Can someone explain this thermodynamics related doubt i have?

Click For Summary
SUMMARY

This discussion centers on the thermodynamic analysis of two scenarios involving a piston-cylinder arrangement under reversible conditions. In Scenario One, heat (q) is added suddenly, leading to a temperature and pressure rise while maintaining constant volume, described by the equation q = n Cv dT. In Scenario Two, heat is added at constant pressure (p1), resulting in expansion and a change in internal energy. The key question posed is whether these scenarios are equivalent in terms of internal energy change, work done, and final volume expansion.

PREREQUISITES
  • Understanding of the first law of thermodynamics
  • Knowledge of adiabatic processes and their equations
  • Familiarity with isobaric processes and their implications
  • Basic concepts of internal energy and work in thermodynamic systems
NEXT STEPS
  • Study the first law of thermodynamics in detail
  • Learn about adiabatic and isobaric processes in thermodynamics
  • Explore the derivation and application of the equation q = n Cv dT
  • Investigate the relationship between work done and internal energy changes in thermodynamic systems
USEFUL FOR

Students and professionals in mechanical engineering, chemical engineering, and physics who are looking to deepen their understanding of thermodynamic processes and their implications in practical applications.

lawofnature
Messages
2
Reaction score
1
Assume a piston cylinder arrangement. External pressure always equals internal pressure ( reversible conditions ). Now consider these two scenarios:

Scenario one: q heat is suddenly added to the arrangement causing a temperature rise and consequently pressure rise, but the volume is yet constant. That implies:
q = n Cv dT
Now let's decrease the pressure adiabatically to initial pressure p1 (before temp. rise). There will be adiabatic work done in expanding to lower pressure at the cost of internal energy:
p dV = n Cv dT

Scenario two: q heat is added at a constant pressure p1 (isobaric process) causing expansion and change in internal energy.

Are the two scenarios equivalent in terms of change in internal energy, work done and final expansion (of volume)?
 
Science news on Phys.org
lawofnature said:
Assume a piston cylinder arrangement. External pressure always equals internal pressure ( reversible conditions ). Now consider these two scenarios:

Scenario one: q heat is suddenly added to the arrangement causing a temperature rise and consequently pressure rise, but the volume is yet constant. That implies:
q = n Cv dT
Now let's decrease the pressure adiabatically to initial pressure p1 (before temp. rise). There will be adiabatic work done in expanding to lower pressure at the cost of internal energy:
p dV = n Cv dT

Scenario two: q heat is added at a constant pressure p1 (isobaric process) causing expansion and change in internal energy.

Are the two scenarios equivalent in terms of change in internal energy, work done and final expansion (of volume)?
Why don't you just run the calculations as see for yourself what happens (rather than guessing)?

Chet
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
6K
  • · Replies 4 ·
Replies
4
Views
897
Replies
5
Views
1K
  • · Replies 60 ·
3
Replies
60
Views
10K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 46 ·
2
Replies
46
Views
6K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
4K