# Is a constant pressure adiabatic irreversible expansion possible?

We know that adiabatic law for an ideal gas is,

$$PV^{\gamma} =C$$

the differential of this under constant pressure is,

$$P \gamma V^{\gamma-1} dV = 0$$

Now, the pressure and volume can't be at all points in process zero, so it must be $$dV$$ that must be zero. So does that suggest such a process is not feasible at all? Why should we expect this physically?

• I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. Sep 7, 2020 at 0:52

You changed the question. I will answer both.

# General

In a closed system of a pure component, we always have a defined relationship for $$p, V, T$$ of the substance in the system. This is called the mechanical equation of state (EoS). It may be the ideal gas law. It may be the van der Waals EoS. It may be the Redlich-Kwong EoS. It may be the relationship of density to $$p, T$$. We always have a mechanical EoS regardless of the type of substance (ideal gas to solid). Therefore, only two of the three parameters in the $$p, V, T$$ set are independent for the substance in the system.

Let us make the starting assumption that we have no "other work" present.

# REVERSIBLE Constant Pressure, Adiabatic Process

A reversible process is one where, by definition, the system is in mechanical, thermal, and chemical equilibrium with the surroundings at all points in time. Here, we have no distinction whether pressure means internal or external. They are the same at all points in time. Ironically, so too are the temperature and chemical potential of the system with the surroundings. We will leave this unexplored (we end with a failure below anyway).

A constant pressure process will have constant and equal pressures inside and outside at all points in time. To change $$V$$ of the system (expand or contract), we are left therefore with only the temperature $$T$$ of the system.

The only way left to change $$T$$ for the system while allowing $$V$$ in the system as an independent parameter is to add or remove heat (allowing that we have no "other work"). The process is specified as being adiabatic. No heat flows.

Conclusion: This process is not a process. It indicates a system at static mechanical state of $$p, V, T$$, presumably in equilibrium with its surroundings at the same $$p, T$$ (based on the term "reversible").

# IRREVERSIBLE Constant Pressure, Adiabatic Process

By default, when we state the word "pressure", it is first understood to represent the pressure of the surroundings. In this case, an irreversible, constant external pressure, adiabatic process is well-represented in undergraduate textbooks on thermodynamics. The first law formulation collapses as below.

$$dU = -p_{ext} dV$$

Once we state irreversible, we drop all connection between pressure in the surroundings $$p_{ext}$$ and pressure in the system (typically just $$p$$). We have no need of the mechanical EoS for the contents of the system.

Suppose instead that you want to consider an irreversible, adiabatic process with a constant pressure $$p$$ in the system. In this case, we revert to the ground rule that $$p, V, T$$ have a defining mechanical EoS for any substance. Holding $$p$$ constant, the only way to change $$V$$ as the independent variable is to change $$T$$ as the one remaining dependent variable. The only way for us to change $$T$$ to see correspondingly what happens to $$V$$ is to add or remove heat (when we have no "other work"). This is true regardless of whether the heat is added/removed reversibly or irreversibly. Since the process is also to be adiabatic, no heat flows in/out of the system. Therefore, the $$T$$ in the system also cannot/does not change.

Conclusions: The process is one of two kinds. Either it is a basic case of expansion/contraction with constant pressure in the surroundings under adiabatic conditions. Or it is not a process, simply a definition again of a static system. The term irreversible in the second case does not give us some hoped-for clever way out of the fundamentals.

# Exception

When we allow other work, such as electrical or photonic (light) energy, to play a role, the above analysis is open to a different approach. We can heat a container with microwaves for example while having an irreversible, adiabatic, constant internal pressure process. In this case, the container volume will inevitably expand when the internal pressure is greater than the external pressure.

• How do we know that P,V and T are always constrained to show behaviour like the one you mentioned? Right, I can understand it for ideal gas law but like how can we claim it generally? Sep 6, 2020 at 19:08
• For what its worth I do have questions about the other answers that I’ll bring up with the autho Sep 6, 2020 at 20:10
• I’m not sure what you mean by “The only way to change $T$ is to add or remove heat”. Perhaps I’m missing something. You can obviously change $T$ for an ideal gas by either a reversible or irreversible adiabatic process without adding or removing heat. For an ideal gas, $\Delta T$ depends $\Delta U$ which in turn equals the work done by or on the gas. The work done by an ideal gas in an adiabatic process, reversible or irreversible, is not due heat transfer. The energy used to perform the work is at the expense of the internal energy of the system. Sep 6, 2020 at 21:00
• “Therefore, the $T$ in the system cannot change by some action from outside the system”. Again, not sure what you mean. An adiabatic compression is an “action from outside the system”. It results in an increase in temperature. The only difference between the reversible and irreversible process is the pressure for the latter is strictly the external pressure except at the piston interface until equilibrium is finally reached. Sep 6, 2020 at 21:01
• @BobD I've removed the ambiguous phrase. I cannot agree that we can change $T$ when $p$ is constant and $V$ is the independent variable. As per the original question, to OP wants to know whether $V$ changes when we do a specific process (reversible/irreversible, adiabatic, constant $p$ process). Arguing with reversible or irreversible seems to be just running in circles or interpreting the title question in a way that makes no sense. Sep 7, 2020 at 0:45

The first equation you provided only applies to a reversible adiabatic expansion of an ideal gas. That can't occur at constant pressure since an expansion or contraction would require a reversible transfer of heat to or from the gas so that the ratio of volume to temperature is a constant. That, of course, can't happen if the process is in fact adiabatic.

On the other hand, a constant pressure irreversible adiabatic expansion or contraction is possible. It requires an initial abrupt decrease or increase in the external pressure, followed by constant external pressure expansion or contraction.

Consider a vertically oriented perfectly thermally insulated cylinder fitted with a piston of negligible mass and area $$A$$. A weight of $$mg$$ is supported by the piston so that the total external pressure is $$mg/A + atm$$. The gas is initially in equilibrium with the surroundings and is therefore at the same pressure.

The weight is abruptly removed so the external pressure abruptly drops to 1 atm and there is no time for the gas to initially expand. Refer to the diagram below which shows both a reversible and irreversible adiabatic expansion. For the irreversible expansion, the pressure associated with the curve is the external pressure only, except at the initial and final equilibrium states 1 and 2a where the gas is in equilibrium with the surroundings. After the abrupt drop in external pressure, the gas is allowed to expand against the constant external atmospheric pressure until equilibrium is attained. The gas pressure only equals the external pressure at the piston interface. For the reversible expansion the pressure represents both the external and gas equilibrium pressure.

Note from the graph that the final equilibrium temperature of the gas is greater for the irreversible process than the reversible process. That means the drop in internal energy is less for the irreversible process than the reversible process, reflecting that less work is done for the irreversible process ($$\Delta U=-W$$). the area under the irreversible process (irreversible work) being less than the reversible process.

Hope this helps. • Ok, I reread this answer like two times and I am fairly certain this contradicts the other answer which was given Sep 6, 2020 at 19:26
• If you can point out the specific contradictions I can address them with you. But I don’t intend to compare their answers with mine. Sep 6, 2020 at 19:39
• Thank you, I read both a few more times then I got it. What I was missing that, he said the temperature can not be changed by external agency, I had misunderstood that to be that the temperature can not be changed at all. Sep 6, 2020 at 19:54