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Chemistry → States of Matter
TG EAPCET 2027
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Chemistry High Weightage β˜…β˜…β˜…β˜… Class 11

States of Matter

Gas laws, kinetic molecular theory, and real gases. PV=nRT underlies everything in physical chemistry. Expect 3–4 questions β€” mostly numerical and conceptual.

3–4Questions in EAPCET
~3%Paper Weightage
10Gas Laws
4Mistake Traps

Concept Core

From Boyle to van der Waals β€” the full gas laws framework.

The Five Gas Laws β€” Unified View
LawRelationshipConstantFormula
Boyle'sP ↑, V ↓T, nPV = k β†’ P₁V₁ = Pβ‚‚Vβ‚‚
Charles'T ↑, V ↑P, nV/T = k β†’ V₁/T₁ = Vβ‚‚/Tβ‚‚
Gay-LussacT ↑, P ↑V, nP/T = k β†’ P₁/T₁ = Pβ‚‚/Tβ‚‚
Avogadron ↑, V ↑T, PV/n = k β†’ equal V = equal n
Ideal GasCombines allβ€”PV = nRT

R = 8.314 J mol⁻¹ K⁻¹ or 0.0821 L atm mol⁻¹ K⁻¹. Temperature must be in Kelvin (K = °C + 273).

Kinetic Molecular Theory (KMT)

Average KE: KE = 3/2 kT per molecule (or 3/2 RT per mole). Proportional to absolute temperature T only β€” independent of gas identity.

rms speed: u_rms = √(3RT/M) β€” root mean square velocity

Average speed: u_avg = √(8RT/Ο€M)

Most probable speed: u_mp = √(2RT/M)

Order: u_rms > u_avg > u_mp

Graham's Law of Effusion

At same T and P, rates of effusion (escape through a tiny hole) are inversely proportional to √(molar mass):

r₁/rβ‚‚ = √(Mβ‚‚/M₁)

Lighter gas effuses faster. Hβ‚‚ effuses 4Γ— faster than Oβ‚‚ (√32/√2 = 4).

Dalton's Law of Partial Pressures

In a mixture of non-reacting gases, total pressure = sum of partial pressures:

P_total = P₁ + Pβ‚‚ + P₃ + ...

Partial pressure of component: P_i = Ο‡_i Γ— P_total where Ο‡_i is mole fraction.

Used in wet gas collection: P_dry = P_total βˆ’ P_water vapour

Real Gases β€” van der Waals Equation

Ideal gas ignores molecular volume and intermolecular attractions. For real gases:

(P + a/VΒ²)(V βˆ’ b) = nRT   (for 1 mole)

a = correction for intermolecular attraction (PaΒ·m⁢/molΒ²). Higher a β†’ stronger attraction.

b = correction for molecular volume (mΒ³/mol). Called excluded volume.

Compressibility Factor Z

Measure of deviation from ideal behaviour:

Z = PV/nRT

Ideal gas: Z = 1. Z < 1: attractive forces dominate (low P, moderate T). Z > 1: repulsive/volume dominates (high P).

At Boyle temperature T_B = a/(Rb), a gas behaves ideally over a wide pressure range.

Formula Vault

Every gas law formula β€” organised for exam use.

Boyle's Law
P₁V₁ = Pβ‚‚Vβ‚‚
Constant T, n
Charles' Law
V₁/T₁ = Vβ‚‚/Tβ‚‚
Constant P, n; T in Kelvin
Gay-Lussac's Law
P₁/T₁ = Pβ‚‚/Tβ‚‚
Constant V, n
Ideal Gas Law
PV = nRT
R = 8.314 J/mol/K
Combined Gas Law
P₁V₁/T₁ = Pβ‚‚Vβ‚‚/Tβ‚‚
Constant n
Average KE
KE = 3/2 RT (per mole)
Independent of gas identity
rms speed
u_rms = √(3RT/M)
M = molar mass in kg/mol
Graham's Law
r₁/rβ‚‚ = √(Mβ‚‚/M₁)
At same T, P
Dalton's Law
P_total = Ξ£P_i
P_i = Ο‡_i Γ— P_total
van der Waals
(P+a/VΒ²)(Vβˆ’b) = RT
For 1 mole real gas
Compressibility Factor
Z = PV/nRT
Z=1 ideal; Z≠1 real
Mole Fraction
Ο‡_A = n_A/(n_A + n_B)
Ο‡_A + Ο‡_B = 1

Worked Examples

5 problems β€” from ideal gas to Graham's law to real gas corrections.

EasyFind volume of 2 mol ideal gas at 27Β°C and 2 atmβ–Ύ
Calculate the volume occupied by 2 mol of an ideal gas at 27Β°C and pressure 2 atm. (R = 0.0821 LΒ·atm/mol/K)
1
T = 27 + 273 = 300 K (always convert to Kelvin)
2
PV = nRT β†’ V = nRT/P = (2 Γ— 0.0821 Γ— 300) / 2 = 49.26/2 = 24.6 L
βœ“ V = 24.6 L
EasyA gas at 27Β°C is heated to 127Β°C at constant pressure β€” new volume?β–Ύ
A gas occupies 4 L at 27Β°C. What is its volume at 127Β°C at constant pressure?
1
T₁ = 300 K, Tβ‚‚ = 400 K (convert!). Charles' Law: V₁/T₁ = Vβ‚‚/Tβ‚‚
2
Vβ‚‚ = V₁ Γ— Tβ‚‚/T₁ = 4 Γ— 400/300 = 16/3 β‰ˆ 5.33 L
βœ“ Vβ‚‚ = 16/3 L β‰ˆ 5.33 L
MediumCompare effusion rates of Hβ‚‚ and Oβ‚‚β–Ύ
Compare the rates of effusion of Hβ‚‚ (M=2) and Oβ‚‚ (M=32) at the same T and P.
1
Graham's Law: r_Hβ‚‚/r_Oβ‚‚ = √(M_Oβ‚‚/M_Hβ‚‚) = √(32/2) = √16 = 4
2
Hβ‚‚ effuses 4 times faster than Oβ‚‚.
βœ“ r_Hβ‚‚ : r_Oβ‚‚ = 4 : 1
EAPCET LevelPartial pressure of Oβ‚‚ in a mixture of 2 mol Oβ‚‚ and 3 mol Nβ‚‚ at 5 atm totalβ–Ύ
A mixture contains 2 mol Oβ‚‚ and 3 mol Nβ‚‚. Total pressure = 5 atm. Find partial pressure of Oβ‚‚.
1
Total moles = 2 + 3 = 5 mol
2
Mole fraction of Oβ‚‚: Ο‡_Oβ‚‚ = 2/5 = 0.4
3
P_Oβ‚‚ = Ο‡_Oβ‚‚ Γ— P_total = 0.4 Γ— 5 = 2 atm
βœ“ P_Oβ‚‚ = 2 atm
Trap QuestionDoes KE depend on the type of gas? What about speed?β–Ύ
At the same temperature, compare average KE and rms speeds of Hβ‚‚ and Oβ‚‚. ⚠️ Classic conceptual trap.
1
KE per mole: KE = 3/2 RT. This depends only on T β€” the same for Hβ‚‚ and Oβ‚‚ at the same temperature.
2
rms speed: u_rms = √(3RT/M). This depends on molar mass M. Hβ‚‚ (M=2) vs Oβ‚‚ (M=32).
3
u_rms(Hβ‚‚)/u_rms(Oβ‚‚) = √(32/2) = 4. Hβ‚‚ molecules move 4Γ— faster.
4
Same KE but different speed β€” because Hβ‚‚ is much lighter (KE = Β½mvΒ²: same KE, less m β†’ more v).
βœ“ Same KE at same T; Hβ‚‚ is 4Γ— faster than Oβ‚‚ (lighter molecules move faster)

Mistake DNA

4 errors from distractor analysis β€” this chapter's traps are mostly unit and concept errors.

🌑️
Using Celsius Instead of Kelvin in Gas Law Calculations
The most frequent numerical error. Charles' Law breaks down completely if you use Β°C.
❌ Wrong
V₁/T₁ = Vβ‚‚/Tβ‚‚
V₁/27 = Vβ‚‚/127 βœ—
(Β°C used directly)
βœ“ Correct
T₁ = 300 K, Tβ‚‚ = 400 K
V₁/300 = Vβ‚‚/400 βœ“
Always convert to Kelvin
Every gas law that involves temperature requires T in Kelvin. K = Β°C + 273. This applies to PV=nRT, Charles', Gay-Lussac, kinetic theory β€” all of them.
πŸ”€
Graham's Law: Taking √(M₁/Mβ‚‚) instead of √(Mβ‚‚/M₁)
The faster (lighter) gas has a larger rate, so the formula inverts the masses.
❌ Wrong
r₁/rβ‚‚ = √(M₁/Mβ‚‚) βœ—
r_Hβ‚‚/r_Oβ‚‚ = √(2/32)
= 1/4 (inverted!)
βœ“ Correct
r₁/rβ‚‚ = √(Mβ‚‚/M₁) βœ“
r_Hβ‚‚/r_Oβ‚‚ = √(32/2)
= 4 (Hβ‚‚ is faster) βœ“
Remember: lighter gas effuses faster β†’ larger ratio. So the heavier mass goes on top in the formula. Check: Hβ‚‚ should be faster than Oβ‚‚, so ratio > 1.
⚑
Thinking Heavier Gases Have Higher KE at Same T
KE = 3/2 RT depends only on temperature, not on molecular mass or identity.
❌ Wrong
"Oβ‚‚ has higher KE than
Hβ‚‚ at 300 K because
Oβ‚‚ is heavier" βœ—
βœ“ Correct
KE = 3/2 RT = same
for all gases at same T.
Different speeds, same KE βœ“
KE is the same; speed differs because KE = Β½mvΒ² β†’ v = √(2KE/m). Heavier gas has lower speed despite equal KE.
🫧
Van der Waals: Confusing which Correction Does What
Students mix up a (attraction) and b (volume) corrections and which direction they push P and V.
❌ Wrong
"a corrects for volume,
b corrects for attraction" βœ—
And: "a reduces V" βœ—
βœ“ Correct
a = attraction correction
(adds to P in equation) βœ“
b = volume correction
(subtracts from V) βœ“
In (P + a/VΒ²)(V βˆ’ b) = RT: the a/VΒ² adds to P because attractions reduce actual pressure below ideal. The b subtracts from V because molecules occupy space.

Chapter Intelligence

Gas laws are heavily numerical β€” know exactly which law to apply when.

EAPCET Topic Weightage (2019–2024)
Ideal gas law PV=nRT
~8
Graham's law of effusion
~6
KMT β€” speeds and KE
~5
Partial pressures
~4
Real gases / van der Waals
~3
High-Yield PYQ Patterns
PV=nRT numerical Graham's law ratio Partial pressure from mole fraction KE at temperature T rms speed comparison Real vs ideal gas Z factor
Exam Strategy
  • ALWAYS write T in Kelvin first. This is the single most common error β€” even in the exam hall.
  • Gas law selection: identify what changes (P? V? T? n?) and what stays constant. This tells you which law to apply.
  • Graham's law: the ratio r₁/rβ‚‚ = √(Mβ‚‚/M₁). Check your answer: lighter gas MUST have higher rate.
  • KMT: KE = 3/2 RT (per mole) depends only on T. Speed = √(3RT/M) depends on M too. These are separate facts tested separately.
  • At STP (0Β°C, 1 atm): 1 mol ideal gas = 22.4 L. At NTP (25Β°C, 1 atm): 1 mol = 24.5 L. These numbers appear in stoichiometry problems across multiple chapters.