Adiabiatic Process: q=0 & Work Done Path Function?

[Rachit Garg]

In an adiabiatic process q=0 so change in internal energy become equal to work. Since internal energy is path function so work done in an adiabiatic process should also be path function then why work done in adibiatic reversible and irreversible different?

 

[PeterDonis]

internal energy is path function so work done in an adiabiatic process should also be path function

This assumes that every state traversed by the process is a thermodynamic equilibrium state. Out of thermodynamic equilibrium, the concept of a "path function" is not well-defined.

why work done in adibiatic reversible and irreversible different?

Because in an irreversible process, the intermediate states are not in thermodynamic equilibrium.

 

[Rachit Garg]

By mistake i have wriitten that i want to say In an adiabiatic process q=0 so change in internal energy become equal to work. Since internal energy is state function so work done in an adiabiatic process should also be state function then why work done in adibiatic reversible and irreversible different?

 

[PeterDonis]

By mistake i have wriitten that i want to say In an adiabiatic process q=0 so change in internal energy become equal to work. Since internal energy is state function so work done in an adiabiatic process should also be state function then why work done in adibiatic reversible and irreversible different?

This is basically the same question you asked in your OP, and I answered it in post #2. (My answer is still valid if you substitute "state function" for "path function".) Do you have questions about my answer?