According to the mathematical definition of entropy, it is only defined for reversible processes only. Then how can it be defined for irreversible processes? Please explain clearly.
Asked
Active
Viewed 1,094 times
1 Answers
6
In thermodynamics, from Clausius' theorem, the definition of entropy change is indeed
$$\Delta S_{A \to B} = S(B)-S(A) = \left(\int_A^B \frac{\delta Q} T\right)_R$$
where the subscript $R$ means that the integral must be evaluated along a reversible path.
It doesn't matter which reversible path we choose: the entropy difference between the states $A$ and $B$ will be $\left(\int_A^B \delta Q/T\right)_R$.
Therefore, $\Delta S_{A\to B}$ is path-independent: it only depends on the states $A$ and $B$. To evaluate it, we just have to choose any reversible path connecting $A$ and $B$. When the change of a quantity only depends on the initial and final states, we say that that quantity is a state function.
Now let's say that we perform an irreversible transformation from $A$ to $B$: what is the entropy change? Easy: it is $\left(\int_A^B \delta Q/T\right)_R$, where the integral is evaluated along any reversible path connecting $A$ and $B$.
Notice that it would be wrong to say that
$$\Delta S_{A \to B} = \left(\int_A^B \frac{\delta Q} T\right)_I \ \ \text{(wrong!)}$$
where $I$ is our irreversible path. In fact, the other part of Clausius' theorem tells us that
$$\Delta S_{A \to B} > \left(\int_A^B \frac{\delta Q} T\right)_I \Rightarrow \left(\int_A^B \frac{\delta Q} T\right)_R > \left(\int_A^B \frac{\delta Q} T\right)_I $$
so if you (erroneously) computed $\Delta S_{A \to B}$ as $\left(\int_A^B \delta Q/T\right)_I$ (along the irreversible path) you would be underestimating it.
Summing up: to compute the entropy change for a generic (reversible or irreversible) transformation from $A$ to $B$, choose any reversible path connecting $A$ to $B$ and compute the intrgral $\left(\int_A^B \delta Q/T\right)_R$.
valerio
- 16,751