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The first law states that

$\text{d}Q = \text{d}W + \text{d}U$

where $\text{d}W$ is the work done by the system on the surroundings.

As far as I remember from my chemistry courses,first law in chemistry is

$\text{d}U= \text{d}Q + \text{d}W$

where $\text{d}W$ is the work done by the surroundings on the system.


My question :

Is the work done by system on surroundings is always equal to work by the surroundings on the system?


P.S.

  1. I believe that the above question must be true as we refer to the same law in Physics and Chemistry.
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    $\begingroup$ Some texts define work done on the system as positive and some texts define work done on the system as negative. This leads to the type of differences seen in your two thermo equations. $\endgroup$ – David White Aug 12 at 20:28
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    $\begingroup$ It does not mean anything in my view, even if I agree that this convention is widely used increasing the confusion. The work of a force is a well defined notion with a sign perfectly defined and not subjected to conventions. $\endgroup$ – Valter Moretti Aug 13 at 6:57
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For the usual simplifying assumptions where the system boundary has negligible inertia, yes, they are the same. For mechanical work by Newton's third law. $$F_\text{system on surroundings}=-F_\text{surroundings on system}$$

and these two forces act at the same location(s). Therefore

$$W_\text{system on surroundings}=-W_\text{surroundings on system}$$

While there are many different systems you can consider (e.g. see FakeMod's answer), in general when the basic form of the first law is described between Chemistry ($\Delta U=Q-W$) and physics ($\Delta U=Q+W$) the two $W$ variables are indeed just related by a negative sign.

To be more precise one would need to be more careful with energy conservation. However, I cannot be more specific without considering more specific scenarios. But at the end of the day work has a precise definition, so when in doubt go back to that.

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  • $\begingroup$ work is not necessarily mechanical work $\endgroup$ – Andrew Steane Aug 12 at 17:23
  • $\begingroup$ @AndrewSteane True. I have tried to edit to be a little clearer. Certainly the equation $\Delta U=Q\pm W$ needs either to be clarified or modified depending on the system / process. $\endgroup$ – BioPhysicist Aug 12 at 17:57
  • $\begingroup$ The weight of the piston is irrelevant as the system is the gas. $\endgroup$ – Valter Moretti Aug 15 at 8:56
  • $\begingroup$ @ValterMoretti I included that because it's not clear what the OP would mean by "surroundings" in the case of a massive piston. Instead of going through all of cases like FakeMod did, I just decided to stay more general. $\endgroup$ – BioPhysicist Aug 15 at 13:00
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Referring to mechanical systems, the answer depends on the considered pair of systems and their interactions. If the work is due to contact forces on pairs of contact points with opposite velocities, then the work done by the system on the external system is equal, up to sign, the the one done by the external system to the system as a trivial consequence of the action-reaction principle. That is the case of some gas contained in a cylinder with piston: the system is the gas and the surronding system is the piston.

There are however cases where the contact points have different velocities. It may happen with friction forces: a building block -- viewed as a single material point for the sake of simplicity -- thrown on a rough table. In this case the total work (till the block stops) done by friction force of the table on the block is negative and equals the initial kinetic energy. Conversely, the work on the table (supposed always at rest in the used reference frame) due to the friction force is zero. Here, the contact point changes at each instant but it has always zero velocity.

In summary, the most general way to state the first principle for a system with thermodynamical energy $U$ is $$\Delta U = \cal L+ \cal Q\:,$$ where $\cal L$ is the work on the system and $\cal Q$ is the heat entering the system. This is also valid for irreversible transformations. This is because the principle, in the ideal mechanical case of a purely mechanical system with conservative internal forces, must reduce to a standard theorem of mechanics: the work done on a mechanical system by external forces equals the variation of kinetical energy plus the variation of internal potential energy.

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Summary

No. In general, it isn't true that the work done by the surrounding on the system is equal to the work done by the system on the surrounding.

Clarifications

In the following answer, I will only be talking about closed systems, which are most commonly encountered in an introductory thermodynamics course. In closed systems, the surroundings and the system are not in contact and are separated by an interface, thus it is meaningless to talk about the work done by the surronding on the system or vice versa. However, we can talk about the work done by the gas on the interface, and the work done by the surroundings on the interface, which are often shortened to the work done by gas and the work done by surroundings. In this answer, I will be talking about the simplest interface possible, a piston.

Explanation

For the sake of simplicity, we will be considering our system as a rigid cylinder (having conducting walls) with a piston, containing a gas. Everything outsid the cylinder shall be considered as the surroundings. The following image gives a visual idea of our system:

Cylinder with a piston

Image source

Now, we will consider the following four different cases:

Reversible process with a massless piston

In this case, since the piston is massless, thus the net force on the piston must be zero (otherwise, the piston's acceleration would be undefined. To see this, apply $\mathbf F_{\rm net}=m\mathbf a$, where $m=0$). For the net force to be zero, the force on the piston by the surroundings, must be equal and opposite to the force on it by the gas. Thus yielding

$$\mathbf F_{\rm surrounding}=-\mathbf F_{\rm system}\Rightarrow F_{\rm surrounding}=F_{\rm system}$$

And since the displacements are the same in both cases, thus we can conclude that the magnitude of the work done is also the same, however the sign must be the opposite, since the forces are in opposite direction. This is the most basic and common case one encounters, however, this isn't the only case.

Reversible process with a heavy piston

Heavy is used to signify that the piston isn't massless.

In this case, since the piston is heavy, there can be a non zero net force acting on the piston. Also, since the process is assumed to be reversible, it is necessary that the process must be carried out quasi-statically (in fact, quasi-static-ness is a characteristic of reversible processes). In other words, the process should be carried out in such a way, that the system is in equilibrium with the surroundings at every. This equilibrium implies that the surrounding and the system must have the same temperature at any moment, and the piston should move infinitesimally slowly, i.e. the acceleration of the piston must be zero. So balancing the forces on the piston, we get

$$\mathbf F_{\rm system}+\mathbf F_{\rm surrounding}+m\mathbf g=0\Rightarrow \mathbf F_{\rm system}\neq \mathbf F_{surroundings}\Rightarrow F_{\rm surrounding}\neq F_{\rm system}$$

Thus, in this case, the force on the piston by the surronding, and by the system are not equal. However, the displacements corresponding to both the forces are equal (in fact this would be the case for all the four scenarios). This clearlyimplies that the work done by the system on the piston, is not equal to the work done by the surroundings on the piston. But as clarified above, the work done on the piston is essentially the work done by that entity. So, in this case, the work done by the system is not equal to the work done by the surroundings.

Irreversible process with a massless piston

This case yields similar conclusions to the first case. It's easy to see why. In our first case, we never really used the characteristic that the process needs to be reversible. All the arguments made in the first case hold even if the process is irreversible. Thus we can also safely conclude here, that the work done by the gas and the surroundings is equal in magnitude, but opposite in signs.

Irreversible process with a heavy piston

Now, we cannot just extend the arguments in the second case for this case, since the arguments in the second case, do make use of the reversibility of the process. In this case, there is no constraint of the process being quasi-static, and neither is there any constraint for the net force on the piston to be zero. This allows uds to conclude that in the general case,

$$F_{\rm surrounding}\neq F_{\rm system}$$

But no matter what, the displacements will always be the same (like I said before). Thus, in general, the work done by the gas will not be equal to the work done by the piston.

Conclusion

The above analysis was not a general analysis (and, in fact, it isn't even possibke to do a general analysis in a single answer). However, the above analysis does a good job in showing that the conclusion $W_{\rm surrounding}=W_{\rm system}$ is, in general, wrong. In my opinion, it's not beneficial to draw out any general conclusion about the work done by the system and the surroundings. It's always better, fool-proof and more appropriate to physically determine the work done by both the agents, whenever needed, instead of generalizing a rule which is bound to fail.

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    $\begingroup$ I don't think I agree with this. The "surrounding" includes all massive bodies exerting the gravitational force, so $\mathbf{F}_{\text{surrounding}}$ should include $m\mathbf{g}$. $\endgroup$ – d_b Aug 12 at 16:27
  • $\begingroup$ @d_b That depends on how you define your system. I find it more intuitive and appropriate to keep the interface, surrounding and system as separate entities. That being said, even if you somehow include $m\mathbf g$ in $\mathbf F_{\rm surrounding}$, still in the fourth case, the work done by the surrounding and the system isn't going to be the same. So the conclusion holds true in both the cases. $\endgroup$ – user258881 Aug 12 at 16:31
  • $\begingroup$ Sir in your opinion,do you think that the First Law,by default,assumes ideal case (frictionless, massless piston,etc) ? I have always found the definition of work by system a little non - intuitive provided that fact that work in mechanics happen by external forces (keeping separation of interacting particles same). I have always wondered if this definition of work by system in the First Law is meant only for ideal cases when it is equal to work by surroundings. $\endgroup$ – Tony Stark Aug 12 at 17:43
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    $\begingroup$ @TonyStark I see no reason why the work term in the first law should only be limited to ideal cases and reversible processes. In fact, I have solve numerous questions in thermodynamics using first law in situations where the conditions aren't "ideal" (heavy piston, irreversible process, real gases). If you haven't yet encountered such a question, then I wncourage you to try and find such questions and solve them to get a better feel for it. $\endgroup$ – user258881 Aug 12 at 17:51
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To add: the equations differ in sign because the conventional sign for work done is reversed between physics and chemistry books. If the work done on the system is considered positive in physics, it becomes negative in chemistry. This is just a matter of convention; the laws remain the same. Be clear and consistent with the sign when using those formulas, and clearly identify the convention you are using.

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