# I Work in a process which is both adiabatic and isobaric?

Tags:
1. Jul 8, 2016

### Soren4

In a adiabatic process (not necessarily reversible) from $V_a$ to $V_b$ the work can be written as $$W=\frac{p_aV_a-p_bV_b}{\gamma-1}$$ Where $\gamma= \frac{c_p}{c_v}$

Suppose that the adiabatic process in question (again, not necessarily reversible, so $pV^{\gamma}$ can also not be constant) is also isobaric: then $p_a=p_b=p$ so
$$W=\frac{p(V_a-V_b)}{\gamma-1}\tag{1}$$
On the other hand in any adiabatic process $$W= p(V_b-V_a)\tag{2}$$
The expressions $(1)$ and $(2)$ are different, so which of the two is to be consider the right one in this case?

2. Jul 8, 2016

### Irene Kaminkowa

We get the first equaltion this way
$$W=\int_{(a)}^{(b)}pdV=p_aV_a^ \gamma\int_{V_a}^{V_b} \frac{dV}{V^ \gamma}=p_aV_a^ \gamma\left ( \frac{V^{1- \gamma}}{1- \gamma} \right )|_a^b =\frac{p_aV_a^ \gamma\left (V_a^{1- \gamma} - V_b^{1- \gamma}\right )}{\gamma -1}=\frac{p_aV_a - p_bV_b}{\gamma -1}$$
As you see, the requirement
$$p_aV_a^ \gamma=p_bV_b^ \gamma=pV^ \gamma = const$$
is necessary.

3. Jul 8, 2016

### Tazerfish

EDIT: I am stupid
As Irene Kaminkowa pointed out, the formula is correct.
First off:
Isn't an adiabatic process DEFINED by not having any heat flow? Then it is reversible and isentropic per definition.right ?
And you can directly derive this $pV^{\gamma}= constant$ from the requirement of no heat flow.

You can compare it to this http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html

PS: If the pressure is the same in the initial and final state there can not be any work done...
Since if $p$ doesn't change $V$also doesn't change due to $pV^{\gamma}= constant$

Last edited: Jul 8, 2016
4. Jul 8, 2016

### Irene Kaminkowa

Tazerfish
formula (1) is right. See the transformations above.

Soren4
Conclusion: if you want a process to be not only adiabatic but also isobaric, then all your system parameters - (p,V,T) - will remain constant.
To prevent speculations about γ, it's also constant, or - again, speaking formally - we cannot get formula 1.

5. Jul 8, 2016

### Staff: Mentor

This is incorrect for an adiabatic irreversible process.

For any expansion or compression process, the work done by the system on the surroundings is always given by $$W=\int{p_{surr}dV}$$, where $p_{surr}$ is the force per unit area exerted by the surroundings on the gas at the moving interface. For a reversible process, the pressure of the gas is uniform and matches the pressure of the surroundings. But for an irreversible process, the gas pressure is non-uniform and, moreover, viscous stresses contribute to the force that the gas exerts on the surroundings (and vice versa). So, for the case where the gas is initially at equilibrium at a pressure $p_1$, and the surrounding pressure at the interface is suddenly dropped to a lower value $p_2$ and held at that value until the gas reaches a new equilibrium (at a new volume), the work done by the system on the surroundings in this irreversible expansion is just $$W=p_2(V_2-V_1)$$
Note that $p_1$ is not even present in this equation, except to the extent that it relates to $V_1$.

For more on how to determine the work done in reversible and irreversible expansions, see my Physics Forums Insights article: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
As I said above, for the case where the expansion is irreversible at a constant lower pressure, the work is $W=p_2(V_2-V_1)$. Thus, Eqn. 2 is correct and Eqn. 1 is incorrect for this irreversible adiabatic expansion.

If you would like to carry out a full analysis of the adiabatic irreversible isobaric expansion case, I will be glad to help you. Unlike what you have been told so far, it will certainly give a different result than the adiabatic reversible isobaric expansion case.

Last edited: Jul 8, 2016
6. Jul 8, 2016

7. Jul 8, 2016

### Irene Kaminkowa

Wrong. For the processes I consider.
Simple logic
$$pV^\gamma = const~ AND ~p=const~AND~ \gamma = const\rightarrow V = const$$

8. Jul 8, 2016

### Aniruddha@94

How did I miss that! Sorry. So it means processes which are adiabatic and isobaric ( and reversible) aren't possible?? ( since no state function changes)

9. Jul 8, 2016

### Irene Kaminkowa

Our equations bind macro-parameters, while we try to describe micro-world.
What is more, we use the model of "ideal gas". Here, one may feel free to grunt that our world is far from "ideal".

How can we implement the "adiabatic and isobaric" processes in practice? To create adiabatic conditions we are to isolate our reservoir, so, the external temperature impact is forbidden. Isobaric conditions demand to fix the pressure. Then, all macro-parameters in our "ideal" system must be stuck.
Since life goes on, the particles continue their chaotic movements, so p,V,T can experience fluctuations, even within the given model. But these fluctuations are negligible.

By the way, Chestermiller,
I wonder, how we can obtain "1) adiabatic, 2) isobaric, and 3) irreversible" process experimentally?

Last edited: Jul 8, 2016
10. Jul 8, 2016

### Staff: Mentor

Yes.

11. Jul 8, 2016

### Staff: Mentor

As I said, if the process is irreversible, the micro parameters vary spatially throughout the system (temperature and pressure), so the ideal gas law (or other equation of state) does not apply macroscopically to the gas. In addition there are viscous stresses present during an irreversible expansion or compression such that, even at the piston face, the force per unit area is not determined by the local version of the ideal gas law $p=\rho RT$ (where $\rho$ is the local molar gas density and T is the local temperature). There are additional (viscous) contributions to the force per unit area related not just to how much the volume has changed but also how fast the volume is changing.
What we mean when we use the term irreversible isobaric process is that we suddenly change the force per unit area at the piston face at time zero and hold it at that value during the entire expansion or compression. So, in terms of pure semantics, it is true that the constant pressure during the process is not equal to that which was present immediately before the initial change. But, be that as it may, this is what we mean by an irreversible isobaric process.
We have complete control over the force that we apply at the piston at any time. All we need to do is move the piston as fast or as slow as we desire while measuring the force per unit area acting on the piston face (say, by using a flush mounted pressure transducer) and adjusting the piston speed as necessary to hold the force constant. It turns out that the work done by the gas this way can be shown to be the same as that obtained if we suddenly removed or added a weight from the top of the piston and allowed the system to re-equilibrate. In this latter case, force of the piston is not constant throughout the expansion, but, in the end, the total amount of work will be exactly the same.

12. Jul 8, 2016

### Irene Kaminkowa

Thus, to illustrate irreversibility we have to carry out two experiments.
In both of them we isolate our reservoir so that δQ = 0 (adiabatic conditions).
Assume the reservoir is a cylinder.
1) reversible process.
- measure the pressure pA when the piston is at the initial Level A;
- move the piston up to Level B smooth and slow. Measure the pressure pB;
- move the piston down, also quite smooth and slow, back to Level A. Measure the pressure pF (final).
CLAIM: pA = pF (the same volume and the same pressure, since during the whole experiment $pV^ \gamma = const$)

2) irreversible process.
- measure the pressure pA when the piston is at the initial Level A;
- move the piston up to Level B sharp and fast. Measure the pressure pB;
- move the piston down, back to Level A (as you like - either slow or fast). Measure the pressure pF.
CLAIM: pA ≠ pF (the same volume but the pressures differ, since $pV^ \gamma \neq const$). We are not back.

Good subject for a lab )

13. Jul 8, 2016

### Staff: Mentor

Correct.
This is not the process I was describing, but it certainly would be irreversible. The final pressure $p_F$ would be higher than $p_A$ because the final temperature would be higher. This would be the result of viscous dissipation of mechanical energy to internal energy during the process.

14. Jul 9, 2016

### Aniruddha@94

What? Now I'm confused again. Don't reversible adiabatic processes follow the curve $PV^\gamma=const$. Then if we impose the restriction that $P$ is constant, $V$ also doesn't change (and neither does $T$).
That means such a process (adiabatic and isobaric) cannot occur. Correct?
ps. I'm talking only about ideal gases and reversible changes.

15. Jul 9, 2016

### mukul

I was about to post the same question, and to my surprize I saw that the same questions is already on the top of the discussion list. Really surprizing. :-O
Actually I know an answer to this question, and my answer is different from what most of the teachers teach.
I want opinion of members here about my working.
My working is as follows.

Lets start by describing the scenario first.

lets say there is some gas in the chamber which has a frictionless piston whose mass is 10kg. The system is in equilibrium ie the pressure inside the chamber matches with the constant pressure outside P0. lets forget gravity for time being for the sake of simplicity (or just consider that the motion of piston is horizontal). Now lets say the piston is compressed so that volume of chamber is reduced to half. The piston is then released. The walls of the chamber and the piston is made of insulating material. Will the process be reversible or irreversible.

According to me (if we are considering frictionless system), the process will be reversible and will be oscillatory.
Following are the equations which were used to derive that the motion is oscillatory

PV = nRT
dQ=0
(P-P0)A = ma (where a is acceleration of piston)
V=V0 + xA (where x is displacement of piston)

16. Jul 9, 2016

### Irene Kaminkowa

1) Go from thermodynamics to statictics.
2) Consider entropy as the measure of order vs chaos.