Home » Ask & Discuss » Physics. » Thermal Physics « Back to Discussion
Thermal Physics
entropy in a mess
None
the following pics are taken from resnick halliday fundamentals of physics.
They clearly say that the entropy change of the system for an irreversible process between two equilibrium states i and f (say) will be same as that for a reversible process between the same two states.This is because entropy being state function depends only on the initial and final states...and for convenience we write it as equal to that for the reversible process..as
dSsystem = [i ]
[f ] (dQr / T)........where dQr is the heat taken by the system in the reversible process.
[f ] (dQr / T)........where dQr is the heat taken by the system in the reversible process.Now they again say that total entropy change of system and surrounding for a reversible process will be zero and that for an irreversible process will be greater than zero.
The first part is clear...
if in a reversible process system absorbs heat dQr ,then surrounding also loses the same heat .
So dSsurrounding for reversible process = - [i ]
[f ] (dQr / T)
[f ] (dQr / T)So dSuniverse(reversible process)
= dSsystem+ dSsurrounding = [i ]
[f ] (dQr / T) - [i ]
[f ] (dQr / T) = 0
[f ] (dQr / T) - [i ]
[f ] (dQr / T) = 0Now how should i define dSsurrounding(for irreversible process) such that
dSuniverse(irreversible process) = dSsystem+ dSsurrounding > 0
__________________________________________________________________
It is to be noted that dSsystem = [i ]
[f ] (dQr / T) for both reversible and irreversible
[f ] (dQr / T) for both reversible and irreversible process as entropy of the system is a state function and does not depend on
the path followed (read the passages)
__________________________________________________________________
Comments (4)
NITESH BASKARAN
Blazing goIITian

Joined: 29 Sep 2007
Posts: 487
11 Jan 2008 18:04:40 IST
Like
0 people liked this
entropy is not there in IIT portions.
Reply










