Does an Irreversible Adiabatic Process Defy the Equation PV^(gamma)=constant?

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SUMMARY

An irreversible adiabatic process does not adhere to the equation PV^(gamma)=constant because it lacks the conditions of thermodynamic equilibrium required for the ideal gas law to apply. Specifically, this equation is valid only for isentropic processes, which are reversible and occur with negligible pressure differences. In contrast, an irreversible adiabatic process occurs too rapidly for the ideal gas law (PV=nRT) to hold, resulting in kinetic energy transfer to the surroundings, thus violating the adiabatic condition.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of thermodynamic equilibrium concepts
  • Familiarity with isentropic and irreversible processes
  • Basic principles of kinetic energy in gas dynamics
NEXT STEPS
  • Study the differences between isentropic and irreversible adiabatic processes
  • Explore the implications of kinetic energy transfer in gas expansions
  • Learn about the conditions for thermodynamic equilibrium in gas systems
  • Investigate the mathematical derivations of PV^(gamma)=constant for reversible processes
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Students and professionals in thermodynamics, physicists, and engineers seeking to deepen their understanding of gas behavior during adiabatic processes.

asdf1
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Why doesn't an irreversible adiabatic process follow the equation,
PV^(gamma)=constant?
 
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asdf1 said:
Why doesn't an irreversible adiabatic process follow the equation,
PV^(gamma)=constant?

Although some people here (i.e. my friend Andrew Mason) are of another "school of knowledge", I must say an irreversible adiabatic process is not an isentropic one, and so it is not described by your equation.
 
Clausius2 said:
Although some people here (i.e. my friend Andrew Mason) are of another "school of knowledge", I must say an irreversible adiabatic process is not an isentropic one, and so it is not described by your equation.
For PV^\gamma = constant to apply, the ideal gas law must apply at all times during the process. But this assumes that the system is at perfect thermodynamic equilibrium at all times during the process.

For an adiabatic gas expansion to be reversible, it must occur with an arbitrarily small pressure difference between the gas pressure and the external pressure. If this is the case, the work done by the gas in expanding (\int P_{gas}dv)is equal to the work done on the gas by the external pressure to return it to its original state (\int P_{ext}dv) - hence it is reversible.

Typically an "irreversible adiabatic process" for an ideal gas is a process that occurs without exchange of heat with the surroundings but too rapidly for the relationship PV=nRT to apply during the process. The reason PV=nRT does not apply is because of the kinetic energy factor in the rapidly expanding or contracting gas.

My (mild) disagreement with friend Clausius2 is in calling all such processes adiabatic where kinetic energy is lost from the gas to the surroundings. If the gas expands rapidly and the resulting kinetic energy of the gas is ultimately transferred to the surroundings, I would say that the process is not adiabatic: heat (molecular kinetic energy) is effectively transferred to the surroundings.

AM
 
hmm... that makes sense~
thanks! :)
 

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