Thermodynamics mainly deals with (Note: more than one of the given options may be correct.) (i) interrelation

NCERT Exemplar · Q15

Q. Thermodynamics mainly deals with (Note: more than one of the given options may be correct.)

(i) interrelation of various forms of energy and their transformation from one form to another.

(ii) energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules.

(iii) how and at what rate these energy transformations are carried out.

(iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.

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Concept understanding — State Function

What is a State Function? The Intuition

Imagine you're standing at the base of a hill. Your height above sea level is a number — say, 100 metres. Now, you walk to the top of the hill. Your height is now 500 metres.

Here's the key question: Does it matter how you got to the top? Did you take the steep path, the winding road, or did you get carried up by a helicopter?

The answer is no. Your height at the top is 500 metres, regardless of the path you took. Height is a state function — it depends only on where you are, not on how you got there.

Now contrast that with distance walked. If you took the winding road, you walked 2 km. If you took the steep path, you walked 500 m. The distance walked depends entirely on the path. That's not a state function — it's a path function.

Note

A state function is a property whose value depends only on the current state of the system, not on the history or path taken to reach that state.

The Precise Statement

In thermodynamics and physics, a state function (or state variable) is any property of a system that is determined entirely by the system's current equilibrium conditions — typically its temperature, pressure, volume, composition, and so on.

If you know the state of the system (say, "1 mole of ideal gas at 300 K and 1 atm"), then every state function has a fixed value. You don't need to know whether the gas was heated slowly, compressed quickly, or cooled then expanded. The value is the same.

Common State Functions in Chemistry & Physics

Property	Symbol	Why it's a state function
Pressure	$P$	Depends only on current $T$ , $V$ , $n$
Volume	$V$	Depends only on current $P$ , $T$ , $n$
Temperature	$T$	A fundamental state variable
Internal Energy	$U$	Depends only on current $P$ , $T$ , composition
Enthalpy	$H$	$H = U + PV$ — a combination of state functions
Entropy	$S$	Depends only on current state
Gibbs Free Energy	$G$	$G = H - TS$ — again, a combination

Common Path Functions (the opposite)

Property	Why it's a path function
Work ( $W$ )	Depends on how you change volume (e.g., fast vs slow)
Heat ( $Q$ )	Depends on how you transfer energy (e.g., conduction vs radiation)

The Mathematical Signature

Here's the crisp, exam-ready way to recognise a state function:

Important

For a state function $f$ , the cyclic integral is zero:

$\oint df = 0$

This means: if you go from state A to state B and back to A by any path, the net change in $f$ is zero. The value of $f$ at A is always the same when you return.

Equivalently, the change in a state function between two states is path-independent:

\Delta f = f_{\text{final}} - f_{\text{initial}}

This is a single number — no integral over a path needed.

A Concrete Example: Internal Energy

Consider a gas in a cylinder. You take it from State 1 ( $T_1 = 300$ K, $P_1 = 1$ atm) to State 2 ( $T_2 = 400$ K, $P_2 = 2$ atm).

Path A: Heat the gas at constant volume, then compress it at constant temperature.
Path B: Compress the gas at constant temperature, then heat it at constant volume.

The work done ( $W$ ) and heat transferred ( $Q$ ) will be different for Path A vs Path B. But the change in internal energy $\Delta U$ will be identical for both paths. That's because $U$ is a state function — it only cares about the starting and ending states. …

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Thermodynamics is about macroscopic energy changes between equilibrium states, ignoring molecular details and rates. The correct options are (i) and (iv).

The question asks what thermodynamics "mainly deals with," and it allows more than one correct answer. This is a conceptual check on the scope and limits of classical thermodynamics — a topic where students often confuse it with kinetics or statistical mechanics.

Let’s examine each option carefully.

Option (i): "interrelation of various forms of energy and their transformation from one form to another."

This is the very heart of thermodynamics. The first law is about energy conservation and transformation (heat to work, work to internal energy, etc.). The second law tells us which transformations are possible. So yes, this is a core concern.
Option (ii): "energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules."

Two problems here. First, thermodynamics deals with macroscopic systems (large numbers of molecules), not microscopic ones with just a few molecules — those require statistical mechanics. Second, while state functions (like internal energy) depend only on initial and final states, thermodynamics also studies path-dependent quantities (heat and work). But the bigger issue is the "microscopic systems" part. This is wrong.
Option (iii): "how and at what rate these energy transformations are carried out."

"At what rate" is the domain of chemical kinetics, not thermodynamics. Thermodynamics tells us whether a process can happen (spontaneity) and the energy changes involved, but it says nothing about how fast it happens. A reaction can be thermodynamically favourable but kinetically slow. So this is outside thermodynamics' scope.
Option (iv): "the system in equilibrium state or moving from one equilibrium state to another equilibrium state." …

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Q.Thermodynamics mainly deals with (Note: more than one of the given options may be correct.)

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