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Condition of equilibrium and criterion for a spontaneous process.
The entropy of a system remains unchanged in a reversible change while it increases in an irreversible change i.e.
0 for reversible change
and 0 for irreversible change
For a system with its surrounding
d_{system }+ dS_{surrounding }= 0 for reversible change
and d_{system }+ dS_{surrounding} > 0 for irreversible change
then we have
d_{system }+ dS_{surrounding }≥ 0
The criteria for reversible and irreversible process in terms. Of entropy as well as in terms U, H, A & G.
By the equation,
We know that
q_{irr} < q_{rev} then
TdS > d_{qrev} (for irreversible process)
Thus we write, TdS = dU + PdV (for reversible process)
TdS > dU + PdV (for irreversible process)
Combining we get, TdS ≥ dU + PdV
(i) Criterion in terms of change of entropy: Let U & V remains constant.
T(dS)U, V ≥ 0 or (dS)U, V ≥ 0
The equal sign refers to reversible and greatersign refers to an irreversible process.
(ii) Criterion in terms of change of internal energy: Let S & V constant then
(dS)U, V ≥ 0
(iii) Criterion in terms of change of enthalpy: Let S & P are constant then
TdS ≥ dU + PdV
⇒ dU + PdV ≤ 0
But dU + PdV = dH
⇒ (dH)S, P ≤ 0
(iv) Criterion in terms of change of work function:
We know that dA = dU – TdS & TdS ≥ dU + PdV
then we get dA ≤ –PdV
at constant volume and temperature (dA)V, T ≤ 0
(v) Criterion in terms of change in free energy:
We know that G = H – TS
⇒ G = (U + PV) – TS
⇒ dG = dU + PdV + VdP – TdS – SdT
& TdS ≥ dU + PdV, we get
⇒ dG ≤ VdP – SdT
Let P & T are constant then
(dG)P, T ≤ 0
Using thermodynamic magic square we can find these criterion
(dU)S, V ≤ 0,
(dH)S, P ≤ 0 (dG)P, T ≤ 0,
(dA)V, T ≤ 0 (dS)U, V ≥ 0
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