# Entropy change in a reversible isothermal process.

1. Sep 18, 2011

### Urmi Roy

1. The problem statement, all variables and given/known data

In a reversible isothermal expansion of an ideal gas, as the gas expands, heat is supplied to it, so that the temperature remains constant. Thus, temperature and hence kinetic energy of the molecules does not change but the 'disorder' of the gas increases as it occupies a greater volume.

But then how can any isothermal process be reversible (since del(S) in reversible process=0)?

Such isothermal process may be quasi-static, but a quasi-static process need not be reversible.

Moreover, the above process is not spontaneous, since it says a reversible isothermal process is not spontaneous (though has delS>0), so increase in entropy does not necessarily imply spontaneity?

2. Relevant equations

delS = delS(transfer) + delS(generated)

3. The attempt at a solution

I asked a professor and he said that I was "mixing up the calculation of the entropy change with the experimental measurement.
The change in entropy in expansion of ideal gas from V to 2 V is R ln(2). This is calculated from the expression of the entropy of ideal gas.
It is the same if the process is done isothermally or in isolated system . All that matters is the initial and the final states. And you do not have to tell how you reached from the initial to the final state.

The story is different when you talk about measuring the change in entropy, then it depends on how you do the process, and whether work is done or not. It also matters if you do it quasi-statically or not."

CAN ANYONE EXPLAIN THIS THING THAT HE SAID??

2. Sep 18, 2011

### vela

Staff Emeritus
The entropy of the gas is a state variable. If you know the state of the gas, you know it's entropy regardless of how it attained that state. So the change in entropy of the gas only depends on the initial and final states.

When you ask if the change is reversible or irreversible, you have to look at how the gas interacts with rest of the universe. If the process is reversible, the change in entropy of the gas is offset exactly by the change in entropy of the environment so that the total entropy of the universe remains constant. For an irreversible process, the total entropy of the universe increases.

3. Sep 19, 2011

### Urmi Roy

So in my isothermal reversible expansion, it is reversible......though the entropy of the
gas increased, the entropy of the environment decreased just that much, so that the net entropy change was zero.

In order to have the entropy increase of the system exactly cancel the entropy decrease of the environment, we must do it quasi-statically, right?

But on the other hand, if we have an isolated system, and say an ideal gas expands against vacuum quasi-statically, its still not reversible.....since net entropy change is >0....so a quasi-static process need not be reversible...is that okay?

4. Sep 19, 2011

### vela

Staff Emeritus
Yes, that's right. Reversible implies quasistatic, but quasistatic doesn't imply reversible.

5. Sep 19, 2011

### Urmi Roy

Thanks, Vela :-)

6. Sep 19, 2011

### Andrew Mason

How would you have a gas expanding quasi-statically against a vacuum? Quasi-static is not the same as a very large number of very small free expansions.

AM

7. Sep 19, 2011

### Urmi Roy

But it says so on wikipedia and other texts I've read...

http://en.wikipedia.org/wiki/Quasistatic_process

8. Sep 20, 2011

### Andrew Mason

I don't see in that article where it says that an expansion against a vacuum can be a quasi-static process.

By the way, this article is misleading if not incorrect in describing a process that is quasi-static but not reversible. It says "An example of a quasistatic process that is not reversible is a compression against a system with a piston subject to friction".

This is incorrect. If the expansion process is reversible if it lifts a weight, it will be reversible if the output work is wasted as friction. The output does not matter. An ideal reversible (Carnot cycle) engine does not cease being reversible simply because the output ends up being wasted as heat through friction. "Reversible" simply means that the engine operates at thermal equilibrium with its surroundings (i.e. the Carnot engine's infinite-heat-capacity hot and cold reservoirs) throughout the process. What the output is used for doesn't matter. The work can be wasted or it can be completely saved and still be a reversible (ie. $\Delta S = 0$).

AM

9. Sep 22, 2011

### Urmi Roy

But then when the work is being done in lifting the weight, its basically increasing the potential energy of the weight, which can be regained in the same magnitude when the weight is let to drop again.
However, in friction, we can never regain the work as work done against friction can never be regained....

10. Sep 22, 2011

### Andrew Mason

You are correct that friction is non-reversible. But "reversible" normally refers to the thermodynamic process by which the useful work is produced. If a thermodynamic process of a system is "reversible" this means that the system and surroundings were arbitrarily close to equilibrium during the entire process. In such a case there is no change in entropy of the system and surroundings. If the reversible process produces useful work and you take the output work and use it irreversibly entropy of the universe will increase overall.

AM

Last edited: Sep 22, 2011
11. Sep 22, 2011

### pabloenigma

Sir, the definition you stated is precisely the definition of a quasi-static process.If in addition to being quasi-static,there are no dissipitative effects like friction or viscosity,only then is a process reversible

I can tell you,you are going against every standard thermodynamics text in the world. IT IS NOT REVERSIBLE IF OUTPUT IS WASTED DUE TO FRICTION.And a carnot cycle ceases being reversible,just because output ends up being wasted as heat through friction.

The Actual definition of reversibility is,the system and the surroundings can be brought back to their exact initial state.Once work is converted as heat,it is impossible to recover the whole of it to work(this is essentially what thermodynamics is about).What you describe as being reversible is the description of a quasi static process.

Whatever is quoted from the wikipedia article is absolutely correct.
And your closing argument that $\Delta$S CAN BE 0 when work is wasted as heat through friction,is the most incorrect thing I have ever heard in thermodynamics

12. Sep 23, 2011

### Andrew Mason

All I am saying is that normally one does not take into account the use that is made of the output work from a thermodynamic process in determining whether that process is reversible.

If I have a Carnot engine and I run it for several cycles and lift a weight with the output work, I think you would agree that I have a reversible engine.

Now after running the engine and lifting the weight, I stop the engine. The engine is not running. I take the weight and let it crash to the ground. Does the fact that the weight crashes to the ground wasting all that output work from the Carnot engine make the (now stopped) Carnot engine a non-Carnot (non-reversible) engine?

I am not saying that the entropy change of the universe is 0 if work is wasted as friction. It clearly is not.

I am just saying that we usually do not take into account the use that is made of the output work when characterizing the efficiency or reversibility of a thermodynamic process. For example, we don't need to know what use is made of the output of a Carnot engine in order to determine its efficiency.

AM

Last edited: Sep 23, 2011
13. Sep 23, 2011

### pabloenigma

oh..then Im sorry, I think i misunderstood you.I apologize.
I thought you said its misleading to say a process can be quasi-static but irreversible