The state of a gas can be described by quoting the relationship between__. (i) pressure, volume, temperature (

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Q. The state of a gas can be described by quoting the relationship between___.

(i) pressure, volume, temperature

(ii) temperature, amount, pressure

(iii) amount, volume, temperature

(iv) pressure, volume, temperature, amount

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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.

The state of a gas is fully defined by the relationship among pressure, volume, temperature, and amount — all four are needed to specify the thermodynamic state uniquely.

The question asks what variables are needed to describe the state of a gas. This is a fundamental idea in thermodynamics: a state function depends only on the current condition of the system, not on how it got there. For a gas, the state is determined by a set of macroscopic properties that are independent of the path taken.

Think of it this way: if you walk into a room and someone asks you "what is the state of the gas in this cylinder?", you need enough information to know everything about it. If you only know pressure and volume, you don't know how much gas there is or its temperature. If you know temperature and amount, you don't know the volume or pressure. The state is only fixed when you specify four interconnected variables: pressure ( $P$ ), volume ( $V$ ), temperature ( $T$ ), and amount ( $n$ ).

Why four? Because these four are linked by the ideal gas equation $PV = nRT$ , but that equation alone doesn't give you a unique state — it's a relationship among them. To know the state, you need to know the values of all four, or equivalently, three independent ones (since the equation gives the fourth). But the question asks for the "relationship between" them, meaning the complete set that defines the state.

Let's break down the options:

Option (i): pressure, volume, temperature — This misses the amount of gas. If you have a fixed amount, these three suffice, but the question doesn't say the amount is fixed. Without $n$ , you can't distinguish between a small amount of gas at high pressure and a large amount at the same pressure and volume.
Option (ii): temperature, amount, pressure — This leaves out volume. Volume is essential because a gas expands to fill its container; knowing $T$ , $n$ , and $P$ doesn't tell you the volume unless you also know the container size.
Option (iii): amount, volume, temperature — This omits pressure. Pressure is a key state variable — without it, you can't describe the force the gas exerts on its surroundings.
Option (iv): pressure, volume, temperature, amount — This includes all four. These are the complete set of macroscopic variables that uniquely determine the thermodynamic state of a gas. The relationship among them is given by the equation of state (e.g., $PV = nRT$ for an ideal gas).

Watch out

A common mistake is to think three variables are enough because the ideal gas law relates them. But the law is a relationship, not a definition of state — you need all four to know the state, or equivalently, three independent ones plus the equation. The question asks for the "relationship between" which variables, and that relationship involves all four.

Tip

In thermodynamics, the state of a simple compressible system is fixed by any two independent intensive properties plus the amount. But here, the question lists the variables themselves — and the complete set is $P$ , $V$ , $T$ , and $n$ . Think of it as: you can't describe a gas without saying how much there is, how hot it is, how much space it takes, and how hard it pushes.

✓Final answer

The correct option is (iv) — pressure, volume, temperature, amount.

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)

Work is explicitly a path function, not a state function, since its value depends on how the process is carried out (e.g. reversibly vs irreversibly), so (C) is false. The work done in an adiabatic process is generally non-zero (e.g. adiabatic compression/expansion does real work; only free expansion into vacuum gives zero work), so (D) is false.

✓Final answer
$The correct option is (B) — internal energy and entropy are state functions.$
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.

All four of these properties (heat, free energy, enthalpy, entropy) DO scale with the amount of matter present (they are extensive properties) — so what distinguishes heat from the other three is specifically its path-dependence.

Therefore heat is the property that depends on BOTH the amount of matter AND the path.

✓Final answer
$(A) Heat$

ANSWER: (A)
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.

Work, however, depends on HOW a process is carried out (e.g., reversible vs. irreversible expansion, or free expansion vs. expansion against constant external pressure) — the same initial and final states can be reached via processes that do different amounts of work. This path dependence makes work a path function, not a state function.

✓Final answer
$(C) Work$

ANSWER: (C)
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

“According to Ellingham diagram, metal oxide with higher $\Delta G^\circ$ is more stable than the oxide with lower $\Delta G^\circ$ .”

This is false. A “higher” $\Delta G^\circ$ means less negative (closer to zero or positive), which indicates a less stable oxide. Stability increases as $\Delta G^\circ$ becomes more negative. So the statement has the relationship backwards.

Conclusion

Statement I is correct; Statement II is incorrect. Therefore the correct option is (B).

Watch out

A common pitfall is to think that a “higher” $\Delta G^\circ$ means “more energy released” — but in thermodynamics, a more negative value means a more spontaneous (and thus more stable) product. Always remember: lower on the Ellingham diagram = more stable oxide.

✓Final answer
$The correct option is (B). ANSWER: B$
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 $\Delta G^\Theta$ ."

This is incorrect. Stability is measured by how negative $\Delta G^\Theta$ is. A more negative $\Delta G^\Theta$ means the oxide formation is more spontaneous, so the oxide is harder to decompose — i.e., it is more stable. A "higher" $\Delta G^\Theta$ (less negative or positive) means the oxide is less stable. So the statement has the relationship backwards.

Watch out

A common mistake is to think "higher" means "more stable" because we often associate "high" with "strong." But in thermodynamics, a lower (more negative) Gibbs free energy means greater stability. Always check the sign.

Thus, Statement I is correct, and Statement II is incorrect.

✓Final answer
$The correct option is (A). ANSWER: A$
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.

Statement III: Lattice enthalpy (energy to form a gaseous ionic solid from its gaseous ions, or its reverse) is not directly measurable; it's calculated using the Born-Haber cycle, applying Hess's Law to a thermochemical cycle of sublimation, ionization, dissociation, electron gain, and formation enthalpies. True.

All three statements are correct, so option (D) is correct.

Common Mistakes

Confusing state functions and path functions — mistakenly calling work a state function.

Thinking lattice enthalpy can be measured directly by calorimetry, missing that Born-Haber cycle is specifically the indirect route used.

✓Final answer
$The correct option is (D) — I, II, III.$

ANSWER: D
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 $U$ , $V$ .

(D) "Enthalpy change can be determined using the first law of thermodynamics" — true. At constant pressure, $\Delta H = \Delta U + P\Delta V = q_p$ , directly derivable from the first law $\Delta U = q + w$ .

Hence the incorrect statement is (A).

Common Mistakes

Assuming since $\Delta H$ is measurable, the absolute $H$ must also be measurable — these are fundamentally different.

Mistaking (D)'s wording as suspicious just because it mentions "first law" — the derivation of $\Delta H=q_p$ genuinely does come from the first law.

✓Final answer
$The correct option is (A) — Its absolute value can be determined accurately.$

ANSWER: A
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).

Statement II: For $\mathrm{H_2O}(l)\to \mathrm{H_2O}(g)$ , gas has far greater disorder than liquid, so $\Delta S > 0$ — true.

Statement III: Entropy is indeed a state function — true.

Correct statements: II and III only.

Common Mistakes

Confusing enthalpy with specific enthalpy (per mole), which would be intensive.

Assuming all thermodynamic quantities are automatically state functions without checking the specific claim.

✓Final answer
$The correct option is (D) — II, III only.$

ANSWER: D
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.

Statement III: $\mathrm{H_2O}(l) \to \mathrm{H_2O}(s)$ is freezing; the solid state is more ordered than the liquid, so $\Delta S < 0$ (entropy decreases). The statement claiming entropy increases is false.

Incorrect statements = II and III only; I remains correct.

Common Mistakes

Confusing intensive vs extensive properties — thinking enthalpy (a total-energy-like quantity) behaves like temperature.

Assuming any phase change increases entropy, without checking the direction (melting/vaporization increase entropy; freezing/condensation decrease it).

✓Final answer
$The correct option is (D) — II, III only.$

ANSWER: D
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).

In free expansion into a vacuum, there is nothing opposing the expansion, so $P_{ext} = 0$ .

Therefore $w = -0\times\Delta V = 0$ , confirming Statement-II is true.

Combining: Statement-I is incorrect, Statement-II is correct.

Common Mistakes

Assuming work is a state function just because it appears in $\Delta U = q + w$ — that equation only guarantees the sum $q+w$ (i.e., $\Delta U$ ) is path-independent, not that $q$ or $w$ individually are.

Thinking free expansion involves work because the volume changes — work requires an opposing force/pressure; with vacuum on the other side, there's nothing to push against.

✓Final answer
$The correct option is (D) — Statement-I is not correct, but statement-II is correct.$

ANSWER: D
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.

Internal energy — This is the total energy stored inside the substance (kinetic + potential at molecular level). Double the number of molecules, double the internal energy. → Extensive.

Number of moles — This is a direct measure of the amount of substance. More substance → more moles. → Extensive.

Watch out

A common mistake is to think “surface tension” is extensive because it involves a surface area. But surface tension is a ratio (force/length or energy/area), and ratios of extensive properties are intensive. For example, density (mass/volume) is also intensive even though mass and volume are extensive.

✓Final answer
$The correct option is (A) . ANSWER: A$
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.

Entropy (D)

Entropy is a state function (it depends only on the state).

But it is extensive: doubling the amount of substance doubles the entropy.

So it fails the intensive requirement.

Watch out

A common mistake is to think that “state function” automatically means intensive, or that “extensive” means not a state function. In fact, most state functions (like internal energy, volume, entropy) are extensive. Temperature and pressure are the classic intensive state functions.

Tip

A quick way to test if a property is intensive: ask yourself, “If I split the system into two equal halves, does the property stay the same in each half?” If yes, it’s intensive. Temperature passes; internal energy, volume, and entropy do not.

✓Final answer
$The correct option is (C) . ANSWER: C$

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Q.The state of a gas can be described by quoting the relationship between___.

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