# I Some Questions about entropy...

1. Feb 25, 2017

### Arman777

I was watching The Feymann online lectures and he talked about the arrow of time.And entrophy etc.
I have some questions.
1-Can ve say non-conservative force are time irreversible, but conservative force are time reversible ?
2-So From one , If a non-conservatice force acts on a system that system will lose energy and that energy has something with entrophy.(I think that way ).And It must be defined by some equation
$ΔS=∫\frac {Q} {T}$
here Q is energy lost or another way, $Δ(ME)=ΔW_{non-conservative force}=Q$
T is the tempeture of the system (I thought this relationship myself so it could be wrong or right )
Here Q be aslo something else like If two atoms colllide some energy will lost.That energy also can be Q.
3-In closed systems ΔS=0 would be zero ?
4-If closed system ΔS=0 then how could we say entrophy always increases (I know its not "always" its "almost" always, but in here is İn a closed system.Entrophy will be same for a long time.So entrophy didnt decreased so.. ?
5- Seems that theres a strict relationship between time and entrophy.Can we say with time some information lost ? (Or hidden information )

2. Feb 25, 2017

### stevendaryl

Staff Emeritus
No, the real definition of $\Delta S$ is: $\frac{\Delta Q_{rev}}{T}$. Entropy is a function of state. So what that means is that the entropy change in going from one state to another is independent of how you made that transition. So to calculate the entropy of a new state, you choose a reversible path to go from the initial state to the final state.

So there is a standard example of how this makes a difference.

Suppose you have 1 liter of a gas, and you release it suddenly into an evacuated 2 liter bottle. It's insulated so that no heat goes in or out. What is the change in entropy?

Well, figure out the new state. Since no energy went in or out, we know that the new state has the same energy as the initial state. For an ideal gas, that means that the temperature didn't change, either. But the volume doubled. Since the temperature didn't change, that means that PV is a constant, so that the pressure is 1/2 what it was.

So we went from a state $P, V$ to a state $\frac{P}{2}, 2V$. To compute the entropy change, we have to figure out a reversible way to go from the initial state to the final state. A reversible way to do it would be to hold the temperature constant and let it slowly expand.

What would be the entropy change in that case? Well, we use:

$\delta U = \delta Q - P \delta V$

Since there is no energy change (constant temperature), then that means $\delta U = 0$ so $\delta Q = P \delta V$

Since $PV = NRT$, that means $P = \frac{NRT}{V}$ so $P \delta V = NRT \frac{delta V}{V} = NRT \delta (log(V))$

So $\delta Q = NRT \delta (log(V))$ and $\delta S = \delta Q/T = NR \delta (log(V))$

Since the volume doubled, then $\delta log(V) = log(V_{final}/V_{initial}) = log(2)$

So $\delta S = NR log(2) > 0$

So this is kind of a weird calculation. You're calculating the entropy change for a sequence of events that didn't actually happen. In this alternate reality, $\delta Q > 0$ (because a gas must take in heat in order to expand at constant temperature). In the real world, $\delta Q = 0$. But the entropy change is the same in the alternate reality as in the real world---the only thing that matters is what state you end up in.

3. Feb 26, 2017

### Arman777

I didnt understand this part.We are going initial to final so entrophy will be positive.But in real world its zero ?

4. Feb 26, 2017

### stevendaryl

Staff Emeritus
No, the change in entropy in the real world for an irreversible expansion of a gas to double its volume is: $NR log(2)$. It only depends on the initial volume and pressure and the final volume and pressure.

If the change had been done reversibly, the entropy change would have been the same, but it would have been compensated by a decrease in entropy of the surrounding system. If the volume of a gas increases reversibly, then the volume of the surrounding gas correspondingly increases. But in an irreversible change, the increase in entropy of one system is not accompanied by the decrease of entropy of another. So the total entropy of the universe increases.

The one-way nature of entropy increase is also illustrated by this example: A gas can irreversibly expand when released into a larger volume, but there isn't a corresponding case where gas spontaneously decreases its volume.

5. Feb 26, 2017

### stevendaryl

Staff Emeritus
Just to clarify once more: In the real-world example of a gas doubling its volume without the input of any heat, $\Delta Q$ is zero, but $\Delta S$ is not. Q and S are only proportional for reversible changes.

6. Feb 26, 2017

### Arman777

I guess I understand.Is my 1 st and 2 nd thinking way are correct ?

7. Feb 26, 2017

### stevendaryl

Staff Emeritus
Yes, nonconservative forces lead to entropy increases. Any irreversible change does, and energy loss due to nonconservative forces are an example of an irreversible change. And the other way around, as well: Motion under conservative forces does not lead to entropy loss.

However, there is a little bit of complication involved in the second statement. Often nonconservative forces arise through an oversimplification of a much more complicated situation. For example, we treat a block sliding on a flat surface using friction, and call it a nonconservative force. However, at the microscopic level, friction is a matter of the particles that make up the block imparting energy to the particles in the surface through electrical interactions. If you could give a detailed analysis of this interaction, you would find that it's perfectly reversible, but extremely complicated. When we simplify the calculation by assuming that the interaction between block and surface is characterized by turning the kinetic energy of the block into heat, then we introduce irreversibility as part of the modeling process.

8. Feb 26, 2017

### Arman777

I understand and it make sense,electromagnetic force is conservative force and friction is part of it.