Animated Solution for Chemistry - Chemical Thermodynamics: Match the thermodynamic processes given under Column-I with the expressions given under Column-II.

Thermodynamics is often perceived as a dense forest of equations, but at its core, it is the study of how nature balances its checkbook. Every process, whether it's ice melting in your glass or a star expanding in the cosmos, obeys the strict rules of energy conservation and entropy. In this classic JEE Advanced problem, we are tasked with analyzing four distinct thermodynamic processes and determining which state variables and path functions vanish or become negative. Let's embark on this journey of logical deduction.

The Freezing of Water

Imagine a beaker of water sitting exactly at $273\text{ K}$ and $1\text{ atm}$. This is the normal freezing point of water. At this precise temperature and pressure, liquid water and solid ice exist in perfect harmony—a state of dynamic equilibrium.

The hallmark of any equilibrium phase transition is that the Gibbs free energy change is zero:

Now, what happens energetically when water freezes? The molecules slow down and lock into a crystalline lattice, releasing heat into the surroundings. Because heat is released, the process is exothermic, meaning $q<0$.

Here is where water throws a curveball: unlike most substances, water expands when it freezes. This expansion means the system pushes against the atmosphere, doing work on the surroundings. By convention, work done by the system is negative, so $w<0$.

Finally, as the chaotic liquid molecules become a highly ordered solid lattice, the randomness of the system decreases. Therefore, the change in entropy of the system is negative:

Thus, Process (A) perfectly matches with (R) and (T).

Expansion into the Void

Next, we consider one mole of an ideal gas expanding into a vacuum under isolated conditions. The word "vacuum" is a massive conceptual trigger. A vacuum means there is absolutely no external pressure ($P_{\text{ext}}=0$). If there is nothing to push against, the gas does no work:

The problem also states the conditions are "isolated." An isolated system cannot exchange heat with its surroundings, which immediately tells us:

According to the First Law of Thermodynamics, the change in internal energy is the sum of heat and work:

Even though the internal energy doesn't change, the gas has expanded to occupy a larger volume. More volume means more available microstates for the gas molecules, leading to an increase in entropy ($\Delta S_{\text{sys}}>0$). Therefore, Process (B) matches with (P), (Q), and (S).

The Chaos of Mixing

In our third scenario, equal volumes of two ideal gases are mixed at constant temperature and pressure inside an isolated container.

Once again, the "isolated" keyword dictates that no heat can enter or leave the system, so $q=0$. Furthermore, because the container is rigid and its total volume remains constant, the system cannot do any expansion work on the outside world, meaning $w=0$.

Applying the First Law just as we did before:

What about entropy? Mixing two distinct gases is a classic example of a spontaneous, irreversible process. The gases diffuse into one another, dramatically increasing the disorder and the number of possible microstates. Consequently, the entropy of the system must increase ($\Delta S_{\text{sys}}>0$). Thus, Process (C) also matches with (P), (Q), and (S).

The Perfect Round Trip

Finally, we look at a cyclic process: reversibly heating hydrogen gas from $300\text{ K}$ to $600\text{ K}$ at constant pressure, and then reversibly cooling it right back to $300\text{ K}$ along the exact same path.

Because the initial and final states of the gas are identical (same temperature, same pressure, same volume), all state functions must have a net change of zero. State functions do not care about the journey; they only care about the destination. Therefore:

But what about the path functions, heat ($q$) and work ($w$)? Normally, in a cyclic process, net heat and net work are not zero (think of a heat engine). However, this specific process retraces its exact steps reversibly. The heat absorbed during the heating phase is perfectly equal and opposite to the heat released during the cooling phase:

Similarly, the expansion work done by the gas during heating is perfectly cancelled by the compression work done on the gas during cooling:

Therefore, Process (D) matches with (P), (Q), (S), and (T).

By carefully dissecting the physical constraints of each process—whether it's an equilibrium phase change, an expansion against zero pressure, or a perfectly reversible cycle—we can confidently map out the thermodynamic consequences. This problem is a beautiful reminder that in thermodynamics, the words "isolated," "vacuum," and "reversible" are not just adjectives; they are strict mathematical constraints.