# Gas Expansion in insulated cylinder (piston & Diaphragm)

1. Sep 13, 2014

### pyroknife

I attached an image of two expansion cases I am analyzing. Both cases involve an insulated cylinder that is divided by a separating element (diaphragm for case A and piston for B). The portion on the right is evacuated. The left contains a calorically perfect gas with an initial pressure and temperature.

Case A is a thin diaphragm and at a certain point in time, it is suddenly removed (removed without friction&flow disturbance) and gas expands to fill the whole cylinder. Case B is a piston which is allowed to move (without friction) while gas expands to fill the whole cylinder.

Would both of these cases be considered free expansion?

#### Attached Files:

• ###### Untitled.png
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2. Sep 14, 2014

### voko

Is the piston massive or massless? What difference does that make?

3. Sep 14, 2014

### pyroknife

I believe it is massless. Being massless would imply it is free expansion?

4. Sep 14, 2014

### voko

A massless piston not backed by anything would not impede the expansion in any way.

5. Sep 14, 2014

### pyroknife

Got it. I was speaking to a professor about case B. He mentioned that the piston is connected to the outside. Does this change anything in the physics of the problem?

6. Sep 14, 2014

### voko

Most likely it does. What are your thoughts?

7. Sep 14, 2014

### pyroknife

I am thinking that if it's connected to the outside it does work on the surroundings.

8. Sep 14, 2014

### voko

That's what I think, too.

9. Sep 14, 2014

### pyroknife

Hmmm, I am trying to analyze which one of the cases would result in a higher final pressure.

For case A, I believe this is irreversible free expansion process, with no heat transfer and work, thus change in internal energy is zero. Since the system involves a perfect gas where internal energy is only a function of temperature, that would mean change in temperature is zero, making the process isothermal. Thus we can use the simple P1*V1=P2V2, where state 1 is the initial state and state 2 is the final state after the gas expands to cover up the whole cylinder.

For case B, I am a bit lost on how I can go about figuring out the final pressure. Would it be at constant process expansion since work is done on the surroundings?
For case B to expand, wouldn't heat need to be applied to the gas?

10. Sep 14, 2014

### voko

I believe you need to assume that the gas can expand in case B without any external heat. I agree with your method for case A.

11. Sep 14, 2014

### pyroknife

Hmmm, I see. So the gas has to create a expansion force large enough to push the cylinder and expand?

12. Sep 14, 2014

### voko

Yes, it will have to do work on the surroundings.

13. Sep 14, 2014

### pyroknife

I made the assumption that case B was an adiabatic expansion process. So this means the pressure inside has to be larger than outside for it to be expanding. Case B is not free expansion.
If I were to obtain the final pressure for case B, would I have to apply the work equation?
I feel there is a much simpler way.

14. Sep 14, 2014

### voko

You won't be able to unless you know exactly what the "surroundings" are. But that is not required if you just need to state whether the final pressure is lower or greater than in case A.

You know that in case A $p_1 V_1 = p_2^A V_2$. You know that in case B you will have some $p_2^B$. What do you know about $p_2^B$ without going into details of the process, and knowing solely that some work is done?

15. Sep 14, 2014

### pyroknife

Does your subscript 2 represent the properties associated with the initially evacuated volume or the final state after the gas expansion?

16. Sep 14, 2014

### pyroknife

Since the initial volume has to have a smaller pressure in order for the gas filled volume to expand, that means in the final state, the pressure of the gas should be lowered because work is done on the surroundings.

17. Sep 14, 2014

### voko

Subscript 2 means "final state" (same as in #9 I believe).

Your logic in #16 is not flawless. Take again you reasoning from #9. We know that $V_2 > V_1$, so $p_1 V_1 = p_2 V_2$ implies that $p_2 < p_ 1$ even though no work has been done (this $p_2$ is what I denoted as $p_2^A$).

18. Sep 14, 2014

### pyroknife

Is there a formatting issue in your last post? If not, your post appears to be displaying wrong for me.

19. Sep 14, 2014

### voko

I do not see any problem there. What seems wrong?

20. Sep 14, 2014

### pyroknife

I refreshed it and it looks fine now. It looked weird earlier (see attached).

I can no longer use the relationship P1V1=P2V2 because this is no longer an isothermal process and thus, there is a change in temperature.

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