Simple clarification regarding the sign convention for work

[JC2000]

Precursor : Sign conventions regarding Work by the system/on the system may even vary among Physics textbooks and among Chemistry textbooks and as a rule of thumb it is better to clearly mark out the convention used by the text being referred to specifically and remain consistent with it. Generally the rules/ resulting implications stated below hold for the respective subjects (Physics and Chemistry):

(Case 1)In Physics : (engineers/physicists tend to be more interested in the work done by the system, i.e the work output being positive)

Work by the system on the surroundings is POSITIVE.
Work by the surroundings on the system is NEGATIVE.

1.Hence, the FLT is generally represented as $Q = \Delta U + \Delta W$ in Physics textbooks generally.(?)
2.This implies that if a system expands (thus work is done BY SYSTEM) the work done is positive and if the system contracts (work done ON SYSTEM) then work is negative. (?)
3. Heat absorbed by the system is $+Q$ and heat evolved is $-Q$. Is this an independent sign convention or is this tied in some way to the sign convention selected for $W$ in FLT (since to me it seems that if work is done by the system then internal energy should decrease and hence it is impossible to say whether $\Delta U + W$ would be positive or negative)?(Case 2)In Chemistry : (experimenters tend to be interested in the work done by the surroundings on the system to drive reactions and so on and hence work input is positive)

Work by the system on the surroundings is NEGATIVE
Work by the surroundings on the system is POSITIVE

4. Hence the FLT is represented as $Q = \Delta U - \Delta W$. (?)
5. This means the if work is done by BY THE SYSTEM in this case then $-(-W)$ in FLT for Case 2 will give FLT stated in Case 1.(?)
6. From (5) we can thus say that FLT is represented in the same way in Case 1 and Case 2, only the implication of the sign for $W$ varies depending on context. (?)

Lastly regarding the correlation between WORK DEFINED IN MECHANICS and WORK DEFINED IN THERMODYNAMICS (picked up this bit of confusion in the answer given here).

In mechanics $dW = F \cdot dx$ from which we can derive $dW = p \cdot dV$. This can be directly used as work in thermodynamics. The advantage being that the definition of work is perfectly consistent here.
OR
We can chose to define thermodynamic work as $dW_{T} = -dW$ whereby we get FLT with the opposite sign for $W$.

7. The answer in the link goes on to explain the advantages of both but I fail to understand the advantages of $dW_{T} = -dW$. (?)
8. Lastly, it is best to check the sign conventions regarding (A) Work done BY SYSTEM vs ON SYSTEM and (B) relation between thermodynamic work and work in mechanics, in the book that one is reading and then consistently follow the same. (?)

Apologies for the long and rambling write-up. Thank you for your time.

 

[kuruman]

The way I sorted it out in my mind is this
1. The words "Work done" are meaningless without prepositions and you are absolutely correct in pointing this out. To ascribe meaning to work, one has to specify the agent who does the work and the system on which the work is done. Thus the correct way to specify work is
Work done by ______ (name of agent) on ________ (name of system). Swapping what's in the blanks flips the sign of the work and that's a consequence of Newton's 3rd law.
2. It is also true that for an ideal gas $dW = p dV$ and this term is positive when the gas expands and negative when the gas is compressed.
3. If the meaning of $dW$ is not specified with prepositions, the application of the FLT is ambiguous. Usually textbooks do that when they introduce it.

Consider the example of a piston representing "the surroundings" pushing down and compressing a gas representing "the system". In this case $p~dV<0$.
(a) Work done by the surroundings on the system
(is positive because the force exerted by the piston is in the same direction as the displacement of the piston)
(b) Work done by the system on the surroundings
(is negative because swapping changes the sign)

Now consider the gas expanding and pushing the piston up. In this case $p~dV>0$.
(c) Work done by the system on the surroundings
(is positive because the force exerted by the gas is in the same direction as the displacement of the piston)
(d) Work done by the surroundings on the system
(is negative because swapping changes the sign)

As for $dQ$, all the definitions I have seen is that when it is positive, heat "goes in" the system and when it is negative, heat leaves the system.

My personal preference is to write the FLT as $dU=dQ-dW$, where $dW$ is the work done by the system on the surroundings. The reason I prefer this formulation is because it's easy to explain to people who see it for the first time in terms of something that's intuitively obvious: "The change in the energy ($dU$) that is in there is equal to what goes in as heat ($dQ$) less what comes out ($dW$) as mechanical work".

This formulation of the FLT is a natural for writing the energy balance in heat engines which take in heat and put out mechanical energy.

 

[JC2000]

Thank you for the detailed response! I think I am clear about these questions now. It would be great if you could shed some light on the advantages of redefining the relation between mechanical work and thermodynamic work. Thanks!

 

[kuruman]

As I understand it, thermodynamic work is a generalization of mechanical work. Say you lift a weight $mg$ off the floor to height $h$ at constant speed. There are two agents in "the surroundings" that interact with the "system" as it is being lifted, the Earth and you, which means that there are two "works" done on the system. Because the weight moves at constant speed, its kinetic energy does not change which, according to the work-energy theorem, means that the sum of the works done on the weight must be zero, $W_{Earth}+W_{you}=0$. You can certainly use the above equation to figure out that $W_{you}=mgh$ only because you know that near the surface of the Earth $\vec F_{Earth}=m\vec{g}$ which enables you to calculate the mechanical work $W_{Earth}$. By contrast, the myriad biological, biochemical, chemical, electrical, hydrodynamic and mechanical processes that take place in your body as you lift the weight produce unknowable transfers of energy within you with the net result that you do combined work $W_{you}$ on the system. It is not as simple to calculate it as it is to calculate $W_{Earth}$, but you can give it a name and call it "thermodynamic work" reflecting the fact that it is the macroscopic manifestation of a large number of unknowable microscopic energy transfers.

Another example is the expanding gas pushing a piston up by adding heat at constant pressure. The gas does work $p~dV$ on the surroundings; this work is the macroscopic manifestation of the many transfers of kinetic energy between gas molecules and piston molecules. The kinetic energy added to the gas as heat is divided so that some of it stays in the gas and some of it goes into raising the piston against gravity. Just like the previous example, the work done by the Earth on the piston (assuming it has mass) is mechanical and the work done by the gas on the piston is thermodynamic. This example is easier to contemplate because, in the ideal gas model, all transfers of energy are due to collisions between gas and container molecules or gas and gas molecules. Nevertheless, it is just as difficult to calculate microscopically and requires a statistical formulation.

 

[George Jones]

(Case 1)In Physics : (engineers/physicists tend to be more interested in the work done by the system, i.e the work output being positive)

Work by the system on the surroundings is POSITIVE.
Work by the surroundings on the system is NEGATIVE.

1.Hence, the FLT is generally represented as $Q = \Delta U + \Delta W$ in Physics textbooks generally.(?)

No, not generally.

These days, the $\Delta U = Q + W$ convention is used quite often in physics texts. For example, the text "Thermal Physics" by Schroeder is widely used in North America and it uses this convention. This convention also is used in the two first-year physics texts by Serway, and the two texts by Knight. I have used all of these as texts for physics courses that I have taught.

The first-year book by Halliday, Resniick, and Walker, however, uses the other convention. So, both conventions are used extensively, and students (and others) should be careful.

At my university, students see thermal/statistical physics in first-year in either Knight or Serway (+W), in second-year in Schroeder (+W), and in fourth-year in Baierlein (-W).

Another place where one has to take care is the term "adiabatic". Different text ascribe different meanings to this term.

 

[Mister T]

7. The answer in the link goes on to explain the advantages of both but I fail to understand the advantages of dWT=−dW (?)

But $dW_T = -dW$ regardless of which convention you use.

 

[JC2000]

No, not generally.

These days, the $\Delta U = Q + W$ convention is used quite often in physics texts. For example, the text "Thermal Physics" by Schroeder is widely used in North America and it uses this convention. This convention also is used in the two first-year physics texts by Serway, and the two texts by Knight. I have used all of these as texts for physics courses that I have taught.

I see so to be on the safe side it is best to figure out what positive and negative work refer to in a given text and then consistently apply that?

The first-year book by Halliday, Resnick, and Walker, however, uses the other convention. So, both conventions are used extensively, and students (and others) should be careful.

Thank you for pointing that out, I did not expect Resnick, Halliday and Walker to be the exception to the trend!

At my university, students see thermal/statistical physics in first-year in either Knight or Serway (+W), in second-year in Schroeder (+W), and in fourth-year in Baierlein (-W).

Another place where one has to take care is the term "adiabatic". Different text ascribe different meanings to this term.

Thanks for pointing this out, I'll keep this in mind. So far I have only had it used to describe adiabatic walls (wall that separates two systems such that they can be at two different equilibriums) and adiabatic process (a process that is rapid or insulated such that no transfer of energy as heat occurs between system and surroundings).

 

[JC2000]

As I understand it, thermodynamic work is a generalization of mechanical work. Say you lift a weight $mg$ off the floor to height $h$ at constant speed. There are two agents in "the surroundings" that interact with the "system" as it is being lifted, the Earth and you, which means that there are two "works" done on the system. Because the weight moves at constant speed, its kinetic energy does not change which, according to the work-energy theorem, means that the sum of the works done on the weight must be zero, $W_{Earth}+W_{you}=0$. You can certainly use the above equation to figure out that $W_{you}=mgh$ only because you know that near the surface of the Earth $\vec F_{Earth}=m\vec{g}$ which enables you to calculate the mechanical work $W_{Earth}$. By contrast, the myriad biological, biochemical, chemical, electrical, hydrodynamic and mechanical processes that take place in your body as you lift the weight produce unknowable transfers of energy within you with the net result that you do combined work $W_{you}$ on the system. It is not as simple to calculate it as it is to calculate $W_{Earth}$, but you can give it a name and call it "thermodynamic work" reflecting the fact that it is the macroscopic manifestation of a large number of unknowable microscopic energy transfers.

Another example is the expanding gas pushing a piston up by adding heat at constant pressure. The gas does work $p~dV$ on the surroundings; this work is the macroscopic manifestation of the many transfers of kinetic energy between gas molecules and piston molecules. The kinetic energy added to the gas as heat is divided so that some of it stays in the gas and some of it goes into raising the piston against gravity. Just like the previous example, the work done by the Earth on the piston (assuming it has mass) is mechanical and the work done by the gas on the piston is thermodynamic. This example is easier to contemplate because, in the ideal gas model, all transfers of energy are due to collisions between gas and container molecules or gas and gas molecules. Nevertheless, it is just as difficult to calculate microscopically and requires a statistical formulation.

But $dW_T = -dW$ regardless of which convention you use.

After reading your answers I understand 'thermodynamic work' to be the work that is difficult to attribute to a single force, that may require statistical methods to calculate or may involve a number of different forces (electric, chemical etc) and hence a single term is used to describe an entire multitude of microscopic forces on the system.

From the examples I gathered that 'thermodynamic work' need not necessarily be equal to mechanical work? (considering a case where say the object has KE)

Is my understanding accurate?
If so, I am at a loss as to what the answer in the link is all about. Admittedly, I stumbled on to it while trying to clarify the sign convention for work. Hence, I will read the whole thing and try to rephrase my questions regarding the link.

Thank you for your detailed response.

 

[Mister T]

The work-energy principle holds only for particle-like objects (meaning they have no internal structure and therefore no internal energy).

When the study of mechanics is expanded to include thermodynamics, a concept of conservation of energy can be developed that includes objects with internal energy.

See, for example,

Development of energy concepts in introductory physics courses
Arnold B. Arons
Citation: Am. J. Phys. 67, 1063 (1999); doi: 10.1119/1.19182
View online: http://dx.doi.org/10.1119/1.19182
View Table of Contents: http://ajp.aapt.org/resource/1/AJPIAS/v67/i12
Published by the American Association of Physics Teachers

 

[kuruman]

The link you provided has, for a quasistatic compression,$$W=p~A~\Delta x~~~~~\mbox{(1.27)}$$ and then says that$$W=-p~\Delta V~~~~~\mbox{(1.28)}$$I note that there is no mention in the link whether $W$ represents the work done on the gas or by the gas. Remember that prepositions are important. However, from this information we can infer which convention the link adheres to. Because we have a compression, $\Delta V~$is negative which makes $W$ positive. If look at post #2, item (a), there is a match: Compression and positive $W$. Therefore the link uses the convention that $W$ represents the work done by the surroundings on the gas.+

On edit: If any two yes/no answers to the following 3 questions are known (or can be found), the answer to the remaining yes/no answer can be determined.
1. Does $W$ represent the work done on the gas by the surroundings?
2. Does the gas expand?
3. Is $W$ positive?

I hope this helps.

Note: It is assumed that the volume of the gas changes. If it doesn't, then $W~$ is zero regardless of what it represents and the convention for $W$ is undefined.