PV^(gamma)=constant?

Clausius2
Gold Member
asdf1 said:
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.

Andrew Mason
Homework Helper
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! :)