1. Thermodynamic work

[Scienc]

Guys, why in physics does the work done by the system have the POSITIVE sign, but in chemistry does the work done by the system have the NEGATIVE sign?

[studiot]

8 hours ago, Scienc said:

Guys, why in physics does the work done by the system have the POSITIVE sign, but in chemistry does the work done by the system have the NEGATIVE sign?

A question about a sign convention that leads to a lot of confusion.
This is not the only instance multiple sign conventions in Science.

In this case it is a result of History.

 

Thermodynamics was originally developed by physical scientists and engineers.
They were concerned with making machines (steam engines) for the industrial revolution.
Steam engines are heat engines. That is they thought in terms of input (heat in the form of fuel) and output (work). Both of these were thought of as 'naturally being' positive quantities.

So they wrote their version of the Law of Conservation of Energy (The First Law of Thermodynamics) as

ΔU  = q - w.

 

Chemists came to the scene from a different point of view.
They wanted all forms of energy to have the same sign, whichever side of the conservation appearance they appeared so they could present the equation as a sum on both sides of the equation.
So they wrote their equation as

ΔU  = q + w.

By then it was also realised that, although all the terms are energies, there is a difference between ΔU , which is a state variable of the system, and q and w which are exchange variables of the energies crossing the system boundary.

So they tidied up by stating that

all energies crossing the boundary from the system to the surroundings are negative
and all energies passing from the surroundings to the system are positive.

Now they could add them up, move them about in equations and between equations in other parts of Science in a consistent manner.

It is an improved system

But it shows the importance of knowing the sign convention in use and the equations that go with it.
This last remark also applies to other such instances of multiple conventions such as those in Electricity, Elasticity and elsewhere.

 

 

 

[swansont]

Because they're chemists. Whaddaya gonna do?

[J.C.MacSwell]

50 minutes ago, swansont said:

Because they're chemists. Whaddaya gonna do?

Some of my best friends (okay, distant acquaintances, but they still are people) are chemists.

It's always the 90 percent of them that give the rest a bad name...

[joigus]

4 hours ago, studiot said:

A question about a sign convention that leads to a lot of confusion.
This is not the only instance multiple sign conventions in Science.

In this case it is a result of History.

 

Thermodynamics was originally developed by physical scientists and engineers.
They were concerned with making machines (steam engines) for the industrial revolution.
Steam engines are heat engines. That is they thought in terms of input (heat in the form of fuel) and output (work). Both of these were thought of as 'naturally being' positive quantities.

So they wrote their version of the Law of Conservation of Energy (The First Law of Thermodynamics) as

ΔU  = q - w.

 

Chemists came to the scene from a different point of view.
They wanted all forms of energy to have the same sign, whichever side of the conservation appearance they appeared so they could present the equation as a sum on both sides of the equation.
So they wrote their equation as

ΔU  = q + w.

By then it was also realised that, although all the terms are energies, there is a difference between ΔU , which is a state variable of the system, and q and w which are exchange variables of the energies crossing the system boundary.

So they tidied up by stating that

all energies crossing the boundary from the system to the surroundings are negative
and all energies passing from the surroundings to the system are positive.

Now they could add them up, move them about in equations and between equations in other parts of Science in a consistent manner.

It is an improved system

But it shows the importance of knowing the sign convention in use and the equations that go with it.
This last remark also applies to other such instances of multiple conventions such as those in Electricity, Elasticity and elsewhere.

 

 

 

Ditto +1. Although I'm in a hurry. Just want to add that physicist's criterion kinda makes sense, because when the system does work, it loses energy. It happens to me all the time.

 

[Carrock]

12 hours ago, Scienc said:

Guys, why in physics does the work done by the system have the POSITIVE sign, but in chemistry does the work done by the system have the NEGATIVE sign?

Gets worse than that. Sometimes in the same system work done in e.g. charging a capacitor is regarded as positive, while work done discharging a capacitor is, er, positive.

You can then say (e.g. from https://www.allaboutcircuits.com/textbook/alternating-current/chpt-6/q-and-bandwidth-resonant-circuit/ )

Quote

More formally, Q is the ratio of power stored to power dissipated in the circuit reactance and resistance, respectively:

Q = Pstored/Pdissipated = I2X/I2R Q = X/R where: X = Capacitive or Inductive reactance at resonance R = Series resistance. 

You can actually buy a power buildup cavity using the same general idea.

From Wiki

Quote

The quality factor of atomic clocks, superconducting RF cavities used in accelerators, and some high-Q lasers can reach as high as 10^11[3] and higher.^[4]

It seems like you could put in 10mW and get 1 MW out. Apparently the concept of imaginary power or reactive power (which is not available for continuous  work) is not used here; 'stored power' is just power.

It's rather like saying torque is energy because the units are dimensionally the same.

There's a simple definition of Q to avoid this confusion*

Q(omega)= omega*(maximum energy stored/power loss)

 

*My confusion anyway - there must be some rationale to the "stored power is not energy" concept.

[swansont]

At the end of the day, you have to conserve energy and define your system. Focusing on whether the work is positive or negative without that context is, IMO, a problem, because you are memorizing a convention without learning the context to properly apply it.

IOW, there is a reason why one would define it as positive or negative, in the context of writing down an energy balance.

 

For example, if you have a constant net force acting on an object through some distance x (along a straight line) the work is W = Fx and we also know that W = ∆KE

But we could rearrange that second equation so that ∆KE - W = 0

And if for some reason we defined the work to be negative then ∆KE + W = 0. That would require that we define the work to be W = -Fx, because we have to be consistent. You could ask why you'd define it to be negative, and the answer would be in some other equation that you want to write a certain way. And the reason for that might be arbitrary.

 

 

[joigus]

Just to give more examples and clarify more, if possible. The matter gets settled when you express your change in internal energy as,

\[dU=\delta Q\pm\Sigma_{i}F_{i}dX_{i}\]

The F_i quantities are intensive parameters (they are the same even if you re-scale the system), and the X are so-called control parameters and are extensive (they scale as the size of the system). The minus or plus sign is just a convention. What's not a convention is what it looks like when you express it in terms of the intensive and extensive parameters.

For an ideal gas, e.g., the eq.

\[dU=\delta Q-PdV\]

does not depend on you criterion. If the gas expands, it looses energy, so, if you define,

\[dU=\delta Q - \delta W\]

then you must define

\[\delta W = PdV\]

but, if you define,

\[\delta U = \delta Q + \delta W\]

then you must define,

\[\delta W = -PdV\]

in such a way that, no matter how you define things, the gas loses energy when it expands, which is what really happens. As Swansont and Carrock suggest, you are on safer grounds if you think of your particular system in terms of energy being a constant and how the different parameters change your system's energy.

I hope that was clear in case there were loose ends.

 

[joigus]

11 hours ago, joigus said:

the gas loses energy when it expands,

Everything else being equal, of course. So no heat exchange or other forms of work.

[studiot]

12 hours ago, joigus said:

 

δU=δQ+δW

 

Aren't you overthinking this a bit ?

The OP asked why is this so ?

He had also correctly realised the all important distinction between work done on the system and work done by the system.

Furthermore what sign do you attach to 'q' in an exothermic chemical reaction ?

And is gravitational potential energy calculated as positive or negative?

[joigus]

5 hours ago, studiot said:

Aren't you overthinking this a bit ?

The OP asked why is this so ?

He had also correctly realised the all important distinction between work done on the system and work done by the system.

Furthermore what sign do you attach to 'q' in an exothermic chemical reaction ?

And is gravitational potential energy calculated as positive or negative?

It is entirely possible that I'm overthinking it, in which case I apologize. But the way I see it, distinctions such as "work done on a system" and "work done by a system" are often confusing, especially when one is dealing with reversible work --in the physicist's sense--. Reversible work is of theoretical interest. And in order to define, e.g., variable pressure for reversible processes, you must make external pressure equilibrate internal pressure, so that the concept of "who" is doing the work becomes more confusing. I may have misinterpreted him, but Swansont's point that,

 

On 5/19/2020 at 5:39 PM, swansont said:

At the end of the day, you have to conserve energy and define your system. Focusing on whether the work is positive or negative without that context is, IMO, a problem, because you are memorizing a convention without learning the context to properly apply it.

IOW, there is a reason why one would define it as positive or negative, in the context of writing down an energy balance.

Is of some importance here, I think. "Define your system," are key words for me in it. It's very easy to get lost in the different steps at which your definition might alter your formulae. For example, suppose somebody defined the different intensive thermodynamic parameters in different ways, while at the same time defined the criterion differently as to "who is doing the work." Example (homogeneous system with variable number of molecules):

\[dU=\delta Q-PdV+\mu dN\]

Maybe some "crazy author" decides to go one step further in an orgy of definitions and set the chemical potential as the negative of the usual one. One possible choice is (I think it's quite standard),

\[P=-\left(\frac{\partial U}{\partial V}\right)_{S,N}\]

\[\mu=\left(\frac{\partial U}{\partial N}\right)_{V,S}\]

\[T=\left(\frac{\partial U}{\partial S}\right)_{V,N}\]

But there could be others. The fact that pressure has a negative sign is just convention. No real meaning attached to that, I think. It's just a thermodynamic potential. And the real question is well away from any ambiguity when you express,

\[dU=\left(\frac{\partial U}{\partial S}\right)_{V,N}dS+\left(\frac{\partial U}{\partial V}\right)_{S,N}dV+\left(\frac{\partial U}{\partial N}\right)_{V,S}dN\]

Now, the signs there in the last eq. do not depend on any criterion. And I do confess I'm something of an unbearable stickler for sound definitions, axioms and language.

12 minutes ago, joigus said:

(I think it's quite standard),

 

 

Actually, I think the "usual" one is the opposite. That's kind of what I mean.

Edited May 20, 2020 by joigus
correct eq.

[studiot]

40 minutes ago, joigus said:

but Swansont's point that,

 

On 5/19/2020 at 4:39 PM, swansont said:

At the end of the day, you have to conserve energy and define your system. Focusing on whether the work is positive or negative without that context is, IMO, a problem, because you are memorizing a convention without learning the context to properly apply it.

IOW, there is a reason why one would define it as positive or negative, in the context of writing down an energy balance.

Is of some importance here, I think. "Define your system," are key words for me in it.

 

Yes indeed, though My take on it is quite different from yours

Swansont said exactly what I said, though obviously in different words.

 

On 5/19/2020 at 10:23 AM, studiot said:

But it shows the importance of knowing the sign convention in use and the equations that go with it.

I was also trying to produce an explanation for the diffrence in the approaches of Physics and Chemistry, not a treatise on thermodynamics.

 

I also agree that knowing and defining the system and its boundary is of fundamental importance, not least because the boundary splits the analysis into two parts viz the system and the surroundings.

So many come here with difficulties in this because they use say the wrong pressure in equations containing  P, regardless of the sign attributed.

Most do not properly realise that q and w are exchange variables, they are not state variables and do not belong exclusively to either the system or the surroundings.

They are particularly important because theya re the only way we can connect changes in the system state variables to changes in the variables of the surroundings.

As a teacher of this in both disciplines you must have noticed this.

Finally you have not addressed my point about the sign of q, which is, confusingly, the other way round.

[joigus]

12 minutes ago, studiot said:

I was also trying to produce an explanation for the diffrence in the approaches of Physics and Chemistry, not a treatise on thermodynamics.

Yes, I know, that's why I said "ditto." But then I realised there are even more reasons for confusion for physicists, so I just tried to throw them in, but there's always a risk that you're bringing more confusion.

 

14 minutes ago, studiot said:

Most do not properly realise that q and w are exchange variables, they are not state variables and do not belong exclusively to either the system or the surroundings.

They are particularly important because theya re the only way we can connect changes in the system state variables to changes in the variables of the surroundings.

As a teacher of this in both disciplines you must have noticed this.

Totally, yes. We don't really know who is asking, though. Whether my explanation or yours have been helpful, we don't really know. I hope both have. I kind of have become a specialist in many common sources of confusion for the students. Believe it or not, sometimes it's in the language. Very recently I remember a problem on hydrostatics, and  the source of the confusion came from how the statement had been worded. It said that the body was "floating," and it really meant that it had sunk in equilibrium with equilibrium equalling buoyancy. I don't know what word the person who wrote the problem would have used for floating like an iceberg. Maybe it would have been quasi-floating.

That's a problem too.

 

23 minutes ago, studiot said:

Finally you have not addressed my point about the sign of q, which is, confusingly, the other way round.

Oh, yes. You're right. I forgot.

6 hours ago, studiot said:

Furthermore what sign do you attach to 'q' in an exothermic chemical reaction ?

To me that would be negative. I also think considering it negative is quite standard, also in molecular biology. But it's not impossible that somebody somewhere could have a different criterion elsewhere. Is it different in any context that you know of in chemistry? It may well be. You're very interdisciplinary, and encyclopedic.

[studiot]

32 minutes ago, joigus said:

To me that would be negative. I also think considering it negative is quite standard, also in molecular biology. But it's not impossible that somebody somewhere could have a different criterion elsewhere. Is it different in any context that you know of in chemistry? It may well be. You're very interdisciplinary, and encyclopedic.

Well exothermic reaction is defined as one in which heat is generated.

And is therefore negative.

I am always reminding posters that this refers to heat, not to energy.

Remember also that Chemists usually work in a lab with open reaction vessels so under constant surroundings pressure.
So work can be calculated on the surroundings side of the boundary.

[joigus]

1 hour ago, joigus said:

with equilibrium equalling buoyancy

I meant "with weight equalling buoyancy."

31 minutes ago, studiot said:

Remember also that Chemists usually work in a lab with open reaction vessels so under constant surroundings pressure.
So work can be calculated on the surroundings side of the boundary.

Yes, I have to teach chemistry sometimes, so have must think about those things. Although my chemistry laboratory practice seems very remote now.