is the formula for Q wrong?
if internal energy = 0 shouldn't Q=W where system performing work on the surroundings = + and heat flow in is + ???
Why does John have Q = - W (i understand this if its in chemistry but this is a physics formula...)
Isothermal process
- Thread starter MrNeuro
- Start date
is the formula for Q wrong?
if internal energy = 0 shouldn't Q=W where system performing work on the surroundings = + and heat flow in is + ???
Why does John have Q = - W (i understand this if its in chemistry but this is a physics formula...)
- Joined
- Dec 30, 2009
- Messages
- 2,640
- Reaction score
- 527
- Points
- 5,171
- Location
- Where the rain grows
- Pre-Medical
Q and W are from the point of view of the system. If you are adding W, Q has to be negative (lose heat) to keep U constant (ΔU=0).
so then how do you answer this question
based upon what your saying if work is done on the system then heat should flow out...the naser though is heat flowed into the reservoir
dooohhhhh i need to read carefully from now on i thought it was saying that heat is flowing into the container not the reservoir.... lol thanks milski...again
Last edited:
- Joined
- Apr 6, 2010
- Messages
- 2,439
- Reaction score
- 4
- Points
- 4,551
- Medical Student
(i missed your last edit. too late)
different books use different equations. namely, W can be work done ON the system or work can be done BY the system. this changes the sign and really just depends on how you prefer to think about it.
The first equation is Q = -W. This means the heat gained is equal to the work done on the system.
You can change it to be Q = W. This means heat gained is equal to work done by system.
different books use different equations. namely, W can be work done ON the system or work can be done BY the system. this changes the sign and really just depends on how you prefer to think about it.
The first equation is Q = -W. This means the heat gained is equal to the work done on the system.
You can change it to be Q = W. This means heat gained is equal to work done by system.
For the first one is it heat gained by the surroundings???
Bc if you work on the system heat should be lost by the system.
Bc if you work on the system heat should be lost by the system.
so then how do you answer this question
based upon what your saying if work is done on the system then heat should flow out...the naser though is heat flowed into the reservoir
dooohhhhh i need to read carefully from now on i thought it was saying that heat is flowing into the container not the reservoir.... lol thanks milski...again
Wait, why would the internal energy not increase? The gas is being compressed so the energy should increase. Since E internal is proportional to temperature, if heat is flowing into the reservior, it indicates that the temperature increased...which is against the question?!
- Joined
- Dec 30, 2009
- Messages
- 2,640
- Reaction score
- 527
- Points
- 5,171
- Location
- Where the rain grows
- Pre-Medical
The reservoir is not part of the system - only the piston and the cilinder are considered here. The energy flows is as increased pressure and out as heat in the reservoir, as a result there is no energy change for the cilinder with the piston itself.
Oh, so the increase in pressure causes internal kinetic energy to rise, causing increased temperature, which flows from the cylinder to the tank (from hot to cold)?
- Joined
- Dec 30, 2009
- Messages
- 2,640
- Reaction score
- 527
- Points
- 5,171
- Location
- Where the rain grows
- Pre-Medical
Yes. You can consider the heat reservoir some sort of a cooler for the cilinder.
(Hm, cilinder seems to be just as correct as cylinder, according to Apple?)
- Joined
- Dec 30, 2009
- Messages
- 2,640
- Reaction score
- 527
- Points
- 5,171
- Location
- Where the rain grows
- Pre-Medical
- Joined
- Apr 8, 2011
- Messages
- 58
- Reaction score
- 1
- Points
- 4,531
- Pre-Medical
Thanks milski. But I'm still a bit confused.
We can arrive to A because we know that change in potential energy = 0, right? And how do we know that (other than by the given info in the passage)? Are all isothermal processes like that? And if so, why?
Oh, is its because deltaU=(3/2)nRdeltaT ???
Last edited:
- Joined
- Dec 30, 2009
- Messages
- 2,640
- Reaction score
- 527
- Points
- 5,171
- Location
- Where the rain grows
- Pre-Medical
It's not really potential energy - it's the internal energy of the gas. You have to ways to change it - by dong (PV) work and by exchanging heat. If the internal energy did not change (since temperature did not change), the energy that was added as work had to be lost somehow and the only way for that is through losing heat.
- Joined
- Apr 8, 2011
- Messages
- 58
- Reaction score
- 1
- Points
- 4,531
- Pre-Medical
Yes, sorry I meant internal energy. I was just trying to understand why the fact that the process is isothermal implies that there is no change in internal energy. But I guess the U=(3/2)nRT explains it!
Thank you! 🙂
