How can an irreversible isothermal process exist?

Question

An isothermal process is one in which temperature of the gas is constant throughout the process.

From my physics textbook:

the state of a gas is described by specifying its pressure $p$, volume $V$ and temperature $T$. if these parameters can be uniquely specified at a time, we say that the gas is in thermodynamic equilibrium.

This is always the case in a reversible process, but if we cannot specify temperature during an irreversible process, how can we say that during an irreversible isothermal process, the temperature of gas remains constant throughout?

Some people, myself included, consider an irreversible isothermal deformation of an ideal gas as one in which the boundary temperature at the interface between the gas and its surroundings is held constant at the initial gas temperature (i.e., gas in contact with ideal constant temperature reservoir) throughout the process until thermodynamic equilibrium is re-achieved. Of course, in such a process, internal temperatures within the gas are not spatially uniform. — Chet Miller, Apr 09 '23 at 20:40

score 2 · Accepted Answer · answered Apr 10 '23 at 12:48

2

how can we say that during an irreversible isothermal process, the temperature of gas remains constant throughout?

As @Chet Miller pointed out in his answer, the term "isothermal", or constant temperature, in connection with an irreversible process refers only to the temperature at the boundary between the system and surroundings, where the surroundings is considered to be an ideal (constant temperature) thermal reservoir. Since there can only be one temperature at a specific location, and given that the temperature of an ideal thermal reservoir at the boundary is considered constant (which in reality is isn't), then so must the temperature of the gas at the boundary be considered constant.

As a practical matter, for want of a better term, one uses the term isothermal for an irreversible process carried out in a constant temperature surroundings to distinguish it from a reversible process carried out in the same constant temperature surroundings. See the figure below.

For the reversible isothermal expansion the external pressure is gradually reduced so that both the pressure and temperature throughout the gas, not just at the boundary with the surroundings, is always in equilibrium with the surroundings.

For the irreversible process the external pressure is suddenly reduced (e.g., suddenly removing a weight from atop a piston) and the gas allowed to expand rapidly (irreversibly) until pressure and temperature equilibrium is reestablished. During the irreversible expansion the temperature and pressure of the gas is only the same as the surroundings at the boundary (pressure because of Newton's 3rd law). Beyond the boundary, however, temperature and pressure gradients exist.

Hope this helps.

answered Apr 10 '23 at 12:48

Bob D

71,527

So if a question doesn't provide external pressure, we can take work done in irreversible isothermal expansion to be equal to work done in isobaric part of the graph? – Aurelius Dec 30 '23 at 21:32
@Aurelius Sorry, but I don't understand your comment. If the pressure is not given how can one speak of the work done by an isobaric (constant pressure) pressure? – Bob D Dec 30 '23 at 22:23
For example there was this question where one mole monoatomic gas underwent irreversible expansion. $$(P_1,V_1,300K)\rightarrow (P_2,V_2,300K)$$ Work done was $-P_2(V_2-V_1)$ Temperature was constant and values were such that $PV=const.$ so it was isothermal. In the figure you gave for irreversible isothermal, only the isobaric blue line does work – Aurelius Jan 01 '24 at 06:43
@Aurelius I don’t know what question you referring to, but $PV = constant$ applies to a reversible isothermal process not an irreversible isothermal process, and the work done for $PV=constant$ is $RT\ln V_{2}/V_1$ not $P(V_{2}- V_{1})$. – Bob D Jan 01 '24 at 07:02
My mistake $PV\neq const$ rather $P_1V_1=P_2V_2$ and $T_1=T_2=300K$ and the answer was like I mentioned. So I assumed it was rapid isochoric change from $P_1$ to $P_2$ followed by isobaric $V_1$ to $V_2$ like your blue graph. – Aurelius Jan 01 '24 at 14:23
@Aurelius Ok, so you are talking about an irreversible isothermal expansion just as I described in my answer. Soo..What exactly is your question about my answer? – Bob D Jan 01 '24 at 16:36
Yep I was just confirming that all work done is by the isobaric part of the blue graph and equal to $-P_2(V_2-V_1)$ – Aurelius Jan 01 '24 at 17:07
@Aurelius If you mean $W=P_{2}(V_{2}-V_{1})$ and not $W=-P_{2}(V_{2}-V_{1})$, then yes. Keep in mind that $P_2$ is the EXTERNAL pressure, and not the pressure of the gas, which varies spatially until equilibrium is re-established at equilibrium state 2. – Bob D Jan 01 '24 at 17:22

Philip Wood · Answer 2 · 2023-04-09T17:34:59.353

(a) the state of a gas is described by specifying its pressure , volume and temperature .

This is not an incorrect statement, but is a little puzzling, because for a gas we need only to specify any 2 of these variables; the third is then known from the equation if state. For example, for an ideal gas the equation of state is $pV=nRT$.

(b) Here is an example of an irreversible isothermal process for a gas... A fan, driven by an electric motor, is set rotating in a leak-proof cylinder of gas. The gas temperature rises, but more and more slowly, because heat escapes at an increasing rate to the surroundings through the walls of the cylinder. Eventually the gas reaches an equilibrium temperature (with negligible variation over the volume of the gas), when work done on the gas by the fan is transferred to heat at a constant rate, and we have an irreversible isothermal process.

Note: I'm using heat in the thermodynamic sense of energy (entering or) leaving as system because of a temperature difference between the system and its surroundings.

If we're not restricted to gases, a very simple example of an irreversible isothermal process is water boiling in an electric kettle. Electrical work is transferred irreversibly to heat and to intermolecular energy in the steam, at a constant temperature of approximately 100°C.

With regard to item (a), the ideal gas law is not valid for non-equilibrium (e.g.,rapidly deforming) states. With regard to item (b), there are temperature gradients within the gas which allow heat flow to the boundary. — Chet Miller, Apr 09 '23 at 20:23
Wrt (a), for a non-equilibrium state, can we even specify a single $p,\ V$ and $T$ ? Wrt (b) (gas and fan example) fan will mix the gas so that the temperature is almost uniform. — Philip Wood, Apr 09 '23 at 20:38
In a rapid irreversible deformation, viscous stresses contribute to the overall compressive stresses, and these are not equal to the pressure, as determined by the ideal gas law, even locally. in the case of the rapid fan mixing, there is still viscous irreversibility/disspation within the gas, but the temperature variations would definitely be much less. — Chet Miller, Apr 09 '23 at 20:46

score 1 · Answer 3 · answered Apr 09 '23 at 18:13

1

To complete the previous answers, we can also cite the example of a chemical reaction. If the kinetic reaction rate is slow enough to allow thermalization processes, the reaction can be carried out isothermally and at constant pressure while being evidently irreversible.

answered Apr 09 '23 at 18:13

Vincent Fraticelli

4,841

This is the only scenario that makes sense to me. – Chet Miller Apr 09 '23 at 20:30

score 1 · Answer 4 · answered Apr 10 '23 at 11:48

Some people, myself included, consider an irreversible isothermal deformation of an ideal gas as one in which the boundary temperature at the interface between the gas and its surroundings is held constant at the initial gas temperature (i.e., gas in contact with ideal constant temperature reservoir) throughout the process until thermodynamic equilibrium is re-achieved. Of course, in such a process, internal temperatures within the gas are not spatially uniform.

score 0 · Answer 5 · answered Apr 09 '23 at 19:17

It is the uniformity of pressure and temperature throughout the system that defines the equilibrium state. To avoid the ambiguity of the meaning of $p$ and $T$ in non-equilibrium states, consider a non-equilibrium partitioning of the system: a rigid insulated is box divided into two parts, each part is in internal equilibrium with its own pressure and its own temperature. The partitioning represents a non-equilibrium state for the entire box in the sense that if the wall between the parts is removed the state of the system will change until it reaches a new equilibrium. By the same argument, if the removal of the partition results in no observable change of state we declare the two parts to be in equilibrium with each other and conclude that they both have the same temperature and pressure. For a multicomponent system we also require uniformity of the chemical potential of all components.

The partitioned system is an idealization of a gradient in temperature and pressure: while each part is individually in thermal and mechanical equilibrium, the overall system is not, because it is characterized by non-uniform temperature and pressure.

score -1 · Answer 6 · answered Apr 09 '23 at 16:50

-1

Whether you can or cannot specify a single temperature is not related to being reversible or not. There are inherently irreversible processes irrespective of its velocity, for example, the hysteretic behavior of a ferromagnet is always dissipative and thus irreversible at any rate of the process. Similarly, the plastic deformation of a metal bar is always dissipative and irreversible. And, of course, if these processes are done very slowly but in thermal contact (diabatic) with its environment then the temperature inside will be uniform and will be the same as that of the environment.

It is true though that if the process is not quasi-static then the local temperature variations may not have time equalize and there is no single temperature in the system and when it equalizes the local heat conduction results in in irreversible behavior. The two examples show a quasi-static process can be irreversible and thus quasi-static $\ne$ reversible, though it is sometimes the two are confused.

Your quote refers to a gas, but even there for an adiabatically isolated ideal gas can have isothermal irreversible process when it expands without work in an empty space. The workless expansion of ideal gas into a vacuum is isothermal, as Joule's experiment showed, but it is obviously irreversible.

answered Apr 09 '23 at 16:50

hyportnex

18,748

For the difference between quasistatic and reversible, it was an endless topic of discussion when I was still a student. In particular, for the hysteresis cycle: even if the cycle is described slowly, the Bloch walls of the domains always move abruptly (Barkhausen effect) and therefore, at this scale, the transformation is not quasistatic ? – Vincent Fraticelli Apr 09 '23 at 18:07
@VincentFraticelli that is true, for small enough (mesoscopic) scale it is "sudden" and not quasi-static, but, of course, if you go to an even smaller scale (single domain and smaller) it is not irreversible either. At small enough scales nothing is homogeneous. All these macroscopic concepts are scale dependent starting with a single molecule whose interactions are always reversible in a thermodynamic sense. But reversibility is an important concept in macroscopic physics and should be definable without referring it to the generally vaguely defined process speed. – hyportnex Apr 09 '23 at 18:26
see the definition of reversible without being quasi-static https://physics.stackexchange.com/questions/727919/defining-reversibility-without-resorting-to-entropy/727922#727922 and https://physics.stackexchange.com/questions/757539/example-of-an-irreversible-process-using-this-formal-definition/757541#757541 – hyportnex Apr 09 '23 at 18:27
In the Joule experiment you describe, during the expansion the temperature is not uniform spatially or constant with time. Only the initial and final temperatures at thermodynamic equilibrium are equal, and then only for an ideal gas. – Chet Miller Apr 09 '23 at 20:29
@ChetMiller as long as internal energy, spatially varying as it maybe, can be defined inside the ideal gas, and the gas stays ideal during the process - an assumption, the caloric equation ensures that the temperature stays constant irrespective of what happens to the homogeneity of the pressure. – hyportnex Apr 09 '23 at 21:15
@hyportnex Are familiar with the 3D tensoriial version of viscous Newtonian fluid behavior? – Chet Miller Apr 09 '23 at 22:12
@ChetMiller I am just a scalar amateur, but regarding viscosity please don't forget I was talking about an ideal gas. – hyportnex Apr 09 '23 at 22:19
See chapter 1, Transport Phenomena, Bird et al. to see how an "ideal gas" behaves in rapid irreversible deformations. It is impossible to describe ideal gas behavior properly for certain kinds of deformations with allowing to to have viscosity. – Chet Miller Apr 09 '23 at 22:38

How can an irreversible isothermal process exist?

6 Answers6