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.

Showing the 12 most recent of 30 on this concept.

KCET 2026Set D31 markMCQ
$Q. Which of the following is a correct statement for a thermodynamic system? (A) The internal energy changes in all processes (B) Internal energy and entropy are state functions (C) Work is a state function (D) The work done in an adiabatic process is always zero$
›Reveal solutionSolution

Distinguishing state functions (path-independent) from path functions (path-dependent) resolves this question.

Step 1 — Statement (A)

Internal energy does not change in every process — for example, in an isothermal expansion of an ideal gas, $\Delta U = 0$ because internal energy depends only on temperature. This statement is false.

Step 2 — Statement (B)

Internal energy $U$ and entropy $S$ are both state functions — their values depend only on the current state of the system (defined by variables like $T$ , $P$ , $V$ , composition), not on the path taken to reach that state. This is a fundamental and correct thermodynamic principle.

Step 3 — Statements (C) and (D) …
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MHT-CET 2026Set pcm-2026-04-15-M1 markMCQ
$Q. Which of the following properties of a system depends upon the amount of matter and path? (A) Heat (B) Free energy (C) Enthalpy (D) Entropy$
›Reveal solutionSolution
Heat (q) is the odd one out: it is both extensive (depends on amount of matter) and a path function (depends on how the process is carried out), unlike enthalpy, free energy, and entropy which are state functions.

Free energy (G), enthalpy (H), and entropy (S) are all thermodynamic STATE functions — their values depend only on the initial and final states of the system, not on the path taken to get there.

Heat (q), however, is a path function: the amount of heat exchanged between a system and surroundings depends on HOW the process is carried out (e.g., reversibly vs irreversibly), even between the same initial and final states. …
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MHT-CET 2026Set pcm-2026-04-19-E1 markMCQ
$Q. Which from following is NOT a state function ? (A) Volume (B) Pressure (C) Work (D) Temperature$
›Reveal solutionSolution
Work is a path function, not a state function, unlike volume, pressure, and temperature.

A state function is a property whose value depends only on the current state of the system (defined by variables like P, V, T, n), not on the path by which that state was reached — examples include volume, pressure, temperature, internal energy, and enthalpy.

Volume, pressure, and temperature are all measurable properties that are fixed once the state of the system is specified — they are true state functions. …
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TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$\Delta G^\circ$
›Reveal solutionSolution
The Ellingham diagram plots $\Delta G^\circ$ vs. T for oxide formation; a lower $\Delta G^\circ$ (more negative) means a more stable oxide, so Statement II is false. Statement I is true. Hence only Statement I is correct.

Concept and Intuition

The Ellingham diagram is a powerful tool in metallurgy. It shows how the standard Gibbs free energy change ( $\Delta G^\circ$ ) for the formation of an oxide varies with temperature. The key idea: the more negative $\Delta G^\circ$ , the more stable the oxide (because a spontaneous formation reaction means the oxide is hard to break apart). A reducing agent (like carbon or aluminium) can reduce an oxide if its own oxide has a more negative $\Delta G^\circ$ at that temperature — that is, if it “outcompetes” the metal for oxygen. Statement I correctly describes this predictive use. Statement II reverses the stability rule, which is a common mistake.

Step-by-step reasoning

Understanding the Ellingham diagram

The diagram plots $\Delta G^\circ$ (in kJ/mol of O₂) on the y-axis against temperature (K) on the x-axis for reactions like:

$\frac{2x}{y}M + O_2 \rightarrow \frac{2}{y}M_xO_y$

A lower (more negative) $\Delta G^\circ$ means the reaction is more spontaneous, so the oxide is thermodynamically more stable.

Evaluating Statement I

“The choice of reducing agent for the reduction of an oxide ore can be predicted by using Ellingham diagram.”

This is correct. For a given metal oxide MO, we look for another element (e.g., C, Al) whose oxide has a more negative $\Delta G^\circ$ at the same temperature. Then that element can reduce MO because its own oxidation is more favourable. The diagram directly shows which lines lie below others, indicating which metal is a stronger reducing agent.

Evaluating Statement II …
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$\Delta G^\Theta$
›Reveal solutionSolution
The Ellingham diagram plots $\Delta G^\Theta$ vs. T for oxide formation; a lower (more negative) $\Delta G^\Theta$ means a more stable oxide, so Statement II is backwards. Only Statement I is correct.

The key concept here is the Ellingham diagram — a graph of the standard Gibbs free energy change ( $\Delta G^\Theta$ ) for the formation of an oxide (or other compound) as a function of temperature. The central idea is that a more negative $\Delta G^\Theta$ indicates a more stable oxide, because the reaction is more spontaneous. The diagram helps predict which metal can reduce another metal's oxide: the metal whose oxide has a lower $\Delta G^\Theta$ will be able to reduce the oxide of a metal with a higher $\Delta G^\Theta$ .

Let’s examine each statement carefully.

Statement I: "The choice of reducing agent for the reduction of an oxide ore can be predicted by using Ellingham diagram, a plot of $\Delta G^\Theta$ Vs T."

This is correct. The Ellingham diagram directly shows which metal (or carbon) can reduce a given oxide at a given temperature. For example, if the line for carbon monoxide formation lies below the line for a metal oxide, then carbon can reduce that oxide. The diagram is a standard tool in metallurgy for selecting reducing agents.

Statement II: "According to Ellingham diagram, metal oxide with higher $\Delta G^\Theta$ is more stable than the oxide with lower $\Delta G^\Theta$ ." …
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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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MHT-CET 2025Set pcm-2025-04-21-E1 markMCQ
$Q. Identify from following an example of intensive property? (A) Surface tension (B) Volume (C) Internal energy (D) Number of moles$
›Reveal solutionSolution
An intensive property does not depend on the amount of substance present. Surface tension is intensive; volume, internal energy, and number of moles are all extensive. The correct option is (A).

Concept & Intuition

In thermodynamics, properties of matter are split into two fundamental categories:

Intensive properties — these are independent of the size or amount of the sample. Think of them as “quality” indicators: temperature, pressure, density, color, melting point, and surface tension. If you take a drop of water or a whole lake, the surface tension is the same.

Extensive properties — these depend on the amount of matter. They are “quantity” indicators: mass, volume, total internal energy, number of moles. If you double the amount of substance, these values double.

The classic pitfall is confusing “intensive” with “always constant.” For example, temperature is intensive but can change; the key is that it doesn’t change just because you have more or less of the substance.

Step-by-step reasoning

Surface tension — This is a force per unit length (or energy per unit area) at the interface of a liquid. Whether you have a tiny droplet or a large puddle, the surface tension of water at 20°C is still about 0.073 N/m. It does not scale with the amount of liquid. → Intensive.

Volume — If you have twice as much substance, you have twice the volume (assuming same conditions). Volume clearly depends on the amount. → Extensive. …
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MHT-CET 2025Set pcm-2025-04-25-E1 markMCQ
$Q. Which from following is an example of both intensive property and state function? (A) Internal energy (B) Volume (C) Temperature (D) Entropy$
›Reveal solutionSolution
An intensive property does not depend on the amount of substance, and a state function depends only on the current state, not the path. Temperature is both intensive and a state function, so the correct option is (C).

To answer this, we need to recall two key classifications of physical properties in thermodynamics: intensive vs. extensive and state functions vs. path functions. The question asks for a property that satisfies both conditions simultaneously.

Intensive property: A property that does not change when the size or amount of the system changes. Examples: temperature, pressure, density, refractive index.

Extensive property: A property that does scale with the size or amount of the system. Examples: mass, volume, internal energy, entropy.

State function: A property whose value depends only on the current state of the system (e.g., temperature, pressure, volume, internal energy, entropy), not on how that state was reached.

Path function: A property that depends on the specific process or path taken (e.g., heat, work).

Now let’s evaluate each option:

Internal energy (A)

Internal energy is a state function (it depends only on the system’s current state).

However, it is an extensive property: if you double the amount of substance, the total internal energy doubles.

So it fails the intensive requirement.

Volume (B)

Volume is a state function (a given state has a definite volume).

But it is extensive: double the system, double the volume.

So it also fails.

Temperature (C)

Temperature is a state function (it is defined by the state of the system).

It is intensive: if you take a cup of water at 30°C and pour half into another cup, both halves are still at 30°C. Temperature does not depend on how much you have.

This satisfies both conditions. …
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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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