Adiabatic Process, Internal Energy vs. Enthelpy.

Click For Summary
SUMMARY

The discussion centers on the relationship between internal energy (ΔU) and enthalpy (ΔH) in adiabatic processes involving ideal gases. The key equations highlighted are ΔH = ΔU + Δ(PV) and ΔU = Q + W, with specific emphasis on the conditions of adiabatic reversible processes where Q = 0. The confusion arises from the assertion that both ΔU and ΔH can equal work (W) in different contexts—specifically, a closed adiabatic system versus an open steady-state adiabatic process. The conclusion clarifies that ΔU relates to constant volume processes while ΔH pertains to constant pressure processes, reinforcing the relationship Cp - Cv = R.

PREREQUISITES
  • Understanding of ideal gas laws and properties
  • Familiarity with thermodynamic concepts such as internal energy and enthalpy
  • Knowledge of adiabatic processes and their characteristics
  • Basic grasp of the first law of thermodynamics
NEXT STEPS
  • Study the derivation of the relationship Cp - Cv = R in detail
  • Explore the implications of adiabatic processes in real-world applications
  • Learn about the differences between closed and open systems in thermodynamics
  • Investigate the role of work in various thermodynamic processes
USEFUL FOR

Students and professionals in thermodynamics, chemical engineering, and physical chemistry who seek to deepen their understanding of energy transformations in adiabatic processes.

Jeremy1789
Messages
2
Reaction score
0
I'm struggling to understand a concept which I assume is basic, but I can't seem to fit the pieces together. When speaking about an ideal gas, I understand that

ΔH = ΔU + Δ(PV) = ΔU + RΔT

So far so good. I also understand the relationship:

ΔU = Q + W... (W here is work being done on the system)

In an adiabatic reversible process, Q = 0, which also makes sense. So,

W = ΔU = nCvΔT

Now, where my confusion lies is in the next part. My book works out a problem, and says:

ΔH = W = nCpΔT

How can both ΔU and ΔH equal W? This doesn't make sense to me, unless Δ(PV) from the first equation was 0. I don't see how this could be 0 unless we were talking about an isothermal case. It also doesn't make sense because W can't equal both nCvΔT and nCpΔT simultaneously, since Cp = Cv + R.

Am I missing something?
 
Science news on Phys.org
Never mind. Just answered my own question. The first instance refers to a closed adiabatic system, and the second refers to an open steady-state adiabatic process.
 
How nCvΔT will equal to W.It is the heat absorbed/released at constant volume only(no work is done at constant volume,the expression
Cv).i think instead of work it is heat,ie ΔU=ΔH=qv at constant volume and ΔH=qp at constant pressuer
actually it is the proof for Cp-Cv=R
 

Similar threads

Undergrad How does ΔU relate to Cv in an adiabatic process?
  • · Replies 13 ·
Replies
13
Views
4K
Graduate Entropy and Enthelpy for adiabatic process
  • · Replies 2 ·
Replies
2
Views
8K
Undergrad Graphical difference between adiabatic and isothermal processes.
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Identification of changes in internal energy with work (in Callen's Thermodynamics)
  • · Replies 5 ·
Replies
5
Views
668
Undergrad Adiabatic cooling in this process involving liquid ammonia
  • · Replies 5 ·
Replies
5
Views
3K
Graduate How to Determine ΔU in Expanding Systems with Changing Pressure?
  • · Replies 39 ·
2
Replies
39
Views
7K
Undergrad Irreversible adiabatic processes & entropy change (clarification needed)
  • · Replies 22 ·
Replies
22
Views
6K
Undergrad Why is the internal energy change of free expansion 0?
  • · Replies 7 ·
Replies
7
Views
5K
High School Change in entropy of reversible isothermal process
  • · Replies 11 ·
Replies
11
Views
3K
High School Question regarding using the expression $dE_{int}$=nC_vdT$
  • · Replies 3 ·
Replies
3
Views
1K