States Of Matter JEE 2025 Lecture Notes PDF

# ARJUNA JEE 2025 - States of Matter - Lecture 11 ## By - Nikhil Saini Sir ## Topics to be covered 1. Eudiometry 2. Collision Theory 3. Virial Equation 4. Equation of State ## Eudiometry: | Gas | Reagents to absorb it | | :------ | :--------------------- | | CO₂, SO₂ (acidic gas) | Alkali -...

# ARJUNA JEE 2025 - States of Matter - Lecture 11 ## By - Nikhil Saini Sir ## Topics to be covered 1. Eudiometry 2. Collision Theory 3. Virial Equation 4. Equation of State ## Eudiometry: | Gas | Reagents to absorb it | | :------ | :--------------------- | | CO₂, SO₂ (acidic gas) | Alkali - NaOH, KOH | | O₂ | Alkaline Pyrogallol | | | Turpentine Oil | | CO, C₂H₂ | Ammonical Cu₂Cl₂ | | NH₃ | CuSO₄ soln in acidic sol | | N₂ | Heated Mg or sometimes inert | | NO | FeSO₄ soln | | H₂O | Neglected (liquid) | **EUDIOMETRY** * 100 ml of O₂, CO₂ and Ne are passed through NaOH and the volume is reduced to 70 ml. * This means the volume of CO₂ absorbed is 30 ml. ie. V_CO₂ = 30 ml. * Then the mixture is passed through alkaline pyrogallol and the volume is further reduced to 20 ml. * So the volume of O₂ must be 50 ml. ie. V_O₂ = 50 ml * The remaining 20 ml is Ne. **Eudiometry: (After Mole Concept)** 1 Cx Hy + (x + 1/4y) O₂ → x CO₂ + 1/2y H₂O 10 ml 10( x + 1/4y) ml 10x ml (10*1/2y) ml **QUES - 16 ml of hydrocarbon is exploded with excess of oxygen. The resulting gases show a volume contraction of 48 ml on treatment with KOH solution. When the gases were cooled to room temperature, a further volume contraction of 48 ml was observed. Find the molecular formula of the hydrocarbon.** * V_H₂O = 48 ml * V_KOH = V_CO₂ = 48 ml * 1 Cx Hy + (x + 1/4y) O₂ → x CO₂ + 1/2y H₂O * 16 ml 48 ml 48 ml * 1 Cx Hy → x CO₂ * 16 ml 48 ml * 1 x 4 = x x x * x = 3 * 1 mol Cx Hy → 1/2y mol H₂O * 16 ml 48 ml * 1 x 48 = 1/2y x 16 * y = 6 * Therefore, the molecular formula of the hydrocarbon is **C₃H₆**. **QUES- 100 ml of CH₄ and C₂H₂ were exploded with excess of O₂. After explosion and cooling, the mixture was treated with KOH where a reduction of 165 ml was observed. Therefore the composition of the mixture is -** * a) CH₄ = 35 ml; C₂H₂ = 65 ml * b) CH₄ = 65 ml; C₂H₂ = 35 ml * c) CH₄ = 75 ml; C₂H₂ = 25 ml * d) CH₄ = 25 ml; C₂H₂ = 75 ml * 1 CH₄ + 2 O₂ → 1 CO₂ + 2 H₂O * x ml x ml * 1 C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O * (100 - x) ml 2(100-x) ml * V_CH₄ = 35 ml * V_C₂H₂ = 100-35 = 65 ml * x + 2(100-x) = 165 * x + 200 - 2x = 165 * 35 = x * Therefore, the correct answer is **a) CH₄ = 35 ml; C₂H₂ = 65 ml** ## Virial Equation of State: **Ques. Given Z = 1 + B/V + C/V² + D/V³ + .....** * 2nd virial coefficient * 3rd virial coefficient * … **Binomial Theorem** **Virial equation of state** **Express B in terms of Vanderwaal's constant under normal condition** . * (P + a/V²) (V - b) = RT * Z = 1 - b/V - a/VRT * Z = (1 - b/V) - a/VRT * Z = [( 1 - b/V)⁻¹] - a/VRT * b/V << 1 * Z = (1 - b/V)⁻¹ - a/VRT * Z = (1 + b/V + (b/V)² + (b/V)³ + ....) - a/VRT * Z = 1 + (b - a/RT)(1/V) + (b²/V²) + (b³/V³) +.... * Z = 1 + B /V + ..... * Under normal condition, b - a / RT = B * **Therefore, B = b - a / RT** * Z = 1 (Real gas behaves ideally) * b - a / RT = 0 * b = a / RT * **T_B = a / Rb = Boyle's Temperature** ## Inversion Temperature: * Temperature at which real gas on expansion neither show cooling nor heating is called inversion temperature (T₁) * T_i = 2a / Rb = 2T_B ## Equation of Law of Corresponding State: * T_c = 8a / 27Rb * P_c = a / 27b² * V_c = 3b * T/T_c - Reduced Temperature * P/P_c - Reduced Pressure * V/V_c - Reduced Volume (P + a / V²) (V - b) = RT * Put T=θT_C * P=πP_C * V= φV_C * [πP_C + a / (φV_C)²] [φV_C - b] = R θT_C * [πa/27b² + a / φ²(3b)²] [φ3b - b] = Rθ8a/ 27Rb * φ/27b² [π + 3 / φ²] [φ3b - b] = 8a/27b * **(π + 3/φ²) (3φ - 1) = 8/φ** ## Law of Corresponding State: * Two gases having same reduced pressure and reduced volume would also have same reduced temperature. ## Collision Theory: * Gas particles are considered as solid spheres. * N = (N*/V) = No. of particles present per unit volume * **V_av** = Average Velocity * Distance covered by the particle in 1 sec. * Area of cross-section = πσ² * Volume swept by the particle in 1 sec = (Area of cross-section) (Distance) = (πσ²) (V_av) * N*(N/V) = Number density * No.of collision suffered by a particle in 1 sec = (πσ²) (V_av) x N* / V_o = No. density = (Statisical Thermo-D) * No.of collisions suffered by a particle in 1 sec = √2 [(πσ²) (V_av) x N*] = Collision Number (Z₁) **Collision Number:** * Number of collisions suffered by one molecule in 1 second. * Z₁ = √2(πσ²) (V_av) (N*) * σ = Radius of gas molecule * V_av = Average Velocity * N* = No. of collision per unit volume **Collision Frequency:** * Default - Binary collision * Ternary collision * We will assume binary collisions only. * Z₁ = √2 (πσ²) (V_av) N* * Z₁ x N* = √2πσ² (V_av x N*) x N* / 2 = Z₂ or Z₁ **Collision Frequency** * Total number of collisions suffered by all molecules per unit time per unit volume * Z₂ = √2 (πσ²) (V_av) N*²/2 = (πσ² (V_av)² (N*² / √2 * (Z₁ x N*)/2 **Factors affecting Collision frequency** * Z₂ = √2 πσ² V_av N*²/2 * σ = Radius of gas molecule * Dependent on size * Dependent on 'b' * For a given gas - Z₂ ∝ V_av N*² * V_av ∝ √(RT/M ) * V_av ∝ √T * PV = nRT * PV/N_A = RT * PNA / (N/V) = N* * N* ∝ P/T * Z₂ ∝ P/T^3/2 * At constant pressure, Z₂ decreases with increasing temperature. * Z₂ ∝ T^1/2 (1/V)² * Z₂ ∝ 1/V² * PV=nRT * V = RT/n * V = RT/P * V ∝ 1/P * 1/V ∝ (P/T) * At constant volume, Z₂ increases with increasing temperature. ## Mean Free Path(λ): * 1 sec = Distance = V_av * λ = l₁ + l₂ + l₃ / 3 **Mean Free Path:** * Average distance travelled by gas molecules between 2 successive collisions. * λ = Distance travelled by the particle in 1 sec/No. of collisions per second by 1 particle = V_av/Z₁ * λ = V_av / √2πσ²N* = 1/ √2πσ²N* = CTM **Factors affecting mean free path** * λ = 1/ √2πσ²N* = dependent on size(b) * λ ∝ 1 / b * PV= N_ART * PNA / (N/V)= N* * λ ∝ 1 / (P / T) * λ ∝ (T / P) * λ ∝ V **Summary** * Z₁ = √2(πσ²) V_av x N* * Z₂ = √2(πσ²) (V_av) x N*²/ 2 * Z₂ ∝ P²/T^3/2 * Z₂ ∝ 1/V² * λ = 1/√2πσ²N* * λ ∝ 1/b * λ ∝ (T/P) * λ ∝ V * σ = Radius of molecule * V_av = Average velocity * N* = NO. of molecules per unit volume # Q. * **D.C. = C_av = √(8RT/πM) = C_av ∝ √T** * The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is... * **D.C ∝ (λC_av) ∝ (T/P)** * T → 4T * P → 2P * D.C → x(D.C) * D.C' ∝ ( (4T)^3/2/2P ) * D.C' ∝ (8T^3/2/ 2P) * D.C' ∝ 4^3/2 T^3/2/ P * D.C' ∝ 4 D.C * **Therefore, x = 4** **THANK YOU**