Although heat is a path function but heat absorbed by the system under certain specific conditions is independ

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Q. Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.

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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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Heat becomes a path-independent quantity when the process is carried out at constant volume (where $q_V = \Delta U$ ) or at constant pressure (where $q_P = \Delta H$ ). Under these conditions, the heat absorbed equals the change in a state function — internal energy or enthalpy — and therefore loses its path dependence.

Heat is a path function because the amount of energy transferred as heat depends on how you go from the initial state to the final state — whether you do it slowly, quickly, in one step, or in many steps. But there is a clever way out: if you constrain the process so that a particular variable (volume or pressure) stays fixed, then the heat absorbed becomes equal to the change in a state function. And a state function depends only on the initial and final states, not on the path.

Let’s see exactly how this works.

Constant volume: $q_V = \Delta U$ From the first law of thermodynamics:

\Delta U = q + W

If the volume is constant, no pressure–volume work is done: $W = -P\Delta V = 0$ . So the first law reduces to:

\Delta U = q_V

Here $q_V$ is the heat absorbed at constant volume. Since $\Delta U$ is a state function (it depends only on the initial and final states), $q_V$ must also be path-independent — it always equals the change in internal energy, no matter how the process is carried out, as long as volume stays constant.

Constant pressure: $q_P = \Delta H$ At constant pressure, the work done is $W = -P\Delta V$ . Substituting into the first law:

\Delta U = q_P - P\Delta V

Rearranging:

q_P = \Delta U + P\Delta V

The right-hand side is exactly the definition of the change in enthalpy: $\Delta H = \Delta U + \Delta(PV)$ . At constant pressure, $\Delta(PV) = P\Delta V$ , so:

q_P = \Delta H

Enthalpy $H$ is a state function, so $q_P$ is path-independent under constant pressure conditions.

Watch out

A common mistake is to think that $q = \Delta H$ always. That is only true when the pressure is constant and only $P$ – $V$ work is done. If the pressure changes during the process, $q$ is not equal to $\Delta H$ , and heat remains path-dependent. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$Q. Identify the correct statements from the following (only) I) Work is a path function. II) Enthalpy is an extensive property III) Lattice enthalpy of ionic compounds can be obtained from Born-Haber cycle. (A) I, II only (B) I, III only (C) II, III only (D) I, II, III$
›Reveal solutionSolution
Tests basic thermodynamics definitions; all three statements (work is a path function, enthalpy is extensive, lattice enthalpy comes from the Born-Haber cycle) are correct.

Concept and Intuition

Thermodynamic quantities are classified as state functions (depend only on initial and final states, e.g., enthalpy, internal energy) or path functions (depend on the path taken, e.g., work, heat). Extensive properties scale with the amount of substance (e.g., enthalpy, volume, mass) while intensive properties don't (e.g., temperature, pressure). Lattice enthalpy cannot be measured directly, so it is obtained using Hess's law via the Born-Haber cycle, which connects it to measurable quantities like ionization energy, electron gain enthalpy, sublimation enthalpy, and enthalpy of formation.

Step-by-Step Solution

Statement I: Work done in a thermodynamic process depends on the path (e.g., reversible vs irreversible expansion give different work for the same initial/final states) — Work is indeed a path function. True.

Statement II: Enthalpy $H = U + PV$ depends on the total amount of substance present, so it scales with quantity — it is an extensive property. True. …
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AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$Q. Which of the following statements is incorrect about enthalpy? (A) Its absolute value can be determined accurately. (B) It is a state function. (C) It is an extensive property. (D) Enthalpy change can be determined using the first law of thermodynamics.$
›Reveal solutionSolution
Enthalpy's absolute value can never be measured accurately (only $\Delta H$ can); the other three statements about enthalpy are all true.

Concept and Intuition

Enthalpy is defined as $H = U + PV$ . Since the absolute internal energy $U$ of a system can't be measured (there's no natural zero-reference for it), the absolute value of $H$ likewise can't be measured — thermodynamics only ever gives us access to changes, $\Delta H$ , measured via calorimetry or derived using Hess's law and the first law of thermodynamics.

Step-by-Step Solution

(A) "Its absolute value can be determined accurately" — false. Just like internal energy, only $\Delta H$ (change) is measurable; the absolute value of $H$ has no accessible reference point.

(B) "It is a state function" — true. $H=U+PV$ depends only on the current state ( $U$ , $P$ , $V$ ), not on the path taken to reach it.

(C) "It is an extensive property" — true. Enthalpy scales with the amount of substance present, just like $U$ , $V$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g)$
›Reveal solutionSolution
This tests basic thermodynamic definitions: enthalpy is extensive (not intensive), while entropy increase on vaporisation and entropy being a state function are both correct facts.

Concept and Intuition

Extensive properties (enthalpy, entropy, internal energy, volume) scale with the amount of substance; intensive properties (temperature, pressure, density, molar/specific quantities) do not. Liquid to gas conversion increases disorder/randomness, so entropy increases. State functions depend only on the current state, not the path taken to reach it — entropy, like enthalpy, is one of these.

Step-by-Step Solution

Statement I: "Enthalpy is an intensive property" — false, enthalpy is an extensive property (it depends on the amount of substance). …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$\Delta U = w_{ad}$
›Reveal solutionSolution
Statement I is a correct thermodynamic fact (adiabatic ⇒ $\Delta U=w$ ); statements II (enthalpy intensive) and III (freezing increases entropy) are both wrong, so the incorrect set is II, III.

Concept and Intuition

The first law of thermodynamics, $\Delta U = q+w$ , becomes $\Delta U=w_{ad}$ exactly when $q=0$ , i.e., for an adiabatic process — so statement I is true by definition. Enthalpy, like internal energy, scales with the amount of substance present, making it an extensive property, not intensive (intensive properties like temperature or density don't depend on the quantity of matter). Entropy tracks disorder: a liquid freezing into a solid becomes more ordered, so its entropy decreases, not increases.

Step-by-Step Solution

Statement I: Adiabatic process ⇒ $q = 0$ ⇒ $\Delta U = q + w = w_{ad}$ . This is thermodynamically correct.

Statement II: Enthalpy $H = U + PV$ scales with the size/amount of the system, so it is an extensive property — the statement calling it intensive is false. …
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AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$Q. Consider the following. Statement-I : Both internal energy (U) and work (w) are state functions. Statement-II : During the free expansion of an ideal gas into vacuum, the work done is zero. The correct answer is (A) Both statement-I and statement-II are correct (B) Both statement-I and statement-II are not correct (C) Statement-I is correct, but statement-II is not correct (D) Statement-I is not correct, but statement-II is correct$
›Reveal solutionSolution
Work is a path function, not a state function (only U, H, S, G etc. are state functions), so Statement-I is wrong; but free expansion against zero external pressure genuinely does zero work, so Statement-II is right.

Concept and Intuition

A state function depends only on the initial and final states of a system, not on the path taken between them (e.g., internal energy $U$ , enthalpy $H$ ). Work and heat, by contrast, are path functions: the amount of work done in going from state A to state B depends on how the process is carried out (reversibly, irreversibly, at constant pressure, etc.) — this is a foundational distinction in thermodynamics, and it's precisely why $q+w=\Delta U$ combines two path-dependent quantities into one state function. Separately, for free expansion into a vacuum, the gas expands against zero opposing (external) pressure, so no mechanical work is done regardless of how large the volume change is.

Step-by-Step Solution

Statement-I claims both $U$ and $w$ are state functions. $U$ is indeed a state function. But $w$ (work) is NOT — the work done between the same two states differs for a reversible vs. an irreversible path (e.g., reversible isothermal expansion does more work than an irreversible one against constant external pressure). So Statement-I is false.

Statement-II: work done on/by a gas during expansion is $w = -P_{ext}\Delta V$ (sign convention: work done BY the system is negative in the IUPAC convention used here). …
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AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
$Q. Which of the following equations are correct? (A) U = H - PV (B) G = H - TS (C) U = q + W (A) A, B and C (B) A and B only (C) A and C only (D) B and C only$
›Reveal solutionSolution
All three equations are valid, so the answer is A, B and C.

Concept and Intuition

Enthalpy is defined by $H = U + PV$ , Gibbs free energy by $G = H - TS$ , and the first law of thermodynamics by $\Delta U = q + W$ (with $W$ the work done on the system). Each of the three listed relations is simply a rearrangement or statement of these definitions.

Step-by-Step Solution

Statement A: from $H = U + PV$ , rearrange to $U = H - PV$ — correct.

Statement B: $G = H - TS$ is the standard definition of Gibbs free energy — correct.

Statement C: the first law is $U = q + W$ (energy change equals heat plus work done on the system) — correct.

All three statements are valid simultaneously.

The option asserting "A, B and C" is the first choice, which maps to option A.

Common Mistakes …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$\Delta U = W_{adiabatic}$
›Reveal solutionSolution
Checks three basic thermodynamics facts: the first law under adiabatic conditions, whether work is a path function, and whether volume is extensive. All three statements are true.

Concept and Intuition

The first law of thermodynamics states $\Delta U = q + w$ (heat absorbed plus work done on the system). In an adiabatic process, no heat is exchanged with the surroundings ( $q=0$ ), so all the change in internal energy shows up as work. Work (like heat) is a path function — its value depends on how the process is carried out (e.g., reversibly vs irreversibly), not just on initial and final states. Volume is an extensive property because it depends on the amount of matter present — double the amount of substance, and (at same conditions) the volume doubles.

Step-by-Step Solution

Statement (i): Adiabatic $\Rightarrow q = 0 \Rightarrow \Delta U = q + w = w_{adiabatic}$ . True.

Statement (ii): Work is not a state function; two processes with the same initial and final states can have different work done depending on the path. True. …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$Q. Which of the followings is not a state function? (A) Internal Energy (B) Work (C) Enthalpy (D) Entropy$
›Reveal solutionSolution
Tests the distinction between state functions (path-independent) and path functions; among the four given quantities, only Work depends on the path taken, not just initial/final states.

Concept and Intuition

A state function's value depends ONLY on the current state of the system (e.g., $P, V, T$ ), not on how that state was reached — internal energy $U$ , enthalpy $H=U+PV$ , and entropy $S$ are all such functions, each with an exact differential. Work and heat, by contrast, are path functions: the amount of work done in going from state 1 to state 2 depends on the specific path (e.g., reversible vs irreversible expansion), even though the initial and final states are identical.

Step-by-Step Solution

Internal Energy ( $U$ ): a state function — $\Delta U$ for a process depends only on initial and final states (first law: $\Delta U = q + w$ , but $\Delta U$ itself is path-independent even though $q$ and $w$ individually are not).

Enthalpy ( $H = U+PV$ ): a state function, since it is built entirely from state variables $U$ , $P$ , $V$ .

Entropy ( $S$ ): a state function, with $dS = \dfrac{dq_{rev}}{T}$ giving an exact differential. …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$Q. Out of molar entropy (I), specific volume (II), heat capacity (III), volume (IV), extensive properties are (A) I , II (B) I , II , IV (C) II , III (D) III , IV$
›Reveal solutionSolution
Tests the extensive-vs-intensive property distinction: extensive properties scale with the size/amount of the system; intensive ones don't. Answer: heat capacity and volume (III, IV) are extensive.

Concept and Intuition

An extensive property doubles if you double the amount of substance (e.g., total volume, total mass, total heat capacity, total internal energy). An intensive property stays the same no matter how much substance you have (e.g., temperature, density, molar or specific — i.e., 'per mole' or 'per unit mass' — quantities), because dividing by the amount of substance cancels out the size-dependence.

Step-by-Step Solution

(I) Molar entropy — 'per mole' quantity, so it doesn't change if you have more or less substance → intensive.

(II) Specific volume — 'per unit mass' quantity → intensive.

(III) Heat capacity (total, not specific/molar as usually meant in bare 'heat capacity') — scales with the amount of substance present → extensive.

(IV) Volume — clearly scales with amount of substance → extensive. …
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Q.Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.

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