NEET Weightage: 3-4%

NEET Chemistry — States of Matter Complete Chapter Guide

States Of Matter for NEET.

4 min read

Chapter Overview & Weightage

States of Matter focuses on gas laws, kinetic theory, and real gas behavior. NEET tests this chapter with formula-based problems — if you know the equations, the marks are yours.

States of Matter carries 3-4% weightage in NEET. Ideal gas law, Graham’s law, and van der Waals equation are the most commonly tested areas.

YearNEET Q CountKey Topics Tested
20252Ideal gas, Graham’s law
20241van der Waals, critical constants
20232KMT, RMS speed
20222Gas laws, compressibility factor
20211Dalton’s law, ideal gas
graph TD
    A[States of Matter] --> B[Gas Laws]
    A --> C[Kinetic Molecular Theory]
    A --> D[Real Gases]
    B --> E[Boyle's Law]
    B --> F[Charles's Law]
    B --> G[Ideal Gas: PV = nRT]
    B --> H[Dalton's Partial Pressures]
    C --> I[RMS, Average, Most Probable Speed]
    C --> J[Graham's Law of Diffusion]
    D --> K[van der Waals Equation]
    D --> L[Compressibility Factor Z]

Key Concepts You Must Know

Tier 1 (Always asked)

  • Ideal gas equation: PV=nRTPV = nRT
  • Graham’s law: rate of diffusion 1/M\propto 1/\sqrt{M}
  • Relationship between molecular speeds: vmp:vavg:vrms=1:1.128:1.224v_{mp} : v_{avg} : v_{rms} = 1 : 1.128 : 1.224
  • Dalton’s law of partial pressures

Tier 2 (Frequently asked)

  • van der Waals equation: (P+a/V2)(Vb)=RT(P + a/V^2)(V - b) = RT
  • Compressibility factor: Z=PV/(nRT)Z = PV/(nRT). Z=1Z = 1 for ideal, Z < 1 for attractive domination, Z > 1 for repulsive domination
  • Critical constants in terms of aa and bb

Tier 3 (Occasional)

  • Boyle temperature
  • Liquefaction conditions
  • Mean free path

Important Formulas

PV=nRTPV = nRT

R=8.314R = 8.314 J/mol/K =0.0821= 0.0821 L atm/mol/K

Boyle’s law: P1V1=P2V2P_1V_1 = P_2V_2 (constant TT, nn)

Charles’s law: V1/T1=V2/T2V_1/T_1 = V_2/T_2 (constant PP, nn)

Dalton’s law: Ptotal=p1+p2+p3+P_{total} = p_1 + p_2 + p_3 + \ldots

RMS speed: vrms=3RTMv_{rms} = \sqrt{\dfrac{3RT}{M}}

Average speed: vavg=8RTπMv_{avg} = \sqrt{\dfrac{8RT}{\pi M}}

Most probable speed: vmp=2RTMv_{mp} = \sqrt{\dfrac{2RT}{M}}

Graham’s law: r1r2=M2M1\dfrac{r_1}{r_2} = \sqrt{\dfrac{M_2}{M_1}}

(P+an2V2)(Vnb)=nRT(P + \frac{an^2}{V^2})(V - nb) = nRT

aa corrects for intermolecular attraction, bb for molecular volume.

Critical constants: Tc=8a27RbT_c = \dfrac{8a}{27Rb}, Pc=a27b2P_c = \dfrac{a}{27b^2}, Vc=3bV_c = 3b

Compressibility factor: Z=PVmRTZ = \dfrac{PV_m}{RT}

For speed ratio problems, remember: vmp:vavg:vrms=2:8/π:31:1.128:1.224v_{mp} : v_{avg} : v_{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1 : 1.128 : 1.224. The most probable speed is the smallest, RMS is the largest. This ordering shows up in NEET conceptual questions.


Solved Previous Year Questions

PYQ 1 — NEET 2023

Problem: The RMS speed of O2_2 at TT K is the same as that of H2_2 at 300 K. Find TT.

Solution:

vrms=3RTMv_{rms} = \sqrt{\frac{3RT}{M}}

Setting RMS speeds equal:

3RT32=3R×3002\sqrt{\frac{3RT}{32}} = \sqrt{\frac{3R \times 300}{2}} T32=3002    T=300×322=4800 K\frac{T}{32} = \frac{300}{2} \implies T = \frac{300 \times 32}{2} = \mathbf{4800 \text{ K}}

PYQ 2 — NEET 2022

Problem: The compressibility factor ZZ for an ideal gas is:

Solution:

For an ideal gas, PV=nRTPV = nRT, so:

Z=PVmRT=1Z = \frac{PV_m}{RT} = \mathbf{1}

Z > 1 means the gas is harder to compress (repulsive forces dominate). Z < 1 means attractive forces dominate.


Difficulty Distribution

Difficulty% of QuestionsWhat to Expect
Easy45%Direct formula, gas law substitution
Medium40%Graham’s law, speed comparisons
Hard15%van der Waals, critical constants

Expert Strategy

Week 1: Master the ideal gas equation and all its sub-laws. Practise unit conversions between atm/Pa/mmHg and L/mL/m3^3.

Week 2: Molecular speeds and Graham’s law. Most problems boil down to ratio-and-proportion with molar masses.

Week 3: van der Waals equation and compressibility factor — these are less frequent but carry higher marks.


Common Traps

Trap 1 — Unit of R. Using R=8.314R = 8.314 when pressure is in atm and volume in litres gives wrong answers. Use R=0.0821R = 0.0821 L atm mol1^{-1} K1^{-1} in that case.

Trap 2 — Graham’s law uses molar mass, not molecular mass. The formula is r1/r2=M2/M1r_1/r_2 = \sqrt{M_2/M_1} where MM is molar mass. Using atomic mass instead of molar mass gives wrong ratios for polyatomic molecules.

Trap 3 — Temperature must be in Kelvin. All gas law calculations require absolute temperature. 0°C=2730°C = 273 K, not 0 K. Adding 273 is a step students skip under pressure.