What does it mean for work to be done ON a system?

What does it mean for work to be done ON a system?

For a given system, e.g. an apple, an external force can be applied to the system, e.g., the force of gravity. We say that gravitational force is a force "on" the apple "by" the earth. We could make this explicit with ordered subscripts like $\vec F_{a,e}$, indicating "on" the apple ("a"), "by" the earth ("e"). When the apple moves a distance $d\vec x$ the force of gravity performs work "on" the apple $$ W_{a,e} = \int d\vec x \cdot \vec F_{a,e} $$

Suppose we have a system consisting of an ideal gas enclosed by a cylinder and a movable piston.

The system of interest in pedagogical questions of this form is usually the gas, not the cylinder. The cylinder is just there to contain the gas and the piston is there to allow the volume of the gas to change.

We say that if we were to compress the gas by letting the piston extert a net force on it (perhaps by increasing the pressure outside the piston), we are doing pressure-volume work ON the system.

Yes, if we apply a force $\vec F$ on the piston, which we assume thereby applies a force to compress the gas by an amount $dV = dx A$, where $A$ is the area of the piston, the net force on the gas is in the direction which causes compression and the work done "on" the gas is $$ dW_{on} = |dx||F| = |dx||A P| = |dV||P|= -PdV\;, $$ where I put absolute value symbols based on the fact that the cylinder and piston cause the force $\vec F$ to be in the exact same direction as the compression displacement $\delta\vec x$ so that $\vec F\cdot d\vec x = |F||dx|$;.

Suppose instead, we were to scrape the surface of the cylinder enclosing the gas with our finger against friction.

You are now changing the thing you are considering the "system." Previously the gas was the system of interest and the cylinder was a convenient way enclose the gas and to apply a force of a given direction on the gas via the piston.

We would still be doing work, and we could calculate its magnitude and all properties we might be interested in, but we have not done work ON the system.

You might still be doing work on some system, but it is not the same system that you initially considered. The usual setup of a pedagogical cylinder full of gas does not specify how frictional heating of the cylinder can cause a change in energy of the gas. Certainly, if the only way for the gas volume to change is via the piston then the scraping of the cylinder has no clear way to do work.

If you want to further specify that the cylinder is not insulating and that the scraping and friction causes heat $d Q$ to flow into the gas, then the energy of the gas can change by $$ dU_{no work} = dQ_{into} $$

If you allow for both heat transfer into the gas and work to be done on the gas then: $$ dU = dQ_{into} + dW_{on} = dQ_{into} - PdV = TdS - PdV\;. $$ where the last equation holds when the heat transfer $dQ$ can be written in terms of the entropy change $dS$ as $dQ=TdS$, and that last equation is called the "fundamental thermodynamics relation."

What is the principal difference between these types of work? How do we know one constitutes work done on the system while one doesn't?

We know based on the problem setup and the fact that we consistently use the word "system" to refer to the same thing in a given problem.

Doing work on the system means increasing the $PV$ state of the gas in the system. Ideal gas law says that $PV=nRT$, and that internal energy $U=C_V T$. Therefore $PV$ is directly proportional to $U$, and increasing the $PV$ value of the gas by mechanical work is the same as increasing the internal energy. The details depend on how the $PV$ work is added:

$$W = \int P~dV $$

e.g. for isentropic:

$$= \int \frac {k}{V^\gamma}~dV$$

The properties and work done on the gas is what is relevant. The piston and cylinder is just common equipment that would allow us to manipulate this. Your example of putting frictional drag into the cylinder might marginally heat the gas, but not affect it otherwise.

What is rather seldom explained, perhaps because it is taken to be so obvious it doesn't need an explanation, is what it means to do work on a system as opposed to just doing work in general.

Indeed, the meaning of work in statistical physics is somewhat different from its meaning in mechanics. In statistical mechanics we distinguish two ways of transferring energy to a system: $$dU=\delta Q + \delta W$$

• energy transfer on the microscopic level - that we call heat, $\delta Q$
• energy transfer via changing macroscopic variables of the system, which we call work, $\delta W$

This distinction is somewhat ambiguous, but usually does not pose much problem in specific cases, where it is agreed which quantities are considered as macroscopic variables. The distinction makes sense, since the goal of statistical physics is to reduce the description of a system with a very large number of degrees of freedom to a description in terms of a few macroscopic variables, so heat pretty much refers to all the energy changes that cannot be accounted for via the variables agreed upon.

Note that, although statistical physics textbooks usually deal with a case of an ideal gas, there may be many other quantities serving as macroscopic variables equivalent to volume and pressure - e.g., magnetization and magnetic field is one commonly used pair: $$ dU = TdS - pdV + \sum_i\mu_idN_i\longrightarrow dU = TdS - \sum_i X_idx_i + \sum_j\mu_jdN_j. $$

Related:
Why do we use different differential notation for heat and work?
What is the differentiating factor between work and heat?

Both the fingernail and piston compression are examples of a force exerted on the system. But the response for both is different. I'm assuming the cylinder remains at rest in some inertial frame.

When pushing on the piston, the gas compresses. When scraping on the cylinder, we can assume that the rigid walls have almost no deformation. By $W = f \times d$ , no work is done by the scraping, since there is no displacement.

Friction does decelerate your finger doing negative work on it, allowing some of that energy to turn to heat. That heat can enter the cylinder, but does no work.

It's possible that instead of the cylinder remaining at rest, that your scratching is sufficient to accelerate it. In that case work is done on the system, but it results in increasing the KE of the system. As that energy is separate from the internal energy of the gas, we aren't normally interested in it for a gas compression problem.

What is rather seldom explained, perhaps because it is taken to be so obvious it doesn't need an explanation, is what it means to do work on a system as opposed to just doing work in general.

As already stated in the other answer, work being done on the system means energy is be transferred to the system from something else outside the system. This needs to be differentiated from work being done by the system on something else outside the system, which means the system is transferring energy to that something else.

The distinction is important to determine the effects of all work being done on or by the system.

An example from mechanics is in applying the work-energy theorem the net work done on the system equals its change in kinetic energy or

$$W_{net}=W_{onsys}-W_{bysys}=\frac{1}{2}mv_{final}^{2}-\frac{1}{2}mv^{2}_{initial}$$

In the case of thermodynamics, work done on the system or by the system increases or decreases the internal energy of the system, respectively, in absence of heat transfer. The difference is critical in applying the first law.

Hope this helps.

Work is always force times displacement. But displacement is not absolute: it refers to changes in the coordinates we use to describe the system. For a gas, the system is described macroscopically by the coordinates of its boundary. If a force acts on the boundary and the boundary moves, then work is done on the system. By comparison, suppose a force acts on an individual molecule and moves it. Work is indeed done on the molecule. But since the macroscopic coordinates do not change, we do not count this as work on the gas. At the macroscopic level, this microscopic work is treated as heat added to the system.

It means that energy is transferred into that system.

As others have pointed out, there is very little difference between work done by part of a system onto another part of the system versus work done on a system by an external force. However, there is a subtle semantic difference. If the work is being done within the system, that work is something that can be reasoned about using the system. If I say there's a piston doing work on a gas, I should be able to trace the physical thing doing the work (the piston) and how it is doing work on the gas.

With work done on a system, the subtle difference is that you're not really worried about how that work is being done, just that it is indeed being done. As an example of how this behaves, one does not worry about the conservation of energy with such outside forces - only the conservations that occur within the system. If one is considering a human acting on a system that is a pile of books, it may lift those books up, adding potential energy to them. If we were to try to book-keep the source of that energy, as we often do with work done within a system, we would have to consider all the nuanced complexities of how the human metabolic system works!

A nice thing about this is that any results you get based on the mere fact that work is done on a system are applicable no matter how that work got applied. It doesn't matter if a complex human uses the Krebs cycle to generate the ATP needed to power muscles to lift the books or if a robot uses electric power to spin a motor. If they do the same work, the consequences of both cases is the same!

I think that you confusion starts with the definition of the system under consideration.
You have a "system" and its "surroundings" and between them there is a boundary.
Usually, in the example that you have given the system is the gas and boundary is demarcated by the cylinder and piston which is part of the surroundings.
So it is the gas which has to "feel"/"be in direct contact with" the force doing the work on the system when that force suffers a displacement.
A moving piston exerts a force on the gas and there is a displacement, thus work is done on the gas.

The first law of thermodynamics relates the change in internal energy of a system to heat transfer though the boundary and external work done on the system.

Suppose we have a system consisting of an ideal gas enclosed by a cylinder and a movable piston.
Then this is a different system from the one mentioned above in that now you have included the piston and the cylinder as part of the system.
This makes the analysis more difficult.
Work done is by the frictional forces on the outside of the cylinder or it could be between the cylinder and the piston.
This results in an increase in the internal energy of the system (gas + cylinder + piston) and I would class it as external work done on the system rather than heat transfer into the system as the heat transfer occurs within the cylinder/piston which is part of the system.

But note that if the gas alone is the system then the increase in internal energy is due to the heat into the system as the cylinder (external) is at a higher temperature than the gas (internal).

Much before your time there was a debate as to link between "work done" measured in ergs/joules/foot-pounds (force) and "transfer of heat" measured in calories/British thermal unit.
A quantity called the mechanical equivalent of heat was measured using a paddle wheel in water apparatus with the paddle wheel made to rotate by a falling weight.

The paddle wheel in water did the "heating" via viscous forces and the falling weight did the external work.

This is your example but with water instead of a gas.

To do work on a system means that in the latter state the system's internal energy has increased. For example: let's say we have a ball with mass $m$ as an entire system on earth. It sits at point $\vec{P}$ some mysterious actors move it at $\vec{P} + \hat{j}h$, i.e they move it upwards. It now has $mgh$ more gravitational potential energy. The mysterious actors have done work on the system. The "mysterious actors" very well could be a kid kicking the ball up where the normal force of his foot would do the work. It wouldn't depend if the ball traveled a curve-like path or a straight path, though. Regardless of path it would have gained the same energy.

Of course what I have described and the concept of work in thermodynamics are not the same thing, but conceptually they are similar and ultimately: intertwined. This is why you are confusing them. Scraping the surface of a piston does not do work on the gas, but it does work on the surface of the piston against the friction on the surface. It did not change the average kinetic energy of the molecules of the gas in the cyclinder, but it did change the state of the surface of the piston and therefore generated, released energy; therefore did work.