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

[member 731016]

Homework Statement

Please below

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!

 

[Steve4Physics]

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

 

[haruspex]

Homework Statement:: Please below
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.

 

[kuruman]

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,
$\mathbf{F}\cdot d\mathbf{r}=(-F\mathbf{\hat {j}})\cdot(dx~\mathbf{\hat {i}}+dy~\mathbf{\hat {j}}+dz~\mathbf{\hat {k}})=-Fdy.$

 

[member 731016]

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

Thank you for your reply @Steve4Physics !

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

Many thanks!

 

[member 731016]

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.

Thank you for your reply @haruspex!

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

Many thanks!

 

[member 731016]

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,
$\mathbf{F}\cdot d\mathbf{r}=(-F\mathbf{\hat {j}})\cdot(dx~\mathbf{\hat {i}}+dy~\mathbf{\hat {j}}+dz~\mathbf{\hat {k}})=-Fdy.$

Thank you for your reply @kuruman!

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

Many thanks!

 

[haruspex]

Thank you for your reply @kuruman!

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.
$\vec r=y\hat j$
$\vec r+\vec dr=(y+dy)\hat j$
$\vec dr=(\vec r+\vec dr)-\vec r=(y+dy)\hat j-y\hat j=dy\hat j$

 

[haruspex]

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.

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.

 

[member 731016]

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

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!

 

[Steve4Physics]

Thank you for your reply @Steve4Physics !

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).

 

[member 731016]

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).

Thank you for your reply @Steve4Physics!

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

Many thanks!

 

[kuruman]

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.

 

[member 731016]

The volume is never negative. The change in volume can be positive or negative.

Thank you for your reply @kuruman!

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!

 

[haruspex]

Thank you for your reply @kuruman!

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.

 

[member 731016]

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!

 

[Steve4Physics]

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

To add a little more...

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.

 

[member 731016]

To add a little more...

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!

 

[member 731016]

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!

 

[kuruman]

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.

 

[member 731016]

Yes they are. It's in the description of the physical that you posted emphasized in bold.

Great, thank you for your help @kuruman!