Keep liquid mole fractions, vapor mole fractions and total composition separate.
JEE Main · JEE Advanced · NEET UG38 min study guide9 indexed sections
BY THE END, YOU CAN
Calculate partial and total vapor pressures for an ideal binary liquid.
Determine vapor composition with Dalton’s law.
Use bubble- and dew-pressure relations and a two-phase material balance.
Interpret positive and negative deviations without overextending simple attraction slogans.
Explain distillation and azeotrope limits with explicit pressure and composition conditions.
BEFORE YOU START
Mole fractions and gas partial pressures.
Phase equilibrium and vapor pressure.
Algebra and material balances.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. The vapor does not usually copy the liquid composition
A volatile liquid in a closed vessel exchanges molecules with its vapor. In a mixture, each component has its own partial pressure above the liquid, and the vapor composition can differ markedly from the liquid composition. Use x_i for liquid mole fraction and y_i for vapor mole fraction. Reserve z_i for the overall composition of both phases combined when needed.
A component’s pure-liquid vapor pressure p_i° depends strongly on temperature. At a specified temperature, the component with greater p_i° is more volatile in the simple ideal comparison. It contributes disproportionately to the vapor relative to its liquid fraction. More volatile does not mean it must have a larger liquid mole fraction.
The equilibrium vapor-pressure problem assumes that liquid and vapor have had time to equilibrate at the stated temperature, with the appropriate phases present. A rapidly evaporating open mixture can have transport limitations and changing composition. Do not apply an equilibrium y_i as an exact composition of every transient gas stream without that qualification.
Both phases have mole fractions summing to one within their own phase. Adding x_A and y_B is not a valid total because they use different denominators. Many vapor-composition errors can be caught by checking that the computed y_A+y_B equals one and that each partial pressure is no larger than total pressure.
02 / UNDERSTAND & APPLY
2. Combine Raoult’s law with Dalton’s law
For an ideal liquid mixture, Raoult’s law gives p_A=x_Ap_A° and p_B=x_Bp_B°. For an ideal vapor, Dalton’s law gives P=p_A+p_B and y_i=p_i/P. In a binary liquid, x_B=1−x_A, so P=x_Ap_A°+(1−x_A)p_B°, a straight-line function of liquid composition at fixed temperature.
The vapor fraction of A is y_A=x_Ap_A°/[x_Ap_A°+(1−x_A)p_B°]. If p_A°>p_B° and both components are present, y_A>x_A. The vapor is enriched in the more volatile component. This inequality follows from the equations rather than from an assumption that the vapor must be pure A.
Use pure-component vapor pressures at the same temperature as the mixture. A table value at another temperature cannot be inserted without correction. Likewise, pressure units must match before addition. The mole fractions are dimensionless, and a partial pressure has the same unit as the pure-component pressure.
The ideal model assumes the liquid activities equal mole fractions and the vapor behaves sufficiently ideally. At higher pressures or strong liquid nonideality, activities and gas fugacities replace these simple expressions. The basic distinction between liquid and vapor composition remains, but the numerical equations change.
p_i=x_i p_i°; P=Σp_i; y_i=p_i/P
Ideal liquid and ideal vapor at a common temperature.
y_A=x_Ap_A°/[x_Ap_A°+(1−x_A)p_B°]
Binary ideal mixture.
ORIGINAL WORKED EXAMPLE
Find the enriched vapor
At a specified temperature, p_A°=80.0 kPa and p_B°=20.0 kPa. An ideal liquid has x_A=0.250. Find partial pressures, total pressure and y_A.
x_B=0.750.
p_A=0.250×80.0=20.0 kPa; p_B=0.750×20.0=15.0 kPa.
P=35.0 kPa.
y_A=20.0/35.0=0.5714, greater than x_A because A is more volatile.
For an ideal binary mixture at one temperature, relative volatility can be written α_AB=p_A°/p_B°. Combining the two phase laws gives y_A/y_B=α_AB x_A/x_B. If α_AB=1, the vapor has the same composition as the liquid in this ideal model; if it differs greatly from one, a single equilibrium contact gives stronger enrichment.
The symbol α here means relative volatility, not degree of dissociation. Context and subscripts matter because chemistry often reuses Greek letters. Relative volatility is dimensionless. For nonideal liquids, a simple extension involves activity coefficients, and the effective relative volatility can depend on composition as well as temperature.
A vapor enriched in A still contains B unless a limiting condition makes B negligible. Repeated vaporization and condensation can enrich the more volatile component further, which is the principle behind fractionating stages. The degree of separation depends on relative volatility, number of effective stages, reflux and operating conditions.
Pure boiling-point comparisons are useful clues but do not replace the actual vapor-pressure data at the stated temperature. Mixtures can deviate from ideality and form azeotropes, changing how enrichment behaves. A global promise that repeated simple distillation always produces both pure components is therefore incorrect.
ORIGINAL WORKED EXAMPLE
Use a volatility ratio
An ideal binary mixture has α_AB=3.00 and x_A=0.400 at the stated temperature. Find y_A.
x_B=0.600, so y_A/y_B=3.00×0.400/0.600=2.00.
The vapor fractions sum to one, so y_A=2y_B and 3y_B=1.
y_A=2/3≈0.667.
The result agrees with the direct expression y_A=αx_A/[1+(α−1)x_A].
Answer y_A≈0.667.
04 / UNDERSTAND & APPLY
4. Starting with a liquid is different from starting with a vapor
At fixed temperature, the bubble pressure is the pressure at which a liquid of specified composition is just in equilibrium with an infinitesimal vapor phase. For an ideal binary liquid, P_bubble=Σx_i p_i°. The first vapor’s composition follows y_i=x_i p_i°/P_bubble.
The dew pressure begins instead with a vapor of specified y_i and asks when the first infinitesimal liquid appears. From x_i=y_iP/p_i° and Σx_i=1, derive 1/P_dew=Σ(y_i/p_i°). The first liquid composition is then x_i=y_iP_dew/p_i°.
The two formulas use different known compositions. Substituting vapor fractions into the bubble-pressure expression treats them as liquid fractions and generally gives the wrong result. The dew equation is not an arbitrary harmonic-mean trick; it follows from enforcing that the emerging liquid fractions sum to one.
At fixed temperature for a specified overall binary composition, pressures above the bubble boundary generally correspond to a liquid state and pressures below the dew boundary to a vapor state in the simple two-component model. Between them, two phases can coexist with distinct x and y. Exact phase behavior outside the model needs the relevant phase diagram.
P_bubble=Σx_i p_i°; 1/P_dew=Σy_i/p_i°
Ideal binary liquid and vapor at fixed T, with x or y specified respectively.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Ideal liquid and vapor, fixed temperature. The horizontal axis is liquid x_A on the bubble line and vapor y_A on the dew curve; the same value is overall z_A for locating a bulk state.
ORIGINAL WORKED EXAMPLE
A dew-pressure calculation
At one temperature p_A°=120 kPa and p_B°=40.0 kPa. A vapor has y_A=0.600. Find its ideal dew pressure and the first liquid’s x_A.
Answer P_dew≈66.7 kPa; x_A≈0.333 in the first liquid.
05 / UNDERSTAND & APPLY
5. Overall composition tells you how much of each phase exists
If both liquid and vapor are present, the overall composition z_A depends on the amounts of both phases. Let f be the vapor mole fraction of the total material, n_vapor/(n_liquid+n_vapor). The A balance is z_A=(1−f)x_A+fy_A, so f=(z_A−x_A)/(y_A−x_A) when the phase compositions differ.
This lever-rule relation requires x_A and y_A to be the equilibrium pair at the same temperature and pressure. You cannot choose a liquid composition from one condition and a vapor composition from another and still use the balance as an equilibrium calculation. The material balance itself is general, but the phase-equilibrium constraints must also hold.
A physically possible two-phase mixture has z_A between x_A and y_A. A calculated f below zero or above one means the assumed phase pair is incompatible with the overall composition or the system is not in the proposed two-phase region. It is a diagnostic, not a fraction to round into the interval.
The lever fraction is a mole fraction of phase amounts, not a volume fraction. Liquid and vapor molar volumes can differ enormously. A small number of vapor moles can occupy most of the container volume, so a sketch showing equal volumes does not establish equal phase mole amounts.
ORIGINAL WORKED EXAMPLE
Split an overall mixture into phase amounts
At a specified equilibrium condition, x_A=0.500 and y_A=0.750. A 2.00 mol overall sample has z_A=0.600. Find vapor and liquid amounts.
A in vapor=0.800×0.750=0.600 mol; A in liquid=1.200×0.500=0.600 mol.
Total A=1.200 mol, equal to 2.00×0.600, verifying the balance.
Answer 0.800 mol vapor and 1.200 mol liquid.
06 / UNDERSTAND & APPLY
6. Real mixtures can lie above or below Raoult’s-law pressures
An ideal liquid solution is one in which each component activity follows its mole fraction over the specified composition range. Its enthalpy and volume changes on mixing are zero in the ideal model, while its mixing entropy is positive for distinguishable components. Similar molecular sizes and interactions can make a real pair approximately ideal, but these are practical clues rather than a formal proof.
Positive deviations mean a component’s vapor pressure exceeds its Raoult prediction at the same liquid composition and temperature; negative deviations mean it falls below. Activity coefficients express this as p_i≈x_iγ_i p_i° in the ideal-vapor approximation. γ_i>1 is positive deviation and γ_i<1 negative deviation for that component relative to the Raoult standard state.
A useful qualitative explanation is that replacing like-neighbor interactions with unlike interactions can make escape from the liquid more or less favorable. Weaker unlike cohesion often accompanies positive deviations and stronger unlike interactions negative deviations. However, entropy, molecular structure and specific association also matter; a single attraction slogan does not quantify an entire activity-coefficient curve.
Examples often used pedagogically include mixtures that disrupt a hydrogen-bonding network for positive deviation and mixtures with particularly favorable unlike interactions for negative deviation. The direction should be supported by the actual system or supplied data. Do not infer every mixture’s deviation merely from whether one component can hydrogen bond.
Nonzero heat or volume of mixing is evidence against the simplest ideal solution model, but a small value of one alone does not prove ideal activity behavior. The ideality criterion concerns chemical potentials and activities across composition, not just one bulk observation at one mixture ratio.
ORIGINAL WORKED EXAMPLE
Read an activity coefficient
At a fixed temperature, x_A=0.300 and p_A°=100 kPa. The measured equilibrium partial pressure is 36.0 kPa in an approximately ideal vapor. Find γ_A and the deviation sign.
Ideal Raoult pressure would be x_Ap_A°=30.0 kPa.
γ_A=p_A/(x_Ap_A°)=36.0/30.0=1.20.
The measured pressure is above the ideal value, so A shows positive deviation at this composition.
This one point does not by itself establish a full azeotrope or an entire phase diagram.
Answer γ_A=1.20; positive deviation at the stated point.
07 / UNDERSTAND & APPLY
7. An azeotrope has equal liquid and vapor compositions
At an azeotropic composition and specified pressure, the equilibrium liquid and vapor compositions are the same, x_i=y_i. The mixture then boils without the usual composition enrichment of its vapor at that condition. It is still a mixture, not necessarily a new pure chemical compound.
Sufficiently strong positive deviations can produce a maximum in total vapor pressure at fixed temperature, corresponding to a minimum-boiling azeotrope at fixed pressure. Sufficiently strong negative deviations can produce a minimum vapor pressure and a maximum-boiling azeotrope. Not every positive or negative deviation is strong enough to create an azeotrope.
Ordinary equilibrium fractional distillation at a fixed pressure cannot pass through an azeotropic composition to obtain arbitrary purity on the other side merely by adding more identical stages. At the azeotrope, vaporization provides no composition change. Other separation strategies can change pressure, add an entrainer or use a different physical principle, but those alter the problem’s conditions.
The azeotropic composition generally depends on pressure. Therefore an azeotrope label should include the relevant operating conditions rather than being treated as one immutable formula ratio. For a supplied graph, locate where the equilibrium vapor and liquid composition curves meet at an interior composition, and distinguish that from the trivial pure-component endpoints.
ORIGINAL WORKED EXAMPLE
Interpret equal phase composition
A nonideal mixture boils at a condition where the liquid has x_A=0.650 and the equilibrium vapor has y_A=0.650. What does one simple distillation stage do to A enrichment at that point?
The vapor and liquid have identical composition.
Condensing that vapor gives the same A fraction 0.650, assuming no other fractionation or reaction.
Thus the stage does not enrich A at that condition.
This is the defining composition behavior of an azeotrope when it occurs at the mixture’s boiling condition.
Answer No enrichment in that equilibrium stage at the azeotropic condition.
08 / UNDERSTAND & APPLY
8. Follow what changes during distillation
In simple batch distillation, vapor is removed and condensed while the remaining liquid composition evolves. For an ideal pair with A more volatile, the early vapor is enriched in A and the liquid becomes relatively richer in B. The equilibrium y value for the initial liquid is not the composition of the entire collected distillate after a substantial fraction has been removed.
Fractional distillation uses repeated vapor–liquid contacts to improve separation. A schematic equilibrium stage changes composition according to the relevant x–y relationship, but real columns have finite efficiency and transport limitations. The elementary theory explains the direction of enrichment; it does not specify a practical apparatus design from vapor pressures alone.
At fixed external pressure, the boiling temperature generally changes as a non-azeotropic batch composition changes. Pure-component vapor pressures therefore change too. A calculation that holds both temperature and pressure fixed while continuously changing arbitrary liquid composition may be imposing incompatible conditions unless a two-phase equilibrium path is explicitly supplied.
Use a short audit: distinguish x,y and z; check whether one or both liquids are volatile; match pure-component pressures to temperature; sum partial pressures; normalize vapor fractions; and verify any phase fraction lies between zero and one. If nonideality is specified, use supplied activity data rather than ideal formulas by habit.
09 / UNDERSTAND & APPLY
9. Original vapor–liquid equilibrium practice
These original problems use ideal binary liquids and ideal vapors unless a deviation is specified. Pure-component vapor pressures refer to the common stated temperature. The questions are instructional constructions, not past-paper claims.
When no numerical temperature is supplied, the given pure-component vapor pressures already encode the relevant temperature for that calculation. Do not import unrelated room-temperature values or convert a vapor composition into a liquid composition without the phase-law relation.
ORIGINAL PRACTICE · 1
For p_A°=90 kPa,p_B°=30 kPa and x_A=0.500, find total pressure and y_A.
Reveal answer & reasoning
P=60.0 kPa;y_A=0.750.
Partial pressures are 45 and 15 kPa. Their sum is 60, and A contributes 45/60 of the vapor.
ORIGINAL PRACTICE · 2
If p_A°=p_B° for an ideal binary mixture, how are x_A and y_A related?
Reveal answer & reasoning
They are equal.
The common pure-component pressure cancels from y_A=x_Ap_A°/[x_Ap_A°+(1−x_A)p_B°]. There is no volatility-based enrichment at that temperature in this model.
ORIGINAL PRACTICE · 3
A nonvolatile solute has liquid mole fraction 0.100 in an ideal solution. What is p_solvent/p_solvent°?
Reveal answer & reasoning
0.900.
The solvent fraction is 1−0.100=0.900. Only the solvent contributes to vapor pressure under the nonvolatile-solute assumption.
ORIGINAL PRACTICE · 4
For p_A°=100 kPa,p_B°=50 kPa and vapor y_A=0.500, find ideal dew pressure.
Reveal answer & reasoning
66.7 kPa.
1/P=0.500/100+0.500/50=0.0150 kPa⁻¹, giving 66.7 kPa. Using the bubble formula with y would be incorrect.
ORIGINAL PRACTICE · 5
An equilibrium pair has x_A=0.200,y_A=0.800 and overall z_A=0.500. Find vapor mole fraction f.
Reveal answer & reasoning
0.500.
f=(z−x)/(y−x)=(0.500−0.200)/(0.800−0.200)=0.500. This is a phase mole fraction, not a phase volume fraction.
ORIGINAL PRACTICE · 6
Does every positively deviating binary mixture necessarily form a minimum-boiling azeotrope?
Reveal answer & reasoning
No.
The deviation must be sufficiently strong and have the required composition dependence to create the relevant extremum and equal phase compositions.
ORIGINAL PRACTICE · 7
At an azeotrope, is the substance necessarily a single pure compound?
Reveal answer & reasoning
No.
It remains a mixture whose liquid and vapor compositions coincide at the specified boiling condition. Constant boiling composition is not proof of a new molecular formula.
ORIGINAL PRACTICE · 8
Why is initial vapor composition not generally the composition of all distillate collected in a batch process?
Reveal answer & reasoning
The remaining liquid and its equilibrium vapor composition change as material is removed.
The total distillate is an accumulation of vapor portions formed at different stages. A material balance over the evolving process is needed.
KEEP THE BIG IDEAS
Quick revision
Liquid x, vapor y and overall z use different mole-fraction denominators.
Raoult gives partial pressures; Dalton converts them into vapor fractions.
The more volatile component is enriched in vapor for an ideal binary mixture with unequal pure pressures.
Bubble and dew formulas start from different known phase compositions.
The lever rule counts phase moles and requires a matched equilibrium x–y pair.
Deviations are activity effects; azeotropes require more than the sign of deviation alone.
Distillation changes composition over time, and azeotropes limit ordinary fixed-pressure fractionation.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.