# Thermodynamics question: Irreversible process

1. Jan 7, 2010

### libelec

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

A gas expands through an adiabatic irreversibly process, from (P1, V1, T1) to (P2, V2, T2). In a PV graphic, localize both states 1 and 2, and the adiabats that pass through each of them.

3. The attempt at a solution

It's a theory question. I can locate 1 anywhere I want, but to locate 2 I have to follow the rules of thermodynamics. If the gas expands, then V2 > V1. This is what I have so far.

Now the question is P and T. I thought that, even though it's an irreversible process, there's only one adiabat that goes from 1 to 2, since adiabats don't cross each other. So, if I tried to go from 1 to 2 through a reversible adiabatic process, I could only take that adiabat. Then P2 < P1.

Is this correct?

2. Jan 7, 2010

### ponjavic

You're saying there's only one adiabat that goes from 1 to 2.

Are you sure this is correct? Because you're saying there is one reversible adiabatic process while the question asks for an irreversible adiabatic process. How would these two processes differ? Think about this for a minute.

3. Jan 7, 2010

### libelec

That is the very same root of my question. What is the difference for this? What's the difference in terms of calculations?

4. Jan 7, 2010

### sigmavirus

well volume is inv. proportional to pressure, so pressure will decrease. temperature, on the other hand, is dir. proportional, so it will also increase.

5. Jan 7, 2010

### ponjavic

Unless it's a trick question or I have misunderstood something you have neglected the term irreversible. To be fair I've never heard of the term adiabat but I assume it means the exponential PV-line for an adiabatic reversible process. Only reason I do this is because I have an idea of what they might be looking for. If I'm wrong just disregard my post unless you left some information out of the question.

The area below a curve in a PV diagram such as the curve during an irreversible (or reversible) expansion is equal to the work done. Draw such a curve for an adiabatic reversible process from point 1 to the volume at 2. Now the question is will the curve for the irreversible process be above or below the previously drawn curve? Here you have to think about what irreversible means and how it relates to work. To finish off just draw a somewhat paralelle curve from point to to the volume at 1 (paralell to the first adiabat you drew).

I think it is an ill posed question and I may certainly have misunderstood.

6. Jan 8, 2010

### libelec

Yes, that's what I meant with adiabat.

I know that in a reversible process, the work done is maximum, so I assume the curve should be below the other one. But still, I don't understand how this helps me with how to locate point 2. It (the location) still lingers from the hypothesis I made at the beginning.

7. Jan 8, 2010

Anybody?

8. Jan 8, 2010

### Andrew Mason

You have to use the relationship between P and V during an adiabatic process:

$$PV^\gamma = K$$

AM

9. Jan 9, 2010

### libelec

So, let me see: PV$$\gamma$$ must remain constant. Even though it's an irreversible process, this stands, right? So, then, since V2 > V1 and $$\gamma$$ remains the same, then P1 > P2. Same with the temperature, because from PV$$\gamma$$ = TV$$\gamma$$-1, and $$\gamma - 1$$ remains the same, then, because of the same as before, T1 > T2. Is this correct?

And how about the adiabats in each point? Is is the same adiabat that goes from 1 to 2 like I said, or are there two different adiabats?

10. Jan 9, 2010

### Andrew Mason

No. It applies to an adiabatic process that is in constant equilibrium - i.e reversible. In irreversible adiabatic process, such as a free expansion of a gas, the relation does not hold.

There is only one reversible adiabatic curve through each point. It is the curve defined by the adiabatic condition $PV^\gamma = K$. So you have to draw the adiabatic curve from point 1. Where is point 2 in relation to the adiabatic curve through point 1?

AM

11. Jan 9, 2010

### libelec

Then I'm in the same situation than before. I know that V2 > V1, so then V2$$\gamma$$ > V1$$\gamma$$ (since both are positive magnitudes). But I can't find a way to relate both Ps.

12. Jan 10, 2010

### Andrew Mason

Use PV=nRT

All you can do in answering this question is to first identify two points (P1,V1,T1) and (P2,V2,T2) connected by a possible irreversible adiabatic expansion path from 1 to 2. To do this you draw the reversible adiabatic path through 1. This is governed by the adiabatic condition:

$$PV^\gamma = K$$

which, since T = PV/nR, can be expressed:

$$TV^{\gamma-1} = PV^\gamma/nR = K/nR$$

If the process is not reversible less than the maximum potential work will be done. Since it is adiabatic, this means that the internal energy will be higher than in the reversible path. Use that to locate the possible points (P2,V2,T2) that result from an adiabatic irreversible path.

AM

13. Jan 10, 2010

### libelec

OK, let me know if this is OK, please:

I choose a random point 1:

Then, I draw the adiabat that goes through 1:

If this was a reversible process, then the blue 2 would be point 2. But now, since the process is irreversible and as a consecuence the work made by the gas is lesser, then the internal energy in 2 will be greater. As a consecuence, the temperature of point 2 must be higher than the temperature of the blue 2. Say the two isotherms Ta and Tb are such that Ta < Tb. Then (moving the blue 2 higher):

Point 2 will be located in the isoterm Ta, since it has a larger amount of internal energy (left?). Then finally this is what I have:

With each point associated with two different isotherms T1 and T2. Then there should be a different adiabat that goes through 2.

Is this correct?

14. Jan 11, 2010

### Andrew Mason

If the final temperature is higher for a given volume, the final pressure will be higher. The path, however, results in less work being done. This means the area below the path is less than the area below the reversible adiabatic path.

AM

15. Jan 11, 2010

### libelec

But there's no way that point 2 is associated to a P2 higher than that of the blue 2, because there's no path from 1 to that 2 that encloses a smaller area than the path from 2 to the blue 2.

16. Jan 11, 2010

### Andrew Mason

Sure there is. For example, if P drops to P2 immediately and expansion occurs at P = P2, the work done will be less. Or, if P drops to 0 immediately and pressure gradually increases as expansion occurs, the work done will be less (this is what occurs when a gas expands into a chamber that is initially at 0 pressure (vaccum).

AM

17. Jan 13, 2010

### libelec

And that can still be done adiabatically?

18. Jan 13, 2010

### Andrew Mason

Sure. Just make the process thermally isolated from the surroundings. As the gas expands, the internal energy will decrease by the amount of work that is done (dQ = 0, so dU = -dW). A reversible expansion (the gas is in thermodynamic equilibrium at all times) results in the maximum amount of work being done. So that is the limiting case. The work done is a function of the external pressure. If the external pressure is between 0 and P1 the gas will expand until the internal and external pressures are equal.

AM

19. Jan 13, 2010

### libelec

I've got it. Thank you very much.