Work done in a reversible adiabatic expansion

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SUMMARY

The work done during a reversible adiabatic expansion can be determined using the relationship \(\Delta U = c_V \Delta T\), where \(\Delta U\) is the change in internal energy, \(c_V\) is the specific heat at constant volume, and \(\Delta T\) is the change in temperature. In this scenario, since the process is adiabatic, the heat transfer \(Q\) is zero, leading to the conclusion that \(\Delta U = W_{rev}\). The discussion emphasizes that this formula applies to ideal gases and does not require constant volume or pressure conditions.

PREREQUISITES
  • Understanding of thermodynamic concepts such as adiabatic processes and internal energy.
  • Familiarity with the ideal gas law and properties of ideal gases.
  • Knowledge of specific heat capacities, \(c_V\) and \(c_P\).
  • Basic algebra for manipulating thermodynamic equations.
NEXT STEPS
  • Study the derivation of the first law of thermodynamics as it applies to adiabatic processes.
  • Learn about the relationship between \(c_V\), \(c_P\), and the gas constant \(R\) for ideal gases.
  • Explore the implications of reversible processes in thermodynamics.
  • Investigate the application of the ideal gas law in various thermodynamic scenarios.
USEFUL FOR

This discussion is beneficial for students of thermodynamics, engineers working with gas systems, and anyone seeking to deepen their understanding of reversible adiabatic processes in ideal gases.

Trowa
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Hi there!

I have to determine the work done of a reversible adiabatic expansion. Becauce the system is adiabatic: Q = 0 so \DeltaU = Wrev

Becauce both the pressure and the volume changes I can't use W = pex\DeltaV.

Homework Statement


Cv, Cp, Ti, Tf, Pi, Pf, Vi, Vf are known

The Attempt at a Solution



I thought at first that I could use \DeltaU = CV.\DeltaT but the volume is not constant so I don't know if I could use it.

Who can help me find the right formula?

Thanx in Advance!
 
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Hi Trowa, welcome to PF.

(1) Because no heat goes in or out of the system, and because the process is reversible, one state variable remains constant. It's one that you haven't listed. What is it?

(2) \Delta U =c_V\Delta T=(c_P-R)\Delta T always holds for an ideal gas, and doesn't require constant volume, constant pressure, or any other condition.
 
Thanx for the fast response.
:smile:
 

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