible path from a to b. If you measure this quantity, you will have a positive number. That is the second law of thermodynamics. If you get a negative number, you are missing part of the system or surroundings.

AM