# Work done on a gas when it is compressed quasi-statically

• ChiralSuperfields
In summary: So...##\vec F ·\vec {dr} = (- F ~\hat j) · (dy ~\hat j) = - F ~dy##Which is what the original expression said. The negative sign is a result of replacing vectors with their components.##\vec F ·\vec {dr} = \vec F · dy ~\hat j = (- F ~\hat j) · (dy ~\hat j) = - F ~dy##In summary, for this derivation, the bit highlighted in orange is not positive due to a notation/convention issue. The force is represented by a negative unit vector, but the displacement is represented by a positive infinitesimal change in position. This results in
ChiralSuperfields
Homework Statement
Relevant Equations
Pleas see below
For this derivation,

I am not sure why the bit highlighted in orange is not positive since the displacement of the piston is downwards in the same direction as the force applied.

Many thanks!

Callumnc1 said:
For this derivation,
View attachment 323568

I am not sure why the bit highlighted in orange is not positive since the displacement of the piston is downwards in the same direction as the force applied.
It appears to be a notation/convention issue.

When we write ##\vec F = -F {\hat j}## the (unadorned) letter ‘##F##’ means the magnitude of ##\vec F##. This is a common convention – shorthand for writing ##|\vec F|##. Of course, ##F## is, by definition always non-negative.

But ##dy##, it is not a magnitude. It is an (infinitesimal) change in ##y##. It can be positive (an increase in ##y##) or negative (a decrease in ##y##).

In the situation described, ##-PAdy## will be positive because ##dy## is negative.

Edit - typo's

ChiralSuperfields and MatinSAR
Callumnc1 said:
Relevant Equations:: Pleas see below

For this derivation,
View attachment 323568

I am not sure why the bit highlighted in orange is not positive since the displacement of the piston is downwards in the same direction as the force applied.

Many thanks!
It appears to me that y is taken to be positive up, so dy will be negative. That is independent of which way the force acts.

ChiralSuperfields and MatinSAR
By definition
##d\mathbf{r}=dx~\mathbf{\hat {i}}+dy~\mathbf{\hat {j}}+dz~\mathbf{\hat {k}}.##
Here, ##\mathbf{F}=-F\mathbf{\hat {j}}##.
Therefore,
##\mathbf{F}\cdot d\mathbf{r}=(-F\mathbf{\hat {j}})\cdot(dx~\mathbf{\hat {i}}+dy~\mathbf{\hat {j}}+dz~\mathbf{\hat {k}})=-Fdy.##

ChiralSuperfields and MatinSAR
Steve4Physics said:
It appears to be a notation/convention issue.

When we write ##\vec F = -F {\hat j}## the (unadorned) letter ‘##F##’ means the magnitude of ##\vec F##. This is a common convention – shorthand for writing ##|\vec F|##. Of course, ##F## is, by definition always non-negative.

But ##dy##, it is not a magnitude. It is an (infinitesimal) change in ##y##. It can be positive (an increase in ##y##) or negative (a decrease in ##y##).

In the situation described, ##-PAdy## will be positive because ##dy## is negative.

Edit - typo's

So are you saying that the textbook is wrong since ##\vec dr = -dy\hat j##?

Many thanks!

haruspex said:
It appears to me that y is taken to be positive up, so dy will be negative. That is independent of which way the force acts.

The force acts downwards and the differential displacement is also in the same direction.

Many thanks!

kuruman said:
By definition
##d\mathbf{r}=dx~\mathbf{\hat {i}}+dy~\mathbf{\hat {j}}+dz~\mathbf{\hat {k}}.##
Here, ##\mathbf{F}=-F\mathbf{\hat {j}}##.
Therefore,
##\mathbf{F}\cdot d\mathbf{r}=(-F\mathbf{\hat {j}})\cdot(dx~\mathbf{\hat {i}}+dy~\mathbf{\hat {j}}+dz~\mathbf{\hat {k}})=-Fdy.##

But why dose ##\vec dr## not equal ##-dy\hat j##?

Many thanks!

Callumnc1 said:

But why dose ##\vec dr## not equal ##-dy\hat j##?

Many thanks!
Because ##\vec r## and ##y## are both positive upwards. ##dy## will have a negative value.
##\vec r=y\hat j##
##\vec r+\vec dr=(y+dy)\hat j##
##\vec dr=(\vec r+\vec dr)-\vec r=(y+dy)\hat j-y\hat j=dy\hat j##

ChiralSuperfields
Callumnc1 said:
View attachment 323595
Many thanks!
That diagram is a little unhelpful since it makes it look as though dy is a magnitude, so positive. It would have been better to illustrate it as an upward displacement.
Callumnc1 said:
The force acts downwards and the differential displacement is also in the same direction.
Why do you keep mentioning that the force is downward? I already replied that that is irrelevant to the sign on dy in the equation.

ChiralSuperfields and MatinSAR
haruspex said:
Because ##\vec r## and ##y## are both positive upwards. ##dy## will have a negative value.
##\vec r=y\hat j##
##\vec r+\vec dr=(y+dy)\hat j##
##\vec dr=(\vec r+\vec dr)-\vec r=(y+dy)\hat j-y\hat j=dy\hat j##
haruspex said:
That diagram is a little unhelpful since it makes it look as though dy is a magnitude, so positive. It would have been better to illustrate it as an upward displacement.

Why do you keep mentioning that the force is downward? I already replied that that is irrelevant to the sign on dy in the equation.
Thank you for your replies @haruspex!

I will do some more thinking.

Many thanks!

Callumnc1 said:

So are you saying that the textbook is wrong since ##\vec dr = -dy\hat j##
You may have reolved this by now, given the recent posts. But if not...

I'm not saying the textbook is wrong. Nor am I saying ##\vec {dr} = -dy~\hat j##. The correct statement is ##\vec {dr} = dy~\hat j## (where ##dy## happens to be negative here).

Your Post #1 attachment says ##\vec F ·\vec {dr} = - F ~\hat j · dy ~\hat j##.

##\vec F## has been replaced by ##- F ~\hat j## (because ##F## is always positive (a magnitude) and ##\vec F## act downwards).

##\vec {dr}## has been replaced by ##dy ~\hat j## (where ##dy## is negative).

ChiralSuperfields
Steve4Physics said:
You may have reolved this by now, given the recent posts. But if not...

I'm not saying the textbook is wrong. Nor am I saying ##\vec {dr} = -dy~\hat j##. The correct statement is ##\vec {dr} = dy~\hat j## (where ##dy## happens to be negative here).

Your Post #1 attachment says ##\vec F ·\vec {dr} = - F ~\hat j · dy ~\hat j##.

##\vec F## has been replaced by ##- F ~\hat j## (because ##F## is always positive (a magnitude) and ##\vec F## act downwards).

##\vec {dr}## has been replaced by ##dy ~\hat j## (where ##dy## is negative).

That helps more. But if ##dy## is negative, then ##dV = A dy## then the volume would be negative?

Many thanks!

Callumnc1 said:
That helps more. But if ##dy## is negative, then ##dV = A dy## then the volume would be negative?
The volume is never negative. The change in volume can be positive or negative.

PhDeezNutz, Steve4Physics and ChiralSuperfields
kuruman said:
The volume is never negative. The change in volume can be positive or negative.

Thank you, you are very right. Why did the textbook not take write ##dy## as ##-dy \hat j## and assume ##dy## was positive? Is this just a matter of convention?

Many thanks!

Callumnc1 said:

Thank you, you are very right. Why did the textbook not take write ##dy## as ##-dy \hat j## and assume ##dy## was positive? Is this just a matter of convention?

Many thanks!
Because they defined y as the height, and dy means the change in y. The height reduced, so dy is negative.

ChiralSuperfields
haruspex said:
Because they defined y as the height, and dy means the change in y. The height reduced, so dy is negative.
Thank you @haruspex! That makes sense!

Callumnc1 said:
Why did the textbook not take write ##dy## as ##-dy \hat j## and assume ##dy## was positive? Is this just a matter of convention?

Yes, it is a convention on the way differentials are used. A differential (e.g. ##dy##) represents a change. We don't limit the change to be only an increase or only a decrease.

For example, you could have a problem where a piston oscillates up and down. You want a single equation - not two equations (one for up and another for down).

I guess if you knew that a differential (e.g. ##dy##) was going to turn-out to represent a decrease in ##y## for a particular problem, to be unambiguous you could write it as '##-|dy|##'. But this seems unnecessarily clumsy. And you loose generality because it would be incorrect if applied to a similar problem where ##y## happens to increase.

ChiralSuperfields
Steve4Physics said:

Yes, it is a convention on the way differentials are used. A differential (e.g. ##dy##) represents a change. We don't limit the change to be only an increase or only a decrease.

For example, you could have a problem where a piston oscillates up and down. You want a single equation - not two equations (one for up and another for down).

I guess if you knew that a differential (e.g. ##dy##) was going to turn-out to represent a decrease in ##y## for a particular problem, to be unambiguous you could write it as '##-|dy|##'. But this seems unnecessarily clumsy. And you loose generality because it would be incorrect if applied to a similar problem where ##y## happens to increase.
Thank for that @Steve4Physics! That is helpful!

Steve4Physics
Hi Everyone,

I am looking into the derivation again,

and I notice that let F = PA when ##dW = - F dy = -PA dy## does this mean that the work is actually slightly greater than ##dW = - P dV## since, initially, the force compressing the piston will be slightly greater than force from the compressed air inside the piston, in order for it to start moving inwards. Are they assuming that the piston is being compressed at a very slow (quasi-static) constant speed so that the forces are balanced?

Many thanks!

ChiralSuperfields said:
Are they assuming that the piston is being compressed at a very slow (quasi-static) constant speed so that the forces are balanced?
Yes they are. It's in the description of the physical that you posted emphasized in bold.

ChiralSuperfields
kuruman said:
Yes they are. It's in the description of the physical that you posted emphasized in bold.
Great, thank you for your help @kuruman!

## 1. What is work done on a gas when it is compressed quasi-statically?

Work done on a gas when it is compressed quasi-statically refers to the energy transferred to the gas as it is compressed slowly and continuously, without any sudden changes or fluctuations in pressure.

## 2. How is work calculated in quasi-static compression of a gas?

The work done on a gas during quasi-static compression can be calculated using the equation W = -PΔV, where W is work, P is pressure, and ΔV is the change in volume of the gas.

## 3. What factors affect the amount of work done on a gas during quasi-static compression?

The amount of work done on a gas during quasi-static compression is affected by the initial and final volumes of the gas, the pressure applied, and the type of gas being compressed.

## 4. Can work done on a gas during quasi-static compression be negative?

Yes, work done on a gas during quasi-static compression can be negative if the gas expands instead of being compressed. This means that the gas is doing work on its surroundings, rather than work being done on the gas.

## 5. What is the significance of quasi-static compression in thermodynamics?

Quasi-static compression is important in thermodynamics because it allows for the accurate calculation of work done on a gas, which is a key factor in determining the change in internal energy of the gas. It also helps to simplify calculations and make them more precise.

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