Internal energy in expansions

In summary, when dealing with thermodynamic expansion, ΔU, and work, it is important to consider if the process is reversible or irreversible and if the initial and final states are equilibrium states. The equation ΔU=q-pΔV is the correct relationship for changes in internal energy, while ΔH=q-pΔV+Δ(pV) is the correct relationship for changes in enthalpy. The external pressure applied to the gas determines the work done in an expansion, and if there is no external pressure, no work is done on the surroundings.
  • #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
 
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  • #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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