- #1
Andrew_
- 13
- 0
According to my thermal physics book , from classical thermodynamics for a reversible change :
[tex]dU = dQ - dW[/tex]
From stat mech ,
[tex]dU = \sum_i \epsilon _i dn_i + \sum_i n_i d \epsilon _i[/tex]
Then by drawing some arguments from quantum mechanics the writer relates the change in volume and therefore work to the change in the energy levels. ( although there are more general treatments that include electric/magnetic fields ... )
The populations , ni , however, are unaltered. Only by supplying heat to the system do the populations change and therefore will there be entropy change to the system ofcourse without altering the energy levels.
But what if the process were irreversible ? For an adiabatic irreversible change dQ =0 , so dS > 0 , and therefore I can safely conclude that the separation of dW and dQ terms cannot be done here. That is , the usual interpretation of work and heat is just wrong for any irreversible change.
Is this correct ?
Thanks,
[tex]dU = dQ - dW[/tex]
From stat mech ,
[tex]dU = \sum_i \epsilon _i dn_i + \sum_i n_i d \epsilon _i[/tex]
Then by drawing some arguments from quantum mechanics the writer relates the change in volume and therefore work to the change in the energy levels. ( although there are more general treatments that include electric/magnetic fields ... )
The populations , ni , however, are unaltered. Only by supplying heat to the system do the populations change and therefore will there be entropy change to the system ofcourse without altering the energy levels.
But what if the process were irreversible ? For an adiabatic irreversible change dQ =0 , so dS > 0 , and therefore I can safely conclude that the separation of dW and dQ terms cannot be done here. That is , the usual interpretation of work and heat is just wrong for any irreversible change.
Is this correct ?
Thanks,