How to Determine ΔU in Expanding Systems with Changing Pressure?

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Discussion Overview

The discussion revolves around the determination of internal energy change (ΔU) in thermodynamic systems during expansion, particularly when pressure is changing. Participants explore the implications of different formulations of the first law of thermodynamics and the conditions under which they apply, including reversible and irreversible processes.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that ΔU=q+w, where work (w) is defined as work done on the system minus work done by the system, leading to questions about how to summarize this for expanding systems.
  • There is a proposal that ΔU=q-Δ(pV) suggests ΔH=q without constant pressure, which raises concerns about its validity.
  • Others argue that ΔU=q-pΔV is correct, but this formulation appears to assume constant pressure, prompting questions about how to evaluate ΔU in isochoric processes with changing pressure.
  • One participant emphasizes that the first law applies only to changes between states of thermodynamic equilibrium, suggesting that the process must be reversible for certain equations to hold.
  • Concerns are raised about the definition of work in non-quasi-static expansions, where the pressure of the gas may not be uniform, and how this affects the calculation of work done on the surroundings.
  • Another participant questions the application of the ideal gas approximation in the context of pressure changes and seeks clarification on the relationship between ∫Vdp and ΔngasRT.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of ΔU and the conditions under which various equations apply. There is no consensus on the validity of ΔU=q-Δ(pV) versus ΔU=q-pΔV, and the discussion remains unresolved regarding the implications of irreversible processes and the ideal gas approximation.

Contextual Notes

Limitations include the dependence on assumptions about reversibility, equilibrium states, and the nature of pressure changes during expansion. The discussion highlights the complexity of applying thermodynamic principles in varying conditions.

  • #31
gracy said:
ΔU=ΔH-PΔV=ΔH-VΔP
chestermiller said:
You mean I should not use delta?
 
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  • #32
Yes. Δ(PV) ≠ PΔV + VΔP
 
  • #33
From your post 11
Chestermiller said:
. But, before we do that, let's first consider the change in the internal energy if the reaction is carried out at constant pressure (and, of course, constant temperature).
Chestermiller said:
ΔU=ΔH−Δ(PV
What about volume?is it constant or not in your above case mentioned.
 
  • #34
gracy said:
From your post 11What about volume?is it constant or not in your above case mentioned.
Not if the total number of moles change as a result of the reaction.

Chet
 
  • #35
gracy said:
Should not it be ΔU=ΔH−PΔV=ΔH−Δn(RT)
Chestermiller said:
No
Chestermiller said:
first consider the change in the internal energy if the reaction is carried out at constant pressure (and, of course, constant temperature)
So why my equation is wrong?I have taken P constant as you have mentioned in above condition (I have underlined) and changing volume as you clarified in last post.
 
  • #36
gracy said:
So why my equation is wrong?I have taken P constant as you have mentioned in above condition (I have underlined) and changing volume as you clarified in last post.
Ugh. It's been a while since I've looked at this thread, so I got a little confused. I thought that what you were asking was whether your equation was true in general. Of course, it is not. But, for the case of an ideal gas mixture undergoing a chemical reaction at constant temperature and pressure, your result is correct. Sorry for the confusion.

Chet
 
  • #37
Chestermiller said:
Sorry for the confusion.
And what about my post 26?Was it a part of confusion or you would still say it is wrong?
 
  • #38
gracy said:
And what about my post 26?Was it a part of confusion or you would still say it is wrong?
We are talking about two different processes here (reaction at constant pressure, and reaction at constant volume), and, in general, each process would have its own ΔU and ΔH. Also we need to recognize that the pressure changes and volume changes for the two processes will, in general, differ.

So far, we have determined the ΔU and ΔH for the constant pressure process (call this step 1), and now we want to find out what these changes would be for the constant volume process. We can do this by adding to step 1 an additional process step (step 2) to compress the products back down to the original volume at constant temperature. Then, all we need to do is add the contributions of the two process steps to obtain the overall changes in U and H. This will give us the ΔU and ΔH for the constant volume process. But, we also know that we are dealing with an ideal gas mixture. We know that for an ideal gas mixture, both the internal energy and the enthalpy are functions only of temperature, and not pressure. So, in step 2, when we compress the products back down to the original volume at constant temperature, there will be no change in either U or H. So, for a chemical reaction involving a mixture of ideal gases at constant temperature, ΔU and ΔH are the same for the reaction carried out at constant volume as they are for the reaction carried out at constant pressure.

Hope this makes sense.

Chet
 
  • #39
Chestermiller said:
We can do this by adding to step 1 an additional process step (step 2) to compress the products back down to the original volume at constant temperature
While doing this the pressure will change,right?
 
  • #40
gracy said:
While doing this the pressure will change,right?
Sure.

Chet
 

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