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The concept of work

by negation
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negation
#1
May8-14, 09:36 AM
P: 819
After a fair bit of time in Physics, I have started to, intuitively, see the big picture. That is, in the area of electricity, mechanics and thermodynamics, there are certain vital concepts that are central such as work and energy. Semantics are involved but they essentially refer to the same thing.

There is one thing (among others) that plague me and that is the whole concept of work. It keeps popping up as a common denominator.

In the context of thermodynamics, the internal energy is the measure of the sum of kinetic energy and the potential energy.

The change in internal energy from state i(initial) to f(final) gives Q(heat) + W(work).

What I am really confused about is working being the negative of the product of pressure and change in volume.
W = -PΔV

I see this in mechanics and as well as in electricity but my understanding of them as regard the NEGATIVE sign continues to be fuzzy.

Let's start with an example:

Suppose I have a cylinder heated (in contact) with a thermal reservoir. At state i, the piston is height Li from the base of the cylinder. At state f. the piston has height Lf from the base of the cylinder.
Li < Lf
Essentially, the piston is being pushed outward as the cylinder is being heated.

Now,

W = -PΔV

As the piston is being pushed outward, the change in v is positive because Lf>Li and pressure decreases and therefore pf < pi and so pressure is negative.
Therefore, in this example, W = -PΔV.
Is my understanding flawed?
If it isn't then would it be correct to state that if the pressure does not change (assuming a certain pressure could be maintained within the system as), then W = PΔV?

work done by the system = negative W
work done on the system = positive W
heat supplied to the system = positive Q
heat supplied from the system = negative Q
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Doc Al
#2
May8-14, 09:48 AM
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Quote Quote by negation View Post
As the piston is being pushed outward, the change in v is positive because Lf>Li and pressure decreases and therefore pf < pi and so pressure is negative.
Therefore, in this example, W = -PΔV.
Is my understanding flawed?
Yes. P is the pressure, which may change but is always positive.

If it isn't then would it be correct to state that if the pressure does not change (assuming a certain pressure could be maintained within the system as), then W = PΔV?
The work done BY the system (by the expanding gas) is positive: The gas exerts an outward force as the piston moves outward. But what you are interested is the work done ON the system (the gas), which is the negative of that.
DrDu
#3
May8-14, 09:51 AM
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Yes, W=-PΔV only applies if P is constant. If P is changing with volume you have to calculate work as ## W=-\int_{V_1}^{V_2} P(V) dV##.

negation
#4
May8-14, 09:59 AM
P: 819
The concept of work

Quote Quote by Doc Al View Post
Yes. P is the pressure, which may change but is always positive.


The work done BY the system (by the expanding gas) is positive: The gas exerts an outward force as the piston moves outward. But what you are interested is the work done ON the system (the gas), which is the negative of that.
I can see that P is factor out as a constant term in the integral. That must mean P is a constant. What is hard to understand is the idea that P does not decrease. If volume increases, P must decrease unless a constant pressure is being maintain hypothetically. So apparently, there must be a certain assumption being made. Do we assume that for certain set up, pressure is constant for simplicity?

Edit:
If work done by a system is positive, then, by corollary the work done on a system is negative. And if work done by a system is negative then the work done on a system is positive, true?
Intuitively, this seems to imply that has to be an overall constant. A conservation of energy of sort such that if one increases, the other decreases and vice versa. I'm trying to link this to potential and kinetic energy in the context of mechanics because there is a certain recurring pattern that I am noticing.

Edit:

I think I might have refined my understanding of work in context of Mechanics. And just to make sure I am on the right path.

U (internal energy) = KE + PE
ΔU = ΔKE + ΔPE = (KEf-KEi) + (PEf-PEi)
But by conservation of energy (vital assumption):
ΔKE + ΔPE = 0
If ΔKE+ΔPE=0 and ΔU = ΔKE+ΔPE then ΔU = 0

ΔKE+ΔPE = (KEf-KEi) + (PEf-PEi) = 0
so,
KEi+PEi=KEf+PEf
This shows that the total Kinetic energy and potential energy at state i equals the total kinetic energy and potential energy at state f.

Going further,
W = ΔKE = KEf-KEi = 0.5mvf2 - 0.5mvi2
IF:
vf>vi → ΔKE=+ve →W = +ve
But we have established that sum total of KE and PE is constant so this necessarily implies that if KE increases, PE decreases.
∴ W = +ΔKE = -ΔPE

If vf2<vi2 → ΔKE = -ve → W = -ve
But we have established that sum total of KE and PE is constant so this necessarily implies that if KE increases, PE decreases.
∴-W = -ΔKE = +ΔPE
jtbell
#5
May8-14, 10:46 AM
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Quote Quote by negation View Post
I can see that P is factor out as a constant term in the integral. That must mean P is a constant. What is hard to understand is the idea that P does not decrease.
If you add heat to the system at the proper rate, while the gas is expanding, the pressure remains constant.
negation
#6
May8-14, 11:00 AM
P: 819
Quote Quote by jtbell View Post
If you add heat to the system at the proper rate, while the gas is expanding, the pressure remains constant.
I see. That explains the function of the thermal reservoir-to maintain a constant ideal pressure. It helped!

I did a second edit in #4. I would be very happy if I finally manage to make sense of the concept of work in Mechanics and Thermodynamics. This would leave just the concept of work in electricity to understand!
dauto
#7
May8-14, 11:14 AM
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Yes the sign of the work may be confusing and to make matters worse different books use different sign conventions. The most common convention seems to be
δW = P δV
Work done by the system is positive
Work done on the system is negative.
But the opposite convention is sometimes used

δW = - P δV
Work done by the system is negative
Work done on the system is positive.

One must keep one's eyes open to avoid confusion.
Chestermiller
#8
May8-14, 03:57 PM
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For additional perspective on this, here is an excerpt from my PF Blog at http://www.physicsforums.com/blog.php?b=4300:

If a process path is irreversible, then the temperature and pressure within the system are typically 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, TI(t) and PI(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)

where P(t) is the pressure calculated from the equilibrium equation of state for the material (e.g., the ideal gas law).

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

Please see my blog for further details.
negation
#9
May8-14, 08:38 PM
P: 819
Quote Quote by Chestermiller View Post
For additional perspective on this, here is an excerpt from my PF Blog at http://www.physicsforums.com/blog.php?b=4300:

If a process path is irreversible, then the temperature and pressure within the system are typically 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, TI(t) and PI(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)

where P(t) is the pressure calculated from the equilibrium equation of state for the material (e.g., the ideal gas law).

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

Please see my blog for further details.
Hi Chester,

my course hasn't delved into such deeper meaning but I'll definitely read up more to achieve a greater understanding.

I was fuzzy in my understanding of work but I think I have understood it. I posted it on post #4. Not sure if it was overlooked but.

"
I think I might have refined my understanding of work in context of Mechanics-at the outset I was unable to understand why work is the negative of PE.
And just to make sure I am on the right path.

U (internal energy) = KE + PE
ΔU = ΔKE + ΔPE = (KEf-KEi) + (PEf-PEi)
But by conservation of energy (vital assumption):
ΔKE + ΔPE = 0
If ΔKE+ΔPE=0 and ΔU = ΔKE+ΔPE then ΔU = 0

ΔKE+ΔPE = (KEf-KEi) + (PEf-PEi) = 0
so,
KEi+PEi=KEf+PEf
This shows that the total Kinetic energy and potential energy at state i equals the total kinetic energy and potential energy at state f.

Going further,
W = ΔKE = KEf-KEi = 0.5mvf2 - 0.5mvi2
IF:
vf>vi → ΔKE=+ve →W = +ve
But we have established that sum total of KE and PE is constant so this necessarily implies that if KE increases, PE decreases.
∴ W = +ΔKE = -ΔPE

If vf2<vi2 → ΔKE = -ve → W = -ve
But we have established that sum total of KE and PE is constant so this necessarily implies that if KE increases, PE decreases.
∴-W = -ΔKE = +ΔPE"

Have I finally got it?
Chestermiller
#10
May8-14, 09:03 PM
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Quote Quote by negation View Post
Hi Chester,

my course hasn't delved into such deeper meaning but I'll definitely read up more to achieve a greater understanding.

I was fuzzy in my understanding of work but I think I have understood it. I posted it on post #4. Not sure if it was overlooked but.

"
I think I might have refined my understanding of work in context of Mechanics-at the outset I was unable to understand why work is the negative of PE.
And just to make sure I am on the right path.

U (internal energy) = KE + PE
ΔU = ΔKE + ΔPE = (KEf-KEi) + (PEf-PEi)
But by conservation of energy (vital assumption):
ΔKE + ΔPE = 0
If ΔKE+ΔPE=0 and ΔU = ΔKE+ΔPE then ΔU = 0

ΔKE+ΔPE = (KEf-KEi) + (PEf-PEi) = 0
so,
KEi+PEi=KEf+PEf
This shows that the total Kinetic energy and potential energy at state i equals the total kinetic energy and potential energy at state f.

Going further,
W = ΔKE = KEf-KEi = 0.5mvf2 - 0.5mvi2
IF:
vf>vi → ΔKE=+ve →W = +ve
But we have established that sum total of KE and PE is constant so this necessarily implies that if KE increases, PE decreases.
∴ W = +ΔKE = -ΔPE

If vf2<vi2 → ΔKE = -ve → W = -ve
But we have established that sum total of KE and PE is constant so this necessarily implies that if KE increases, PE decreases.
∴-W = -ΔKE = +ΔPE"

Have I finally got it?
For the case of mechanical interchange between kinetic energy and potential energy, pretty much yes. If there is an additional force applied to the mass, this force does work on the mass and increases the sum of potential energy and kinetic energy. Extending this further into the realm of thermodynamics requires consideration of thermal energy (heat) and the contributions of mechanical energy flow (work) and heat flow to changes in the internal energy of a system.
negation
#11
May8-14, 09:16 PM
P: 819
Quote Quote by Chestermiller View Post
For the case of mechanical interchange between kinetic energy and potential energy, pretty much yes. If there is an additional force applied to the mass, this force does work on the mass and increases the sum of potential energy and kinetic energy. Extending this further into the realm of thermodynamics requires consideration of thermal energy (heat) and the contributions of mechanical energy flow (work) and heat flow to changes in the internal energy of a system.
Awesome! Finally!
For some reason I thought thermodynamics should be the first thing that's taught. It's somehow easier to grasp.


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