A gas is taken from state A to state B by two different paths. Along Path 1 it absorbs $400\ \text{J}$ of heat and $100\ \text{J}$ of work is done on the surroundings by the gas. Along Path 2 it absorbs only $250\ \text{J}$ of heat and $50\ \text{J}$ of work is done on the gas by the surroundings. Show that $\Delta U$ is the same along both paths, and explain what this demonstrates about internal energy.

A gas is taken from state A to state B by two different paths. Along Path 1 it absorbs 400J of heat and 100J o

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400\ \text{J}

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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.

Watch out

A common mistake: students think "energy is conserved, so $\Delta U$ is always zero." No — $\Delta U$ is zero only for a cyclic process (returning to the same state). For a change between two different states, $\Delta U$ is non-zero but path-independent.

Why This Matters for Exams

When you see a problem asking for $\Delta H$ or $\Delta U$ or $\Delta S$ , you do not need to know the path. You only need the initial and final states. That's why we can use tables of standard enthalpies, entropies, and Gibbs energies — they are state functions, so their values are fixed for a given substance at a given temperature and pressure.

When you see a problem asking for $W$ or $Q$ , you must know the path (isothermal, adiabatic, isobaric, etc.). These are path functions.

Final takeaway: A state function is like your bank account balance at a given moment. It doesn't matter if you earned the money, inherited it, or found it on the street — the balance is what it is. A path function is like the total amount of money that moved in and out of your account over a month — that depends entirely on how you earned and spent.

This topic is commonly searched as "State Function 11 chemistry important questions" or "State Function formula and examples", and it maps cleanly onto the Class 11 Chemistry portion of the NCERT/CBSE syllabus. Because state function shows up repeatedly in JEE Main, NEET and state CET Chemistry papers, mastering the underlying idea (not just the formula) is genuinely worth the extra time.

Along Path 1: the gas absorbs $q_1 = +400\ \text{J}$ of heat, and since $100\ \text{J}$ of work is done ON the surroundings BY the gas, the work done ON the system is $w_1 = -100\ \text{J}$ . So $\Delta U_1 = q_1 + w_1 = 400 + (-100) = 300\ \text{J}$ . Along Path 2: the gas absorbs $q_2 = +250\ \text{J}$ of heat, and $50\ \text{J}$ of work is done ON the gas BY the surroundings, so $w_2 = +50\ \text{J}$ . So $\Delta U_2 = q_2 + w_2 = 250 + 50 = 300\ \text{J}$ . Both paths give exactly the same $\Delta U = 300\ \text{J}$ , even though $q_1 eq q_2$ and $w_1 eq w_2$ individually. This is a direct demonstration that internal energy, $U$ , is a state function — its change depends only on the identity of the initial state A and final state B, never on the specific path (or mechanism) connecting them — while $q$ and $w$ individually are path functions, free to differ from one path to another as long as their sum, $\Delta U$ , remains fixed. [!ANSWER] $\Delta U = +300\ \text{J}$ along both Path 1 and Path 2, confirming that internal energy depends only on the initial and final states, not on the path taken.

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