In my 12th standard book, the formula for entropy change is given as $\Delta S = \frac{q_\text{reversible}}{T}$. What is the importance of absorbing heat reversibly and not irreversibly? What does absorbing heat reversibly mean? Does it mean that the process should be reversible, and if it does, then how would we define change in entropy of a spontaneous process i.e an irreversible process? Would its formula also include $q_\text{reversible}$?
Answer
A process is thermodynamically reversible if it is essentially at equilibrium. Specifically, the system and its surroundings stay infinitesimally close to equilibrium with each other throughout a reversible process. Small changes in intensive variables of the system are perfectly balanced by changes in those variables in the surroundings. For example, $T_{\text{system}} = T_{\text{surroundings}}$ if heat can be exchanged reversibly; any temperature difference will result in an irreversible exchange of heat.
What's special about reversible heat transfer? When a tiny amount of heat $dq_{\text{rev}}$ is released by the system during a reversible process, the surroundings will absorb exactly that amount of heat, no matter what path the process takes.
If you have an irreversible process, the amount of heat (and work) depends on how the system and surroundings change during the process.
What's the difference between reversible and irreversible heat? The heat transferred by an irreversible process $dq_{\text{irrev}}$ is less than $dq_{\text{rev}}$, where $dq_{\text{rev}}$ is the heat that would have been transferred if the process had been performed reversibly. The difference between the two is the amount of work you'd have to do on the system to reverse the effects of the irreversible process.
How do we know that? Energy $U$ is a state function, so it'll be the same whichever way we do the process: $$dU = dq_{\text{irrev}} + dw_{\text{irrev}} = dq_{\text{rev}} + dw_{\text{rev}}$$ so the difference between the two heats is $$dq_{\text{rev}} - dq_{\text{irrev}} = dw_{\text{irrev}} - dw_{\text{rev}}$$
The differences on either side are positive, so $$dq_{rev} > dq_{irrev}$$ This means that an irreversible process will always absorb less heat from the surroundings and perform less work on the surroundings than it would have if the process had been done reversibly.
Why $dS = dq_{\text{rev}}/T$? With a reversible process, $$\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{system}} = -\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{surroundings}}$$ so if we define $dS \equiv dq_{\text{rev}}/T$, then we have $dS_{\text{universe}} = dS_{\text{system}} + dS_{\text{surroundings}} = 0$.
Consider what would happen in an irreversible process that took the system from state A to state B. The inequality $dq_{\text{rev}} > dq_{\text{irrev}}$ together with the definition of entropy in terms of reversible heats gives $$\Delta S = S_B - S_A = \int_A^B\frac{dq_{\text{rev}}}{T} > \int_A^B\frac{dq_{\text{irrev}}}{T}$$ for any such process. This also implies that for any cyclic process (e. g. A to B back to A), we have $$\oint\frac{dq_{\text{irrev}}}{T} \lt 0, \ \text{and}\ \Delta S = \oint\frac{dq_{\text{rev}}}{T} = 0$$ This shows that $q_{\text{rev}}/T$ is a state function (it's zero for any cyclic process) but $q_{\text{irrev}}/T$ is not a state function. If entropy is to be a state function, then, you can't define it as simply $q/T$; you've got to use $\Delta S = q_{\text{rev}}/T$, or for the differential form $dS = dq_{\text{rev}}/T$.
How could we relate the entropy change for an irreversible process to $dq_{\text{irrev}}$? From the difference between $q_{rev}$ and $dq_{irrev}$ above and the definition of entropy, we have $$dS = \frac{dq_{\text{rev}}}{T} = \frac{dq_{\text{irrev}}}{T} + \frac{dw_{\text{irrev}}-dw_{\text{rev}}}{T}$$
TL;DR: There IS no TL;DR; you're asking some important questions here, and the answers need to be made carefully.
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