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Physics - Work done

Hi, so the bits highlighted are what I don't get.

compression means system volume is decreasing so how is it positive for compression?

Would really like some help!! Thanks!! :smile:

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They are just saying, if dV and p dV are negative, then - p dV will be positive.
(I think - I always got in a tangle over = & - signs)

Reply 3

h26

Original post by old_teach

They are just saying, if dV and p dV are negative, then - p dV will be positive.
(I think - I always got in a tangle over = & - signs)

Thanks for the help! But tbh, I am still confused :/ I don't understand why that is the case and how that relates to my previous question :/

Reply 4

username3442196

Ok - I'll write some random stuff and see if any helps.
dV: if the volume is getting smaller, the change of volume is negative, i.e. dV is negative during compression.
work done by gas: I prefer to think of a gas expanding, perhaps a steam engine pushing the piston out and driving the wheels - this is +ve dV and the gas is doing work.
So if an expanding gas does work, a compressed gas has work done on it, so you'd say the work done by the gas is negative.
Helpful?

Reply 5

h26

Original post by old_teach

Thank you very much! :smile:

This was very helpful.

But for your last point, isn't it the opposite way round?
So for the case of compression: The surroundings(the piston i think ) does work on the gas (system) , and work done on a gas is negative.

This is given here: the bit in red

Could you please kindly explain this? Thanks very much!!

Reply 6

username3442196

I think we've both said it the same way?
You're a good student to worry so much - but I think we agree - sorry - I'm babysitting and she's crying!

Reply 7

h26

Original post by old_teach

I think we've both said it the same way?
You're a good student to worry so much - but I think we agree - sorry - I'm babysitting and she's crying!

I am not too sure - I have just quoted below what you kindly mentioned "So if an expanding gas does work, a compressed gas has work done on it, so you'd say the work done by the gas is negative."

I said work done on the gas is negative. Just quoted what I said below:
"But for your last point, isn't it the opposite way round?
So for the case of compression: The surroundings(the piston i think ) does work on the gas (system) , and work done on a gas is negative. "
work done by the gas on the surroundings is not really the same as work done on the gas by the surroundings.

This is what is provided by the book:

So I am not sure if we are saying the same thing haha.

Apologies for all the headache - I am quite a worrier and try to understand things perfectly and not memorise it and that got me in trouble at school! I understand you must be very busy with other reposibilities :smile:

so no worries at all if you can't reply as soon enough

(edited 5 years ago)

Reply 8

h26

Thanks! But what what do you mean by "against" in this context so work being done by the system on the surrounding or work being done on the system by the surroundings?

Reply 9

h26

Oh wait to edit my reply back:

so work being done on the sytem (air/gas) by the surroundings is the same thing as saying work is being done against the system and this is negative.

so therfore compression must be negative but it is apparently positive.

Could you please kindly let me know your thoughts on this whenever possible? Thanks!

Reply 10

h26

Haha we replied at exactly the same time :biggrin:

Thanks, so compression would be an example.

Could you please kindly let me know your thoughts about my other reply if you have time?

Really appreciate the help

Reply 11

h26

I think so! So the first one makes sense I think but for the second part"On the other hand, when the system expands, the force vector ( which is caused by the surroundings,in this case still pushing the air) is moving in the direction opposite to itself as the piston is pushed by the air. Hence work done by the surroundings is negative in this case."

so the force vector which is caused by the surroundings is instead compressing air on the other side of the piston? Wait nevermind I don't really udnerstand the second part

I am also trying to relate it to the formulas in my notes but that isn't going too great at the moment

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(edited 5 years ago)

Reply 12

h26

I mean the force by the surroundings is still in the same direction but the direction of motion has changed as the particles are expanding and getting bigger so the direction of motion is opposite the pushing force by the surroundings so therefore the pushing force by the surroundings is in the negative direction and so work done by the surroundings is negative?

Dunno if I am correct - assumed that cause of as mechanics :biggrin:

(edited 5 years ago)

Reply 13

h26

Oh wow, thanks for helping me get somewhere haha!

Ok, so now I am trying to relate them to the formulas in these notes and just doing that now.

Just thought to let you know so you don't think I just vanished after you said I am correct lol.

Reply 14

h26

So after reading the second paragraph, I just don't get the bit in red.

I fully understand that work done on the system by the surrounding is positive during compression and negative during expansion btw. The bit in red though slightly confuses me.

Btw THank you so much for all your help - it means a lot!!!

Reply 15

h26

Oh yeah, sorry one more thing:

Why is the change in volume negative during compression? Isn't it a change and therefore it cannot be a negative value.

Reply 16

h26

Oh, so due to the fact dv will be negative in compression, they have a random negative sign in the formula

so that they can make the work done by the surroundings on the system positive. This is because compression should result in positive work done by the surroundings on the system and the negative sign in the formula allows for that to happen.

And for expansion, dv is positive so the negative sign in the formula makes the work done by the surrounding on the system negative. This is because expansion should result in negative work done by the surroundings on the system and the negative sign in the formula allows for that to happen.

Wait. Is this correct? :biggrin:

Reply 17

h26

Thank you so much!! :smile:

I really really appreciate it!

Sorry for the slightly late replies - there's something seriously wrong with my laptop as it tends to freeze a lot and I lose what I wrote to you.

I am just going over the equations in redd at the moment .

If anything I don't get just on this page , is it ok if I ask you as you are really good at explaining (as if I have not given you enough hesadache lol)? But no worries if you don't have time.

Edit: wow it doesn't even let me upload the file. ok it worked finally :biggrin:

Attachment not found

(edited 5 years ago)

Reply 18

h26

In first paragrapph
"If the system increases its volume by counteracting external forces it does mechanical work equal to the product of force and distance along which the force is exerted..."

What counteracting of external forces are they talking about here - kinda confused me.

In last last paragraph
"In systems that are more complicated , we may also encounter chemical work, electrcial work etc..the formula is as follows"

I don't quite understand teh formula they have provided.

Could you please kindly help me with these as that's pretty much it of this chapter :biggrin: