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Gaseous State

NEET > Chemistry > States Of Matter

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Overview content

Chapter Snapshot - Gaseous State

A foundational physical chemistry chapter covering the behaviour of gases under varying temperature, pressure, and volume. Boyle's Law, Charles's Law, Gay-Lussac's Law, Avogadro's Law, Ideal Gas Equation (PV = nRT), Van der Waal's equation, kinetic theory of gases, molecular velocities (RMS, average, most probable), Dalton's law of partial pressure, Graham's law of diffusion, and the critical state form the core pillars tested in NEET. The chapter demands fluency with gas equation manipulations, density calculations from PV = nRT, and correct substitution of R in appropriate units. Van der Waal's corrections and compressibility factor Z problems require clear distinction between ideal and real gas behaviour.

āœ“ Use This To Plan Your First 2–3 Hours
Expected Questions (Typical)
Q
2-3
NEET draws 2 to 3 questions from this chapter. One numerical on ideal gas equation (density, molecular weight, or combined gas law calculation), one on kinetic theory or molecular speeds (RMS velocity, KE per mole), and occasionally one on Graham's law of diffusion or Van der Waal's equation/compressibility factor.
Time Required (Practical)
ā±
10-12 hrs
Gas laws and ideal gas equation 2 hrs; Van der Waal's equation and real gas behaviour 2 hrs; kinetic theory and molecular velocities 2.5 hrs; Dalton's law and Graham's law 2 hrs; critical state 1 hr; MCQ practice across all topics 2.5 hrs.
Difficulty Level
⚔
Moderate
Conceptual framework is straightforward but numerical problems demand careful unit handling. Using R = 0.0821 L atm/(mol K) when pressure is in bar, or confusing RMS velocity with most probable velocity, are the two most frequent scoring errors. Van der Waal's equation problems require algebraic care with the correction terms.
Most Asked Style: Numerical MCQ: calculate density or molecular weight from PV = nRT; find RMS or most probable velocity at a given temperature; apply Graham's law to find molecular weight from diffusion rate ratio; determine compressibility factor Z and identify whether gas shows positive or negative deviation; calculate partial pressure using Dalton's law.Biggest Trap: Confusing the three molecular velocities. The ratio u_rms : u_avg : u_mp = sqrt(3) : sqrt(8/pi) : sqrt(2), and u_rms > u_avg > u_mp always. NEET places the wrong velocity formula as a distractor. A second major trap is using R = 0.0821 when pressure is in Pa or bar instead of atm.Fast Win: Memorise five results: (1) PV = nRT with d = PM/RT for density. (2) u_rms = sqrt(3RT/M). (3) u_avg = sqrt(8RT/pi M). (4) u_mp = sqrt(2RT/M). (5) KE per mole = 3RT/2, per molecule = 3kT/2. (6) r1/r2 = sqrt(M2/M1) for Graham's law. These six formulas cover 90% of NEET numericals from this chapter.Revision-Friendly: Yes. Core formulas fit on one card: four gas laws, ideal gas equation, density from gas equation, three velocity expressions, KE formula, Dalton's law, Graham's law, Van der Waal's equation, and critical constants (Tc = 8a/27Rb, Pc = a/27b^2). A 30-minute sweep of these plus one worked numerical per topic covers the full scoring range.

Subtopics - Gaseous State (NEET)

Nine topic blocks: four gas laws (Boyle, Charles, Gay-Lussac, Avogadro), ideal gas equation with gas constant R and Boltzmann constant, deviation from ideality and compressibility factor, Van der Waal's equation with virial reduction, kinetic theory of gases with the kinetic gas equation, molecular velocities (RMS, average, most probable) and kinetic energy, Dalton's law of partial pressure with Amagat's law, Graham's law of diffusion and effusion, and the critical state with critical constants.

Revision tip: Before solving any gas numerical: (1) identify which gas law or equation applies (combined gas law, ideal gas equation, or Van der Waal's), (2) convert all temperatures to Kelvin, (3) match R units to pressure units (0.0821 for atm, 8.314 for Pa or J, 0.083 for bar), (4) check whether the question asks for RMS, average, or most probable velocity. This four-step check eliminates the three most common NEET errors in this chapter.
NCERT LinesMCQsQuick Test

1) Gas Laws

Four fundamental laws relating pressure, volume, temperature, and amount of gas. Boyle's Law: V is inversely proportional to P at constant T (PV = constant). Charles's Law: V is directly proportional to T at constant P (V/T = constant). Gay-Lussac's Law: P is directly proportional to T at constant V (P/T = constant). Avogadro's Law: V is directly proportional to n at constant T and P (V/n = constant).

PV = constant (Boyle)V/T = constant (Charles)P/T = constant (Gay-Lussac)V/n = constant (Avogadro)
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Boyle's LawAt constant temperature, the volume of a definite mass of gas is inversely proportional to pressure: V is proportional to 1/P, so PV = constant, or P1V1 = P2V2. The PV vs P plot is a horizontal line for an ideal gas. At high pressures real gases deviate because molecular volume and intermolecular forces become significant.
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Charle's LawAt constant pressure, the volume of a given mass of gas is directly proportional to its Kelvin temperature: V = V0(T/273.15), so V/T = constant, or V1/T1 = V2/T2. The volume-temperature plot is a straight line passing through the origin on the Kelvin scale. Extrapolation to zero volume gives absolute zero (minus 273.15 degrees C).
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Gay-Lussac's LawAt constant volume and fixed amount of gas, pressure is directly proportional to Kelvin temperature: P is proportional to T, so P/T = constant, or P1/T1 = P2/T2. This law explains why a sealed container of gas shows increased pressure on heating.
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Avogadro's LawEqual volumes of all gases at the same temperature and pressure contain the same number of molecules. V is proportional to n at constant T and P, so V/n = constant, or V1/n1 = V2/n2. One mole of any gas at STP (0 degrees C, 1 atm) occupies 22.4 L. This law establishes the molar volume concept central to stoichiometric gas calculations.

2) Ideal Gas Equation and Gas Constant

Combining Boyle's, Charles's, and Avogadro's laws gives PV = nRT. The universal gas constant R has dimensions of energy per mole per Kelvin. R = 0.0821 L atm/(mol K) = 8.314 J/(mol K) = 1.99 cal/(mol K). The Boltzmann constant k = R/NA = 1.38 x 10^-23 J/K. The gas equation enables calculation of molecular weight (M = mRT/PV) and density (d = PM/RT) of any gas.

PV = nRTR = 0.0821 L atm/(mol K)k = R/NAd = PM/RT
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Ideal gas equationPV = nRT combines all four gas laws into one equation. For n moles: PV = nRT. For mass m of gas with molar mass M: PV = (m/M)RT, so M = mRT/(PV). This equation holds exactly only for an ideal gas; real gases approximate it at high temperature and low pressure.
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Nature and values of RR = PV/(nT) has units of work or energy per mole per Kelvin. Key values: R = 0.0821 L atm/(mol K), 8.314 J/(mol K), 8.314 kPa dm^3/(mol K), 1.99 cal/(mol K), 5.189 x 10^19 eV/(mol K). Matching R to the pressure unit in a problem is the single most important step in any gas calculation.
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Boltzmann ConstantThe gas constant R for a single molecule is the Boltzmann constant k = R/NA = 1.38 x 10^-23 J/K. It connects macroscopic gas behaviour (PV = nRT) to molecular-level kinetic energy (KE = 3kT/2 per molecule). Boltzmann constant appears in all molecular-level energy and velocity expressions.
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Calculation of mass, molecular weight and density of the gas by gas equationFrom PV = (m/M)RT: M = mRT/(PV) and d = PM/RT. Since M and R are constants for a given gas, dT/P = M/R = constant, giving d1T1/P1 = d2T2/P2 for different conditions. Gas densities are in g/L, strongly depend on P and T (d is proportional to P and inversely proportional to T), and are directly proportional to molar mass. At STP, density = molar mass / 22.4.

3) Deviation from ideality [Ideal Behaviour]

No real gas is truly ideal. The compressibility factor Z = PV/(nRT) measures deviation: Z = 1 for ideal gas, Z < 1 when attractive forces dominate (gas more compressible than ideal), Z > 1 when repulsive forces or molecular volume dominate (gas less compressible). At high temperature and low pressure gases approach ideal behaviour. Near liquefaction, deviation is greatest.

Z = PV/nRTZ = 1 for ideal gasHigh T, low P: ideal behaviourH2, He: Z > 1 always
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Compressibility Factor and Behaviour of Real GasesZ = PV/(RT) for one mole. For most gases (CO2, N2, CH4), Z first decreases below 1 at low-to-moderate pressures (attractive forces dominate) then increases above 1 at high pressures (molecular volume dominates). For H2 and He, Z > 1 at all pressures because their intermolecular attraction constant a is extremely small. At higher temperatures, the Z vs P curve flattens toward Z = 1 over a wider pressure range. Gases behave ideally at high temperature and low pressure where intermolecular forces and molecular volume are both negligible relative to thermal energy and container volume.

4) Vander Waal's Equation

Van der Waal's corrected the ideal gas equation for molecular volume (b correction) and intermolecular attraction (a/V^2 correction): (P + a/Vm^2)(Vm minus b) = RT for one mole. The constant a measures intermolecular attraction; b is related to molecular volume. At low pressure, Z = 1 minus a/(VmRT). At high pressure, Z = 1 + Pb/(RT). At Boyle temperature TB = a/(Rb), real gas behaves ideally over a wide pressure range.

(P + a/V^2)(V - b) = RTa: attraction correctionb: volume correctionTB = a/(Rb)
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Vander Waal's corrections and equationTwo corrections to ideal gas equation: (1) Volume correction: actual free volume = Vm minus b, where b = NA[4(4 pi r^3/3)] is four times the actual molecular volume per mole. (2) Pressure correction: actual pressure on walls is reduced by intermolecular attraction, so corrected pressure = P + a/Vm^2. Combining: (P + a/Vm^2)(Vm minus b) = RT for one mole. For n moles: (P + n^2a/V^2)(V minus nb) = nRT.
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Behaviour of Vander Waal's gas at different pressuresAt low pressure, volume is large so b is negligible relative to Vm. The equation simplifies to (P + a/Vm^2)Vm = RT, giving Z = 1 minus a/(VmRT). Here Z < 1 and decreases with increasing P. At high pressure, a/Vm^2 is negligible relative to P. The equation simplifies to P(Vm minus b) = RT, giving Z = 1 + Pb/(RT). Here Z > 1 and increases with P. For H2 and He, a is so small that even at low pressure the high-pressure approximation applies, so Z > 1 at all pressures.
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Reduction of Vander Waal's gas to Virial EquationThe virial equation of state: Z = PVm/(RT) = 1 + B/Vm + C/Vm^2 + ... where B, C are second, third virial coefficients. Expanding the Van der Waal's equation in powers of 1/Vm: Z = 1 + (b minus a/RT)(1/Vm) + (b/Vm)^2 + ... The second virial coefficient B = b minus a/(RT). At Boyle temperature TB = a/(Rb), B = 0, so Z approximately equals 1 over a wide pressure range and the real gas mimics ideal behaviour.
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Other Equations of StateBerthelot equation: P = RT/(V minus b) minus a/(TV^2). Dieterici equation: P = [RT/(V minus b)] times exp(minus a/(VRT)). These alternative equations attempt to improve on Van der Waal's accuracy at extreme conditions. NEET rarely tests these directly but awareness of their existence is expected.

5) Kinetic theory of Gases

Eight postulates form the kinetic molecular model: gas molecules are tiny elastic spheres with negligible volume, no intermolecular forces, random motion, and average kinetic energy proportional to temperature. The kinetic gas equation PV = (1/3)mNu^2 where m is molecular mass, N is number of molecules, and u^2 is mean square speed. All gas laws can be derived from this equation.

Molecules: elastic, random motionNo intermolecular forces assumedPV = (1/3)mNu^2KE proportional to T
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Postulates of kinetic theory of gasesEight postulates: (1) Gas consists of tiny spherical molecules identical in size, shape, and mass. (2) Molecular volume is negligible compared to total gas volume. (3) No intermolecular forces of attraction or repulsion. (4) Molecules are perfectly elastic; no energy lost in collisions. (5) Molecular motion is completely random. (6) Gas pressure results from molecular collisions with container walls. (7) Newton's laws govern molecular motion. (8) At any instant, molecular energies range from small to large values, but average KE is directly proportional to Kelvin temperature.
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Kinetic gas equationDerived from the postulates: PV = (1/3)mNu^2, where m is mass of one molecule, N is total number of molecules, and u^2 is the mean square speed. Rewriting: PV = (2/3)N times (1/2)mu^2 = (2/3) times total kinetic energy. For one mole (N = NA): PVm = (1/3)Mu_rms^2, connecting macroscopic PV to molecular speeds.

6) Expression of some useful physical quantities

Three molecular speed expressions derived from kinetic theory: RMS velocity u_rms = sqrt(3RT/M), average velocity u_avg = sqrt(8RT/(pi M)), and most probable velocity u_mp = sqrt(2RT/M). Their ratio is sqrt(3) : sqrt(8/pi) : sqrt(2), so u_rms > u_avg > u_mp. Average kinetic energy per molecule = (3/2)kT; per mole = (3/2)RT. KE depends only on temperature, not on molecular identity.

u_rms = sqrt(3RT/M)u_avg = sqrt(8RT/piM)u_mp = sqrt(2RT/M)KE/mol = 3RT/2
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Root mean square velocityu_rms = sqrt(u1^2 + u2^2 + ... + uN^2)/N)^(1/2) = sqrt(3RT/M) = sqrt(3P/rho), where M is molar mass in kg/mol, rho is gas density. RMS speed depends only on temperature for a given gas. It is independent of pressure and volume at constant T because PV and P/rho are constant at a given temperature.
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Average velocityu_avg = (u1 + u2 + ... + uN)/N = sqrt(8RT/(pi M)). The average speed is the arithmetic mean of all molecular speeds. It is always less than u_rms because squaring gives more weight to faster molecules.
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Most Probable velocityu_mp = sqrt(2RT/M). This is the speed corresponding to the maximum in the Maxwell-Boltzmann speed distribution curve. More molecules have this speed than any other. The ordering u_rms > u_avg > u_mp holds at all temperatures. Numerically: u_rms : u_avg : u_mp = 1.732 : 1.596 : 1.414.
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Average Kinetic EnergyFrom PV = (1/3)mNu^2, average KE per molecule = (1/2)mu^2 = (3/2)kT, where k = R/NA is the Boltzmann constant. Per mole: KE = (3/2)RT. Average KE depends only on temperature, not on molecular mass or identity. At the same temperature, 1 mole of H2 and 1 mole of O2 have the same total kinetic energy, but H2 molecules move faster because of lower mass.

7) Law of partial pressure

Dalton's law: total pressure of a mixture of non-reacting gases equals the sum of their partial pressures. Partial pressure of a component = mole fraction times total pressure. Amagat's law: total volume equals the sum of partial volumes. Both laws hold for ideal gas mixtures. Dalton's law is used to calculate dry gas pressure by subtracting aqueous tension.

P_total = P1 + P2 + ...Pi = xi times P_totalP_dry = P_atm minus aqueous tensionAmagat: V = V1 + V2
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Dalton's law of partial pressureThe total pressure of a mixture of non-reacting gases is equal to the sum of their partial pressures: P = P1 + P2 + ... By definition, partial pressure is the pressure each gas would exert if it alone occupied the entire volume at the same temperature. From PV = nRT for each component at the same V and T: P1 = (n1/(n1+n2))P = x1 P, where x1 is the mole fraction of gas 1. Partial pressure also equals (volume of gas / total volume) times P. Does not apply to reacting gas mixtures (e.g., NH3 and HCl which form NH4Cl). Used to find dry gas pressure: P_dry = P_atmospheric minus aqueous tension.
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Amagat's law of Partial volumesTotal volume of a mixture of non-reacting gases equals the sum of partial volumes: V = VA + VB + VC + ... where partial volume VA is the volume gas A would occupy alone at the same total pressure and temperature. VA/V = mole fraction of A. Dalton's law and Amagat's law are mathematically equivalent for ideal gases; they differ only in whether you hold volume or pressure constant in the thought experiment.

8) Diffusion of gases

Diffusion is spontaneous intermixing of gas molecules. Effusion is escape through a tiny hole. Graham's law: at constant T and P, rate of diffusion is inversely proportional to the square root of vapour density (or molar mass). r1/r2 = sqrt(M2/M1). Lighter gases diffuse faster; hydrogen has the highest rate. Used in separation of isotopes (atmolysis) and determining molecular weights.

r proportional to 1/sqrt(M)r1/r2 = sqrt(M2/M1)H2 diffuses fastestAtmolysis: isotope separation
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Graham's law of diffusion and EffusionAt constant T and P: r is proportional to 1/sqrt(d), where d is vapour density. For two gases: r1/r2 = sqrt(d2/d1) = sqrt(M2/M1), since 2 times VD = molar mass. Rate can be measured as volume diffused per unit time, mass diffused per unit time, or distance travelled in a tube per unit time. When equal volumes diffuse: r1/r2 = t2/t1. When equal times are given: r1/r2 = V1/V2. When pressure differs: r1/r2 = (P1/P2) times sqrt(M2/M1). Application in atmolysis: separation of U-235 F6 from U-238 F6 during World War II. The law is strictly valid for gases diffusing under low pressure gradients.

9) The critical state

A state where vapour and liquid phases become indistinguishable. Critical temperature Tc is the temperature above which a gas cannot be liquefied regardless of pressure: Tc = 8a/(27Rb). Critical pressure Pc is the minimum pressure needed to liquefy at Tc: Pc = a/(27b^2). Critical volume Vc = 3b. These constants are derived from Van der Waal's equation by setting the first and second derivatives of P with respect to V equal to zero.

Tc = 8a/(27Rb)Pc = a/(27b^2)Vc = 3bAbove Tc: no liquefaction
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Critical state, critical temperature and critical pressureThe critical state is defined by critical temperature Tc, critical pressure Pc, and critical volume Vc. Above Tc, no amount of pressure can liquefy the gas. At the critical point, the meniscus between liquid and gas disappears as both phases merge. From Van der Waal's constants: Tc = 8a/(27Rb), Pc = a/(27b^2), Vc = 3b. The ratio PcVc/(RTc) = 3/8 = 0.375 for a Van der Waal's gas. Gases with high Tc (like CO2, NH3) are easily liquefied; gases with low Tc (like H2, He) require extreme cooling. A gas above its Tc is called a supercritical fluid.

Gaseous State Download Notes & Weightage Plan

For each topic in the Gaseous State chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Gas Laws

Four foundational laws connecting P, V, T, and n. Boyle's, Charles's, Gay-Lussac's, and Avogadro's Laws form the basis for the ideal gas equation.

Boyle: PV = constCharles: V/T = constGay-Lussac: P/T = constAvogadro: V/n = const

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Boyle's Law: V inversely proportional to P at constant T and n; P1V1 = P2V2. Charles's Law: V directly proportional to T(K) at constant P; V1/T1 = V2/T2. Gay-Lussac's Law: P directly proportional to T(K) at constant V; P1/T1 = P2/T2. Avogadro's Law: V proportional to n at constant T and P; V/n = constant; 1 mol at STP = 22.4 L. All four laws hold strictly for ideal gases only.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write each law in words, as a proportionality, and as a ratio equation. Draw the characteristic graph for each: PV vs P (horizontal line for Boyle), V vs T (straight through origin for Charles), P vs T (straight through origin for Gay-Lussac). Solve 3 combined gas law problems converting between different T and P conditions.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Individual gas laws are rarely tested as standalone NEET questions. They appear as part of combined gas law or ideal gas equation problems. Charles's Law graphical interpretation occasionally appears.
Time Required1.5 hrs30 min on Boyle's and Charles's Laws with graphs; 20 min on Gay-Lussac's and Avogadro's Laws; 40 min solving combined gas law numericals.
DifficultyEasyDirect proportionality and inverse proportionality relationships. The only challenge is remembering to convert Celsius to Kelvin before applying Charles's or Gay-Lussac's Law.
  • Scoring Focus: Always convert temperature to Kelvin first. T(K) = t(C) + 273.15. Using Celsius in V/T or P/T ratios gives wrong answers. NEET distractors are calculated using Celsius temperatures.
  • High-risk Area: Forgetting to convert Celsius to Kelvin. A problem at 27 degrees C requires T = 300 K, not 27. This error changes the answer by an order of magnitude in ratio problems.
  • Best Practice Style: First step in every gas problem: write T in Kelvin. Then identify which variables are constant and which change. Apply the appropriate gas law ratio.
Priority rule: Medium priority. Gas laws are the building blocks; the ideal gas equation subsumes all four. Spend 1.5 hours then move to PV = nRT-based problems.

Ideal Gas Equation and Gas Constant

The unified equation PV = nRT with gas constant R in various units. Boltzmann constant k = R/NA. Density and molecular weight calculations from the gas equation.

PV = nRTd = PM/RTM = mRT/(PV)STP density = M/22.4

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)PV = nRT: n = m/M so PV = (m/M)RT. Rearranging: M = mRT/(PV) and d = PM/RT. dT/P = M/R = constant for a given gas, so d1T1/P1 = d2T2/P2. R = 0.0821 L atm/(mol K) = 8.314 J/(mol K) = 8.314 kPa dm^3/(mol K) = 1.99 cal/(mol K). Boltzmann constant k = R/NA = 1.38 x 10^-23 J/K. Gas density in g/L; depends strongly on P and T (d proportional to P, inversely proportional to T). Density at STP = M/22.4 g/L.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write PV = nRT and derive d = PM/RT and M = mRT/(PV) from it. Make a table of R values with matching pressure units. Solve 5 numericals: 2 on molecular weight from PV data, 2 on density at given T and P, 1 on density ratio at different conditions using d1T1/P1 = d2T2/P2.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One numerical per NEET paper on density calculation (d = PM/RT) or molecular weight determination from given P, V, T data. The question tests correct unit matching for R.
Time Required2 hrs30 min on ideal gas equation derivation and R values; 30 min on density and molecular weight formulas; 1 hr MCQ practice with unit-matching exercises.
DifficultyEasy-ModerateThe formulas are direct algebraic manipulations of PV = nRT. The challenge is purely in unit matching: R = 0.0821 with atm, R = 8.314 with Pa or kPa, R = 0.083 with bar.
  • Scoring Focus: d = PM/RT is the most frequently tested formula. Know that M must be in g/mol and d comes out in g/L when R = 0.0821 and P is in atm. Cross-check: N2 at STP has d = 28/22.4 = 1.25 g/L.
  • High-risk Area: Using R = 0.0821 when pressure is given in bar or Pa. Mismatched units give wrong density by 1.3% (bar vs atm) or by orders of magnitude (Pa vs atm).
  • Best Practice Style: Circle the pressure unit in the question. Pick R accordingly: atm uses 0.0821, kPa uses 8.314, bar uses 0.083. Then substitute with all units written out.
Priority rule: High priority. PV = nRT and d = PM/RT are the most tested formulas. Master these before any other topic in this chapter.

Deviation from ideality [Ideal Behaviour]

Real gases deviate from PV = nRT. Compressibility factor Z quantifies deviation: Z = PV/(nRT). Z = 1 ideal; Z < 1 attractive forces dominate; Z > 1 repulsion/volume dominates.

Z = PV/(nRT)Z = 1: idealHigh T, low P: idealH2/He: Z > 1 at all P

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Z = PV/(RT) for 1 mol. For ideal gas: Z = 1 at all P. For real gases: at low P (up to ~10 atm) Z approximately 1. For CO2, N2, CH4: Z first dips below 1 (attractions dominate) then rises above 1 (molecular volume dominates) as P increases. For H2 and He: Z > 1 at all pressures because a is negligibly small. At higher T, Z vs P curve flattens toward 1. Gases behave ideally at high temperature and low pressure. Near liquefaction conditions (low T, high P), deviation is maximum.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Sketch Z vs P curves for an ideal gas (horizontal line at Z = 1), CO2, N2, and H2. Mark where Z < 1 and Z > 1 on each curve. Understand why H2 and He show only Z > 1: their a constant is nearly zero.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1One conceptual MCQ testing Z interpretation: identify which gas shows Z > 1 at all pressures (H2/He), or state conditions for ideal behaviour (high T, low P).
Time Required1 hr30 min on Z concept and graph interpretation; 30 min MCQ practice on real vs ideal gas behaviour.
DifficultyEasyConceptual recall. Know the Z vs P graph shapes and the two conditions for ideal behaviour. No calculations needed.
  • Scoring Focus: Two key facts: (1) High temperature and low pressure give ideal behaviour. (2) H2 and He always show Z > 1. NEET assertion-reason questions frequently test these.
  • High-risk Area: Confusing Z > 1 (less compressible, repulsive) with Z < 1 (more compressible, attractive). Students sometimes reverse the physical meaning.
  • Best Practice Style: Z < 1 means the real gas occupies LESS volume than ideal prediction (attractive forces pull molecules closer). Z > 1 means the real gas occupies MORE volume than ideal (molecular volume matters).
Priority rule: Medium priority. Quick conceptual topic. Spend 1 hour then move to Van der Waal's equation for the mathematical treatment.

Vander Waal's Equation

Mathematical correction of ideal gas equation for intermolecular attractions (a/V^2) and molecular volume (b). Behaviour at low and high pressures. Virial equation and Boyle temperature.

(P + a/V^2)(V - b) = RTLow P: Z = 1 - a/(VRT)High P: Z = 1 + Pb/(RT)TB = a/(Rb)

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Van der Waal's equation: (P + a/Vm^2)(Vm minus b) = RT for 1 mol. Two corrections: (1) Pressure correction +a/Vm^2 for intermolecular attraction; (2) Volume correction: actual free volume = Vm minus b. For n moles: (P + n^2a/V^2)(V minus nb) = nRT. At low P: b negligible, Z = 1 minus a/(VmRT), Z < 1. At high P: a/V^2 negligible, Z = 1 + Pb/(RT), Z > 1. Virial equation: Z = 1 + B/Vm + C/Vm^2 + ... Second virial coefficient B = b minus a/(RT). At Boyle temperature TB = a/(Rb), B = 0 and gas behaves ideally. Berthelot and Dieterici equations are alternative state equations.
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ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write the Van der Waal's equation for n moles and for 1 mole. Derive Z at low P and at high P by dropping the appropriate correction. Write B = b minus a/(RT) and set it to zero to derive TB. Solve 3 numericals: one calculating Z from given a and b, one on Boyle temperature, one on assertion-reason using Z = 1 + Pb/(RT).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1One question on Van der Waal's equation: assertion-reason on Z at high pressure, or numerical calculating Z from a, b, P values. Sometimes combined with compressibility factor concept.
Time Required2 hrs45 min on Van der Waal's equation derivation; 30 min on low P and high P approximations; 20 min on virial equation and Boyle temperature; 25 min MCQ practice.
DifficultyModerateAlgebraic manipulation is required. Knowing when to drop b (low P) vs when to drop a/V^2 (high P) is the key decision. Virial expansion involves series approximation.
  • Scoring Focus: At high pressure: Z = 1 + Pb/(RT). At low pressure: Z = 1 minus a/(VmRT). NEET assertion-reason questions directly test these two approximate forms.
  • High-risk Area: Confusing which correction to drop at which pressure. At LOW pressure, molecular volume b is negligible (drop b). At HIGH pressure, attraction a/V^2 is negligible (drop a/V^2). Students often reverse this.
  • Best Practice Style: Low P means large V, so the b term (small compared to V) vanishes. High P means V is compressed small, so attractive correction a/V^2 becomes negligible compared to the large P.
Priority rule: Medium priority. One question every 2-3 NEET papers. Focus on the two approximate Z expressions rather than complex algebra.

Kinetic theory of Gases

Eight postulates of the kinetic molecular model. The kinetic gas equation PV = (1/3)mNu^2 connects macroscopic gas properties to molecular motion.

8 postulatesPV = (1/3)mNu^2All gas laws derivableKE proportional to T

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Topic Notes (Condensed)Postulates: (1) Gas = tiny spherical molecules. (2) Volume negligible. (3) No intermolecular forces. (4) Perfectly elastic collisions. (5) Random motion. (6) Pressure from wall collisions. (7) Newton's laws apply. (8) Average KE proportional to temperature. Kinetic gas equation: PV = (1/3)mNu^2 where m = mass of one molecule, N = total molecules, u^2 = mean square speed. Rewriting: PV = (2/3) times total KE. All gas laws derive from this equation. Boyle's Law follows from constant T (constant KE). Charles's Law follows from constant P.
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ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: List all 8 postulates from memory. Write the kinetic gas equation and show how PV = (2/3) times total KE. Derive Boyle's Law and Charles's Law from the kinetic equation. Focus on which postulates fail for real gases (postulates 2 and 3).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1One conceptual question: identify false postulate, state which relationship follows from kinetic theory, or verify that mean KE depends only on temperature.
Time Required1.5 hrs30 min memorising postulates; 30 min on kinetic gas equation derivation; 30 min MCQ practice on conceptual statements.
DifficultyEasyConceptual recall. The postulates and the kinetic equation are directly examined. No complex calculations.
  • Scoring Focus: Most tested fact: average KE per molecule = (3/2)kT depends only on temperature. At the same temperature, all ideal gases have the same average KE per molecule regardless of molecular mass.
  • High-risk Area: Confusing 'average KE per molecule is the same for all gases at same T' with 'molecular speed is the same'. Speed depends on mass; KE at same T does not.
  • Best Practice Style: When a question says 'same temperature, different gases', KE per molecule is identical, but heavier molecules move slower. u_rms = sqrt(3RT/M): M in denominator means heavier gas has lower speed.
Priority rule: Medium priority. Postulates are conceptual building blocks for velocity expressions which carry more marks.

Expression of some useful physical quantities

Three molecular speed expressions (RMS, average, most probable) and average kinetic energy per molecule and per mole. The speed ratio sqrt(3) : sqrt(8/pi) : sqrt(2) is a NEET favourite.

u_rms = sqrt(3RT/M)u_avg = sqrt(8RT/piM)u_mp = sqrt(2RT/M)KE = 3kT/2 per molecule

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Topic Notes (Condensed)RMS velocity: u_rms = sqrt(3RT/M) = sqrt(3P/rho). Average velocity: u_avg = sqrt(8RT/(pi M)). Most probable velocity: u_mp = sqrt(2RT/M). Ratio u_rms : u_avg : u_mp = sqrt(3) : sqrt(8/pi) : sqrt(2) = 1.732 : 1.596 : 1.414. Always u_rms > u_avg > u_mp. Average KE per molecule = (3/2)kT; per mole = (3/2)RT. KE depends on T only; speed depends on T and M. Doubling T increases u_rms by factor sqrt(2), not 2.
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ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write all three velocity formulas side by side. Note the numerator changes: 3, 8/pi, 2. Memorise the ratio 1.732 : 1.596 : 1.414. Solve 5 numericals: 2 comparing speeds of different gases, 2 finding temperature for a given speed change, 1 finding KE per mole or per molecule.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One numerical on RMS or most probable velocity: find the speed at a given temperature, or find temperature for speed to double, or compare speeds of two different gases. The velocity ratio MCQ is a NEET favourite.
Time Required2 hrs30 min on three velocity formulas; 30 min on KE per molecule and per mole; 1 hr MCQ practice with temperature and mass variations.
DifficultyModerateFormulas are similar in structure (sqrt(constant times RT/M)). The trap is using the wrong constant. Also, doubling temperature makes speed increase by sqrt(2), not 2.
  • Scoring Focus: u_rms = sqrt(3RT/M) has M in kg/mol when R = 8.314 J/(mol K). The result is in m/s. For NEET calculations, use M in g/mol with appropriate unit conversion. Speed ratio comparisons between gases at same T: u1/u2 = sqrt(M2/M1).
  • High-risk Area: Confusing the three speeds. Students pick u_mp formula when u_rms is asked, or vice versa. The coefficient under the square root is the distinguishing factor: 3 for RMS, 8/pi for average, 2 for most probable.
  • Best Practice Style: Tag each formula with its coefficient: RMS = 3, AVG = 8/pi approximately 2.55, MP = 2. In MCQs, check which coefficient is used. If the numerator is 2RT: it is most probable, not RMS.
Priority rule: High priority. Molecular velocity questions appear almost every year. Master the three formulas and their ratio.

Law of partial pressure

Dalton's law for pressure additivity in gas mixtures. Partial pressure equals mole fraction times total pressure. Amagat's law for volume additivity.

P = P1 + P2 + ...Pi = xi times PP_dry = P_atm - aqueous tensionVi/V = xi

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Topic Notes (Condensed)Dalton's law: P = P1 + P2 + ... for non-reacting gases. Pi = (ni/(n1+n2+...))P = xi P. Also Pi = (Vi/V_total) times P if volumes are known. Applied to find dry gas pressure: P_dry = P_atmospheric minus aqueous tension (vapour pressure of water). Does NOT apply to reacting gases (e.g., NH3 + HCl). Amagat's law: V = VA + VB + ... where VA/V = mole fraction of A. Mathematically equivalent to Dalton's law for ideal gas mixtures.
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ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write Dalton's law and the mole fraction formula for partial pressure. Solve 3 problems: one mixing gases in a container, one collecting gas over water (subtract aqueous tension), one finding partial pressure from mass ratios. Know: equal moles means equal partial pressures regardless of molecular weight.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1One question on partial pressure calculation from given moles or masses of gases in a mixture. The aqueous tension subtraction question appears periodically.
Time Required1.5 hrs30 min on Dalton's law and mole fraction derivation; 20 min on Amagat's law; 40 min MCQ practice on gas mixtures.
DifficultyEasyStraightforward mole fraction calculation and pressure summation. The only trap is confusing mass ratio with mole ratio.
  • Scoring Focus: Pi = xi times P. Always convert masses to moles first, then find mole fractions. Equal masses of different gases do NOT give equal partial pressures because their moles differ.
  • High-risk Area: Using mass fractions instead of mole fractions. If 56 g N2 and 44 g CO2 are mixed, moles are 2 and 1, so mole fractions are 2/3 and 1/3, not 56/100 and 44/100.
  • Best Practice Style: Step 1: convert all masses to moles (divide by molar mass). Step 2: find mole fraction = moles of component / total moles. Step 3: Pi = xi times P_total.
Priority rule: Medium priority. Quick scoring topic. Master the mole fraction approach and move on.

Diffusion of gases

Graham's law: rate of diffusion inversely proportional to sqrt(molar mass). Multiple forms for volume, time, and pressure variations. Applied in atmolysis for isotope separation.

r1/r2 = sqrt(M2/M1)Equal V: r1/r2 = t2/t1Equal t: r1/r2 = V1/V2H2 diffuses fastest

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Topic Notes (Condensed)Graham's law: r proportional to 1/sqrt(d) at constant T and P, where d = vapour density. r1/r2 = sqrt(d2/d1) = sqrt(M2/M1). Rate = V/t or w/t. Equal volumes: r1/r2 = t2/t1. Equal times: r1/r2 = V1/V2. When pressures differ: r1/r2 = (P1/P2)sqrt(M2/M1). H2 (M = 2) has highest diffusion rate. Atmolysis: separation of U-235 F6 from U-238 F6 by diffusion through porous barriers. Law is accurate for gases diffusing under low pressure gradients. Vapour density is unitless and independent of temperature; absolute density has units (g/L) and depends on T.
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ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write Graham's law in all three forms (volume ratio, time ratio, mass ratio). Solve 3 problems: one finding unknown MW from diffusion rate ratio, one with equal volumes (time ratio), one with pressure correction. Memorise: diffusion rate of X compared to H2: rX/rH2 = sqrt(2/MX).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One numerical using Graham's law: find molecular weight from rate ratio, or find time ratio for diffusion of equal volumes of two gases. The sqrt relationship is key.
Time Required1.5 hrs30 min on Graham's law statement and derivation; 20 min on various forms (volume, time, pressure); 40 min MCQ practice.
DifficultyEasy-ModerateThe formula is simple but NEET varies the form: sometimes rate as V/t, sometimes as t2/t1. Keeping track of which ratio is direct and which is inverse requires care.
  • Scoring Focus: r1/r2 = sqrt(M2/M1). When times are given for equal volumes: r1/r2 = t2/t1 (inverse). When a gas takes 3 times longer than He to effuse: r_gas/r_He = 1/3, so M_gas = 9 times M_He = 36 u.
  • High-risk Area: Inverting the ratio. If gas X takes twice as long as H2 to diffuse the same volume, rX/rH2 = 1/2, NOT 2. Squaring gives M_X/M_H2 = 4, so M_X = 8. Students who set the ratio as 2 get M_X = 0.5, which is nonsensical.
  • Best Practice Style: Identify which gas is faster from the problem context. Faster gas has lower M. Set up r_fast/r_slow = sqrt(M_slow/M_fast). Confirm your ratio gives the faster gas a higher rate.
Priority rule: High priority. Graham's law questions appear frequently and are quick to solve once the formula is memorised.

The critical state

Critical temperature, pressure, and volume define the point where gas-liquid distinction vanishes. Derived from Van der Waal's equation. Determines ease of gas liquefaction.

Tc = 8a/(27Rb)Pc = a/(27b^2)Vc = 3bPcVc/RTc = 3/8

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Topic Notes (Condensed)Critical state: vapour and liquid are indistinguishable. Tc = 8a/(27Rb): temperature above which no liquefaction is possible regardless of pressure. Pc = a/(27b^2): minimum pressure to liquefy at Tc. Vc = 3b: molar volume at the critical point. The critical compressibility factor Zc = PcVc/(RTc) = 3/8 for a Van der Waal's gas. Gases with high Tc (CO2, NH3, SO2) are easily liquefied; gases with very low Tc (H2, He, N2) require extreme cooling. A substance above Tc is a supercritical fluid.
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ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Memorise the three critical constant formulas: Tc = 8a/(27Rb), Pc = a/(27b^2), Vc = 3b. Verify Zc = PcVc/(RTc) = 3/8. Know one example: CO2 has Tc = 304 K (easy to liquefy); He has Tc = 5.2 K (very hard). Solve 2 problems deriving a and b from given Tc and Pc values.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Occasionally one conceptual question on critical temperature: which gas cannot be liquefied at room temperature (answer: gases with Tc below room temperature). Numerical on Tc from a and b is rare but possible.
Time Required1 hr20 min on critical constants and their derivation from Van der Waal's equation; 20 min on Zc calculation; 20 min MCQ practice.
DifficultyEasy-ModerateFormulas are direct. The difficulty lies in remembering the 27 and 8 coefficients correctly. Conceptual understanding of why gases with low Tc cannot be liquefied at room temperature is essential.
  • Scoring Focus: Tc determines whether a gas can be liquefied at a given temperature. If T > Tc, no amount of pressure works. NEET tests this concept directly when asking about gas liquefaction.
  • High-risk Area: Confusing Tc with boiling point. Critical temperature is the maximum T for liquefaction; boiling point is the T where liquid vapour pressure equals atmospheric pressure. They are different quantities.
  • Best Practice Style: Link critical state to Van der Waal's constants: high a (strong attraction) means high Tc (easier to liquefy). Low a (weak attraction like He) means very low Tc (extremely hard to liquefy).
Priority rule: Low priority. One question every 3-4 NEET papers. Memorise formulas and one comparison example. Do not spend excessive time here.

Gaseous State Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Gaseous State chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives
Molecular Velocities and Temperature
NEETRMS velocityMost probable velocityTemperature scaling

Mistake Snapshot (What Students Do Wrong)

  • Using the wrong velocity formula (RMS vs average vs most probable): The three velocity expressions differ only in the coefficient under the square root: 3RT/M for RMS, 8RT/(pi M) for average, 2RT/M for most probable. NEET distractors substitute one coefficient for another. Picking u_mp = sqrt(2RT/M) when the question asks for u_rms = sqrt(3RT/M) gives an answer that is sqrt(2/3) = 0.816 times too low.
  • Assuming doubling temperature doubles the speed: Since u = sqrt(constant times T/M), doubling T multiplies speed by sqrt(2), not 2. To double the speed, temperature must be quadrupled (T2 = 4T1). NEET expects you to know this sqrt relationship.
2–3 Line Example (Typical Error)

RMS velocity of N2 at 300 K: u_rms = sqrt(3 x 8.314 x 300 / 0.028) = sqrt(267214) = 517 m/s. At 1200 K: u = 517 x sqrt(1200/300) = 517 x 2 = 1034 m/s. A student who says 'temperature quadrupled so speed quadrupled' gets 2068 m/s, which is wrong.

How NEET Frames The Trap

NEET asks: 'RMS velocity of a gas at 27 degrees C is v. At what temperature will the RMS velocity become 2v?' The correct answer is 1200 K (4 times 300 K), not 600 K (2 times 300 K). NEET places 600 K as a distractor.

NEET-Style Trap Question Format

Q. The RMS velocity of a gas at 300 K is v. The temperature at which the RMS velocity becomes 2v is
A. 1200 K   B. 600 K   C. 900 K   D. 150 K  
Trick: u_rms is proportional to sqrt(T). To double u, T must be quadrupled: T2 = 4 x 300 = 1200 K (Option A). Option B (600 K) is the 'doubled temperature' distractor, giving speed = v times sqrt(2), not 2v. Option C (900 K) gives sqrt(3) times v. Option D (150 K) halves the speed.

Quick rule: Speed is proportional to sqrt(T). Doubling speed requires 4x temperature. Halving speed requires T/4. For different gases at same T: u1/u2 = sqrt(M2/M1).
Unit Mismatch in Gas Equation
NEETGas constant RPressure unitsDensity calculation

Mistake Snapshot (What Students Do Wrong)

  • Using R = 0.0821 when pressure is in Pa or bar: R = 0.0821 L atm/(mol K) is valid only when P is in atm and V is in litres. When P is in Pa, use R = 8.314. When P is in bar, use R = 0.083. Using mismatched units gives density or molecular weight off by a factor of 101.325 (Pa vs atm) or 1.013 (bar vs atm).
  • Forgetting to convert temperature to Kelvin: PV = nRT requires T in Kelvin. Substituting T = 27 instead of T = 300 gives an answer 11 times too large (300/27). NEET routinely gives temperature in Celsius and places the Celsius-substitution answer as a distractor.
2–3 Line Example (Typical Error)

Density of N2 at 227 degrees C and 5 atm. d = PM/(RT) = 5 x 28 / (0.0821 x 500) = 140/41.05 = 3.41 g/L. If student uses T = 227: d = 140/(0.0821 x 227) = 7.51 g/L, which is the distractor.

How NEET Frames The Trap

NEET gives temperature as 227 degrees C and expects conversion to 500 K. When a student forgets to add 273, the wrong answer matches a distractor option exactly.

NEET-Style Trap Question Format

Q. What is the density of N2 gas at 227 degrees C and 5.00 atm pressure? (R = 0.0821 L atm/(mol K), M = 28 g/mol)
A. 3.41 g/L   B. 7.51 g/L   C. 1.71 g/L   D. 6.84 g/L  
Trick: T = 227 + 273 = 500 K. d = PM/(RT) = (5 x 28)/(0.0821 x 500) = 3.41 g/L (Option A). Option B (7.51) uses T = 227. Option C halves the answer. Option D doubles it.

Quick rule: Three checks before substituting into PV = nRT: (1) T in Kelvin? (2) R matches P units? (3) V in litres? If any check fails, fix it before calculating.
Graham's Law Rate Ratios
NEETDiffusionEffusionRate ratio inversion

Mistake Snapshot (What Students Do Wrong)

  • Inverting the rate-time relationship for equal volumes: When equal volumes of two gases diffuse, the rate ratio r1/r2 = t2/t1 (times are inversely proportional to rates). Students who write r1/r2 = t1/t2 get the inverse ratio. If gas X takes 3 times longer than H2 for the same volume, rX/rH2 = 1/3, not 3.
  • Confusing vapour density with absolute density in Graham's law: Graham's law uses vapour density (unitless, temperature-independent) or molecular weight. Absolute density (g/L) depends on T and P. Substituting absolute density at non-STP conditions without adjusting for T and P gives incorrect molecular weight ratios.
2–3 Line Example (Typical Error)

A gas takes 3 times as long as He (M = 4) to effuse. r_gas/r_He = t_He/t_gas = 1/3. M_gas/M_He = (r_He/r_gas)^2 = 9. M_gas = 9 x 4 = 36 u. If student writes r_gas/r_He = 3 (inverting): M_gas = M_He/9 = 0.44 u, which is physically impossible.

How NEET Frames The Trap

NEET states a gas takes N times longer than a reference gas. The rate of the slower gas is 1/N of the reference. Squaring gives the MW ratio. Students who set rate equal to N get M_gas = M_ref/N^2, yielding an impossibly small molecular weight that appears as a distractor.

NEET-Style Trap Question Format

Q. A certain gas takes three times as long to effuse out as helium (M = 4). Its molecular mass is
A. 36 u   B. 12 u   C. 27 u   D. 9 u  
Trick: r_gas/r_He = 1/3 (slower gas, lower rate). (r_He/r_gas)^2 = 9 = M_gas/M_He. M_gas = 9 x 4 = 36 u (Option A). Option B (12) uses ratio = 3 instead of 1/3. Option C (27) cubes instead of squaring. Option D (9) forgets to multiply by M_He.

Quick rule: Longer time = slower rate. If gas X takes N times longer than gas Y for same volume: rX/rY = 1/N. Then MX = MY x N^2. Always verify: the slower (heavier) gas should have the larger molecular weight.
Dalton's Law and Mole Fraction
NEETPartial pressureMole fractionMass vs moles

Mistake Snapshot (What Students Do Wrong)

  • Using mass fraction instead of mole fraction for partial pressure: Partial pressure = mole fraction x total pressure. Mole fraction requires converting mass to moles first. Equal masses of O2 (M = 32) and CH4 (M = 16) give mole ratios of 1:2, not 1:1. Using mass ratios directly as mole fractions gives wrong partial pressures.
  • Applying Dalton's law to reacting gas mixtures: Dalton's law applies only to non-reacting gases. NH3 and HCl react to form NH4Cl, so their mixture does not obey Dalton's law. NEET tests this exception as a standalone MCQ.
2–3 Line Example (Typical Error)

Equal masses of O2 and CH4 mixed at total pressure 1 atm. Moles O2 = m/32, moles CH4 = m/16 = 2(m/32). Total moles = 3(m/32). x_O2 = 1/3, P_O2 = 1/3 atm. x_CH4 = 2/3, P_CH4 = 2/3 atm. Student using mass fractions: each is 1/2, giving P = 0.5 atm for each, which is wrong.

How NEET Frames The Trap

NEET gives 'equal masses' of two gases with different molecular weights. This signals that moles are unequal. The distractor option uses mass fraction = 0.5 for each gas.

NEET-Style Trap Question Format

Q. Equal masses of methane and oxygen are mixed in a container at 25 degrees C. The fraction of total pressure exerted by oxygen is
A. 1/3   B. 2/3   C. 1/2   D. 8/9  
Trick: Let mass = m each. Moles CH4 = m/16, moles O2 = m/32. x_O2 = (m/32)/((m/16)+(m/32)) = 1/3. Fraction = 1/3 (Option A). Option C (1/2) uses mass fraction instead of mole fraction. Option B (2/3) is the CH4 fraction.

Quick rule: Always convert mass to moles before finding mole fraction. Partial pressure = mole fraction x total pressure. 'Equal mass' does NOT mean equal moles unless the gases have the same molecular weight.
Van der Waal's Equation Approximations
NEETVan der WaalCompressibility factorH2 He behaviour

Mistake Snapshot (What Students Do Wrong)

  • Dropping the wrong correction term at a given pressure: At low pressure, molecular volume b is negligible (large V, so Vm >> b). Drop b to get Z = 1 minus a/(VmRT). At high pressure, V is small, so a/Vm^2 is negligible compared to large P. Drop a/V^2 to get Z = 1 + Pb/(RT). Dropping the wrong term reverses the Z prediction.
  • Claiming H2 shows Z < 1 at low pressure: For H2 and He, the intermolecular attraction constant a is negligibly small. The volume correction b dominates even at low pressure, so Z > 1 at all pressures. Students who apply the general pattern (Z < 1 at low P) to H2 and He answer assertion-reason questions incorrectly.
2–3 Line Example (Typical Error)

For N2 at low pressure: Z = 1 minus a/(VmRT). Since a is significant for N2, Z dips below 1. For H2 at low pressure: a is nearly zero, so Z = 1 + Pb/(RT) > 1. Student who says 'all gases show Z < 1 at low P' picks the wrong assertion-reason option for H2.

How NEET Frames The Trap

NEET assertion-reason: 'Assertion: Z for H2 is always greater than 1. Reason: At low pressure, repulsive forces dominate for H2.' The correct response requires knowing that for H2, a is so small that even the low-pressure Z expression gives Z > 1.

NEET-Style Trap Question Format

Q. Dominance of strong repulsive forces among the molecules of a gas (Z = compressibility factor)
A. Depends on Z and indicated by Z = 1   B. Depends on Z and indicated by Z > 1   C. Depends on Z and indicated by Z < 1   D. Is independent of Z  
Trick: When repulsive forces (or molecular volume) dominate, real gas molecules occupy more space than ideal gas prediction. Z = PV_real/(nRT) > 1 (Option B). Z = 1 is ideal. Z < 1 means attractions dominate. Option D is incorrect because Z directly measures the force balance.

Quick rule: Z > 1: repulsion/volume dominates (gas harder to compress than ideal). Z < 1: attraction dominates (gas easier to compress). H2 and He: always Z > 1. Most other gases: Z < 1 at moderate P, then Z > 1 at high P.
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