Internal energy in expansions

  • Thread starter Big-Daddy
  • Start date
  • #26
2,486
83
Therefore, ΔU for carrying out the reaction at constant volume is the same as ΔU for carrying out the reaction at constant pressure.
Do you want to say that
ΔU=ΔH-PΔV=ΔH-VΔP
If yes ,then how?I understood that from your post #11
PΔV=VΔP=ΔnRT
 
  • #28
20,767
4,494
Do you want to say that
ΔU=ΔH-PΔV=ΔH-VΔP
No. This is not correct mathematically. You are confusing Δ's with d's.

Chet
 
  • #29
2,486
83
The definition of H is U + PV.
I did not understand.
You mean I should not use delta?
 
  • #30
20,767
4,494
I did not understand.
What is it that you don't understand? Enthalpy H is a physical property of a material at thermodynamic equilibrium, defined by the equation H ≡ U + PV, where U is the internal energy of the material, P is the pressure and V is the volume of material.

Chet
 
  • #33
2,486
83
From your post 11
. 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).
ΔU=ΔH−Δ(PV
What about volume?is it constant or not in your above case mentioned.
 
  • #34
20,767
4,494
From your post 11


What 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
2,486
83
Should not it be ΔU=ΔH−PΔV=ΔH−Δn(RT)
No
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
20,767
4,494
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
2,486
83
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
20,767
4,494
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
2,486
83
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
20,767
4,494
While doing this the pressure will change,right?
Sure.

Chet
 

Related Threads on Internal energy in expansions

  • Last Post
Replies
15
Views
797
  • Last Post
Replies
3
Views
696
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
4K
  • Last Post
Replies
1
Views
1K
Replies
3
Views
134
  • Last Post
Replies
2
Views
876
  • Last Post
Replies
2
Views
16K
  • Last Post
Replies
3
Views
2K
Replies
8
Views
344
Top