In what circumstances does a gas do positive work?

[leroyjenkens]

Let's say we have gas and it expands against a constant external pressure. Is it always doing work, regardless if it's an adiobatic or isothermal or isobaric or reversible or irreversible, etc, process? I'm having trouble with this. It seems like if it expands, it must be doing positive work. But if you take the equation that says work is the negative external pressure times the change in volume, you're going to get a negative answer for the work done by the gas, because the volume goes from low to high.
Thanks.

 

[mfb]

If a gas expands against external pressure, it always needs energy for that - usually coming from the gas. This is independent of the details of the process.
If a gas gets compressed, it gets this energy back.

 

[Chestermiller]

Let's say we have gas and it expands against a constant external pressure. Is it always doing work, regardless if it's an adiobatic or isothermal or isobaric or reversible or irreversible, etc, process?

Yes.

I'm having trouble with this. It seems like if it expands, it must be doing positive work.

Yes.

But if you take the equation that says work is the negative external pressure times the change in volume, you're going to get a negative answer for the work done by the gas, because the volume goes from low to high.
Thanks.

The pressure of the gas at the interface always matches the external pressure. The work done by the gas is always:

##W=\int{P_{ext}dV}##

Here is a write up that I prepared to help students struggling with these confusing issues (in this write up, I call the pressure at the interface P_I, rather than P_ext). Hope it helps:

FIRST LAW OF THERMODYNAMICS

Suppose that we have a closed system that at initial time t_i is in an initial equilibrium state, with internal energy U_i, and at a later time t_f, it is in a new equilibrium state with internal energy U_f. The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let [itex]\dot{q}(t)[/itex] represent the rate of heat addition across the interface between the system and the surroundings at time t, and let [itex]\dot{w}(t)[/itex] represent the rate at which the system does work on the surroundings at the interface at time t. According to the first law (basically conservation of energy),
[tex]\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W[/tex]
where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

The time variation of [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

If a process path is irreversible, then the temperature and pressure within the system are inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperature at the interface can be measured and controlled using the surroundings to impose the temperature and pressure boundary conditions that we desire. Thus, T_I(t) and P_I(t) can be used to impose the process path that we desire. Alternately, and even more fundamentally, we can directly control, by well established methods, the rate of heat flow and the rate of doing work at the interface [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex]).

Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
[tex]\dot{w}(t)=P_I(t)\dot{V}(t)[/tex]
where [itex]\dot{V}(t)[/itex] is the rate of change of system volume at time t. However, if the process path is reversible, the pressure P within the system is uniform, and

[itex]P_I(t)=P(t)[/itex] (reversible process path)

Therefore, [itex]\dot{w}(t)=P(t)\dot{V}(t)[/itex] (reversible process path)

Another feature of reversible process paths is that they are carried out very slowly, so that [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] are both very close to zero over then entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (t_f-t_i) becomes exceedingly large. In this way, Q-W remains constant and finite.

SECOND LAW OF THERMODYNAMICS

In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate [itex]\dot{q}(t)[/itex] and the rate of doing work [itex]\dot{w}(t)[/itex] as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:
[tex]Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}[/tex]
[tex]W=\int_{t_i}^{t_f}{\dot{w}(t)dt}[/tex]
In the present section, we will be introducing a third integral of this type (involving the heat transfer rate [itex]\dot{q}(t)[/itex]) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
[tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}[/tex]
where T_I(t) is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing. He found that, for any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) was not arbitrary; instead, there was a unique upper bound (maximum) to the value of the integral. Clausius also found that this result was consistent with all the "word definitions" of the Second Law.

Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call today the reversible process paths. So, to determine the change in entropy between two equilibrium states, one must first conceive of a reversible path between the states and then evaluate the integral. Any other process path will give a value for the integral lower than the entropy change.

So, mathematically, we can now state the Second Law as follows:

[tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq\Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}[/tex]
where [itex]\dot{q}_{rev}(t)[/itex] is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the system temperature at time t (which, for a reversible path, is equal to the temperature at the interface with the surroundings). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.

 

[Andrew Mason]

Let's say we have gas and it expands against a constant external pressure. Is it always doing work, regardless if it's an adiobatic or isothermal or isobaric or reversible or irreversible, etc, process? I'm having trouble with this. It seems like if it expands, it must be doing positive work.

It depends on how you define the work. Unfortunately, there are two conventional ways of stating the first law:

1) [itex]Q = \Delta U + W[/itex] where W is work done BY the gas and Q is the heat flow INTO the gas

2) [itex]Q = \Delta U - W[/itex] where W is work done ON the gas and Q is the heat flow INTO the gas

If you define work as work done ON the gas as in 1), then negative work is work done (ON the gas) when the gas expands.
If you define work as work done BY the gas as in 2), then positive work is work done (BY the gas) when the gas expands.

AM

 

[PJC01]

I can clarify that the circumstances in which a gas does positive work depend on the type of process it undergoes. In general, work is considered positive when it is done by the gas on its surroundings, meaning the gas is expanding against an external pressure.

In an adiabatic process, where there is no heat exchange between the gas and its surroundings, the gas does positive work as it expands against the external pressure. This is because the gas is increasing its volume and pushing against the external pressure, resulting in positive work.

In an isothermal process, where the temperature remains constant, the gas also does positive work as it expands against the external pressure. However, the work done in this case is less than in an adiabatic process because some of the energy is being used to maintain the constant temperature.

In a reversible process, the gas does positive work as it expands against the external pressure, but the work done is equal to the external pressure times the change in volume, resulting in a positive value.

On the other hand, in an irreversible process, the work done by the gas may not always be positive. This is because in an irreversible process, some of the energy is lost as heat, and the work done by the gas may be less than the work done in a reversible process.

In summary, the gas does positive work in most cases, but the value of the work done may vary depending on the type of process and the energy changes involved. I hope this clarifies your confusion.