BPSC 70th CCE Re-exam Prelims Q2: For two liquids A and B to form an ideal

Q: What is the answer to BPSC 70th CCE Re-exam Prelims Q2?

A. Correct — A, Enthalpy of mixing should be zero. An ideal solution is one that obeys Raoult's law over the entire composition range, which means the partial vapour pressure of each component equals its mole fraction times the vapour pressure of that component in the pure state: p(A) = x(A)·p°(A) and p(B) = x(B)·p°(B). Physically, that can only hold if the A–B attraction is the same in strength as the A–A and B–B attractions the molecules already felt in the pure liquids. If nothing changes in the energy of interaction when the molecules are shuffled together, then no heat has to be supplied or given out, and no contraction or expansion occurs — so the two measurable signatures of ideality are ΔH(mixing) = 0 and ΔV(mixing) = 0. That is exactly what option (a) states. The other two thermodynamic quantities in the options cannot be zero, and knowing why is what makes this question quick. Mixing always increases the number of ways the molecules can be arranged, so the entropy of mixing is strictly positive: ΔS(mixing) = −nR[x(A) ln x(A) + x(B) ln x(B)], which for an equimolar binary mixture works out to R ln 2 = 5.76 joules per kelvin per mole of solution. And because ΔG = ΔH − TΔS, an ideal solution has ΔG(mixing) = 0 − TΔS(mixing) = −TΔS(mixing), a negative number — about −1.7 kilojoules per mole of solution at 298 K for the equimolar case. A negative free energy is precisely why two such liquids mix spontaneously when you pour them together. So only the enthalpy vanishes, and (a) is the answer.

Q: How is this topic asked in BPSC & UPSC?

BPSC states the criterion flat and asks you to pick the quantity, rewarding one clean line of NCERT Class 12 Chemistry. UPSC rarely tests solution thermodynamics in this bare definitional form; when it comes to prelims at all it arrives applied — why a solute raises the boiling point, why antifreeze works, or why a mixture cannot be separated past a certain purity — so the criterion has to be usable as an explanation rather than only as a recalled sentence.

BPSC 70th CCE Re-exam Prelims General Studies · Question 2 of 142

Verified against the official question booklet. Answer derived — no official key exists for this edition.

Answers derived, not from the official key. BPSC published an answer key for BPSC 70th CCE Re-exam at the time, but it is no longer available from the Commission or in any web archive. The answers here are our best-supported answers — solved from the question itself and cross-checked against primary sources — and each card says how certain it is. The question text is verbatim from the official booklet. These are not the Commission's answer key.

Question order, option order and answers follow the official BPSC Booklet Series I, paper code 01/BPP/GO/CC/PT-2025(Y).

For two liquids A and B to form an ideal solution

(a)Enthalpy of mixing should be zero
(b)Entropy of mixing must be zero
(c)Free energy of mixing should be zero
(d)None of the above

Answer

Why

Correct — A, Enthalpy of mixing should be zero. An ideal solution is one that obeys Raoult's law over the entire composition range, which means the partial vapour pressure of each component equals its mole fraction times the vapour pressure of that component in the pure state: p(A) = x(A)·p°(A) and p(B) = x(B)·p°(B). Physically, that can only hold if the A–B attraction is the same in strength as the A–A and B–B attractions the molecules already felt in the pure liquids. If nothing changes in the energy of interaction when the molecules are shuffled together, then no heat has to be supplied or given out, and no contraction or expansion occurs — so the two measurable signatures of ideality are ΔH(mixing) = 0 and ΔV(mixing) = 0. That is exactly what option (a) states. The other two thermodynamic quantities in the options cannot be zero, and knowing why is what makes this question quick. Mixing always increases the number of ways the molecules can be arranged, so the entropy of mixing is strictly positive: ΔS(mixing) = −nR[x(A) ln x(A) + x(B) ln x(B)], which for an equimolar binary mixture works out to R ln 2 = 5.76 joules per kelvin per mole of solution. And because ΔG = ΔH − TΔS, an ideal solution has ΔG(mixing) = 0 − TΔS(mixing) = −TΔS(mixing), a negative number — about −1.7 kilojoules per mole of solution at 298 K for the equimolar case. A negative free energy is precisely why two such liquids mix spontaneously when you pour them together. So only the enthalpy vanishes, and (a) is the answer.

Why the others are wrong

(b)Entropy of mixing must be zero — The exact opposite of the truth, and the most attractive wrong answer because it sits next to the right one in the same 'should be zero' pattern. Entropy of mixing is positive for every real mixture, ideal or not — that is the whole statistical meaning of mixing, since a mixed state can be realised in vastly more molecular arrangements than a separated one. If ΔS(mixing) were zero there would be no thermodynamic driving force to mix at all.
(c)Free energy of mixing should be zero — A zero free-energy change would mean the mixed and unmixed states are equally stable — the liquids would sit in their separate layers with no tendency to combine. Since ΔG(mixing) = −TΔS(mixing) for an ideal solution and ΔS(mixing) is positive, ΔG(mixing) is always negative. Mixing is spontaneous, and its spontaneity here is entirely entropy-driven.
(d)None of the above — Fails because option (a) is a correct and complete statement of the defining condition. An escape option only wins when each of the other three can be shown to fail, and here the first of them is the standard textbook criterion for ideality.

Concept

Raoult's law, published by François-Marie Raoult in 1887, says that in a solution of volatile liquids the partial vapour pressure of each component is proportional to its mole fraction, the constant of proportionality being the vapour pressure of the pure component. A solution that obeys it at every composition and temperature is called ideal. Ideality is a statement about forces: the molecules must not be able to tell whether their neighbours are of their own kind or the other kind. Real pairs come close when the two liquids are structurally and chemically similar — benzene with toluene, n-hexane with n-heptane, chlorobenzene with bromobenzene, ethyl bromide with ethyl iodide. Where the forces differ, the solution deviates. If A–B attraction is weaker than the average of A–A and B–B, escaping tendency rises, vapour pressure exceeds the Raoult prediction and ΔH(mixing) is positive — a positive deviation, as in ethanol and water. If A–B attraction is stronger, as when chloroform hydrogen-bonds to acetone, vapour pressure falls below the prediction, ΔH(mixing) is negative and the mixture warms up.

You do not have to remember the definition to answer this. Run the second law across the three options instead. Free energy of mixing must be negative, otherwise nothing would mix — so (c) is out. Entropy of mixing must be positive, because mixing is the classic disorder-increasing process — so (b) is out. That leaves enthalpy as the only one of the three that is free to be zero, and the elimination is complete before any chemistry is recalled. The one caution is that ΔH(mixing) = 0 is necessary but not by itself the whole definition: an ideal solution also has ΔV(mixing) = 0, and the primary definition is obedience to Raoult's law, of which the two zeros are consequences. BPSC offers only one of them, so (a) stands. It is also worth noticing what the zero enthalpy predicts in the laboratory: pour benzene into toluene and the mixture stays at room temperature, whereas pouring concentrated sulphuric acid into water — a strongly non-ideal mixing — makes the beaker too hot to hold.

Key facts

Raoult's law for a binary solution of volatile liquids: p(total) = x(A)·p°(A) + x(B)·p°(B); a solution obeying it over the whole composition range is ideal.
Signatures of an ideal solution: ΔH(mixing) = 0 and ΔV(mixing) = 0, with A–A, B–B and A–B intermolecular forces of equal strength.
ΔS(mixing) = −nR[x(A) ln x(A) + x(B) ln x(B)] is always positive; for an equimolar binary mixture it is R ln 2 ≈ 5.76 J K⁻¹ per mole of solution, giving ΔG(mixing) ≈ −1.7 kJ per mole at 298 K.
Near-ideal pairs: benzene–toluene, n-hexane–n-heptane, chlorobenzene–bromobenzene, ethyl bromide–ethyl iodide.
Positive deviation (ΔH > 0, e.g. ethanol–water) gives a minimum-boiling azeotrope — about 95% ethanol boiling at 351.1 K; negative deviation (ΔH < 0, e.g. nitric acid–water) gives a maximum-boiling azeotrope — about 68% HNO₃ boiling at 393.5 K. Neither mixture can be separated further by simple fractional distillation.

Mixing two liquids — which thermodynamic quantity is zero for an ideal solution

ΔH(mixing) — heat absorbed or released on mixing→ZERO for an ideal solution — A–B forces equal A–A and B–B forces, so no energy is gained or lost

ΔV(mixing) — volume change on mixing→ZERO for an ideal solution — the second signature of ideality, not offered in the options

ΔS(mixing) — entropy change on mixing→ALWAYS POSITIVE, never zero — R ln 2 ≈ 5.76 J K⁻¹ per mole for an equimolar mixture

ΔG(mixing) — free-energy change on mixing→ALWAYS NEGATIVE — equals −TΔS(mixing) for an ideal solution, which is why the liquids mix on their own

Only the enthalpy is free to vanish. Entropy of mixing must rise and free energy must fall, or nothing would mix at all — so options (b) and (c) are ruled out by the second law before any definition is recalled.

Study next

Raoult's law and Henry's law — where each applies, and why Raoult's is the special case of Henry's when the solute is the same substance as the solventPositive and negative deviations from Raoult's law, the molecular reason for each, and the azeotropes they produceColligative properties — relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure, and the van't Hoff factor for electrolytesFractional distillation and why an azeotrope sets the practical limit for separating ethanol from water at about 95%Gibbs free energy as the criterion of spontaneity: ΔG = ΔH − TΔS, and processes that are entropy-driven rather than enthalpy-driven

Common traps

Reading 'ideal' as 'nothing changes' and therefore ticking whichever quantity is offered as zero. Only the enthalpy and the volume change are zero; entropy and free energy are not.
Assuming an exothermic mixture must be ideal. Chloroform and acetone release heat because they hydrogen-bond to each other, which is a negative deviation from ideality, not a case of it.
Forgetting that an ideal solution's Raoult behaviour must hold over the whole composition range and at all temperatures — many real pairs are near-ideal only in a narrow range.