States of Matter 11 Class Notes - Arjuna JEE 2025 PDF

# ARJUNA JEE 2025 ## States Of Matter * **Chemistry** * **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 | | O3 | Turpentine Oil | | CO, C2H2, NH3 | Ammonical Cu2Cl2, CuSO4 solution in acidic solution | | N2 | Heated Mg or sometimes inert | | NO | FeSO4 solution | | H2O | Neglected (Liquid) | ### Eudiometry example * A mixture of 100 ml of O2, CO2 and Ne is treated with NaOH, resulting in a decrease of 30 ml. * This indicates that the volume of CO2 was 30 ml. * The remaining 70 ml of O2 and Ne is then treated with alkaline pyrogallol, resulting in a decrease of 50 ml. * This indicates that the volume of O2 was 50 ml and the remaining 20 ml is Ne. ## Eudiometry (After Mole Concept) - 1 CxHy + (x + y/4) O2 -> xCO2 + (y/2) H2O - 10 ml -> 10(x + y/4) ml -> 10x ml -> (10 x y/2) 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. - 1 CxHy + (x + y/4) O2-> xCO2 + (y/2) H2O - 16 ml -> 48 ml -> 48 ml - 1 CxHy -> xCO2 - 16 ml -> 48 ml - x = 3 - 1 mol CxHy -> (y/2) mol H2O - 16 ml -> 48 ml - y = 6 - **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 -> 2CO2 + H2O - (100 -x) ml -> 2(100 - x) ml - x + 2(100 - x) = 165 - x = 35 - VCH4 = 35 ml - VC2H2 = 100 - 35 = 65 ml - **a) CH4 = 35 ml; C2H2 = 65 ml** ## Virial Equation of State Ques. Given zth virial coefficient Z = 1 + B/V + C/V2 + D/V3 + ----- * **Virial equation of state** * **Express B in terms of Vanderwaal's constant under normal conditions** (P + a/V^2)(V - b) = RT Z = 1 - a/VRT - b/V Z = [1 - b/V] - a/VRT Z = [1 - b/V]-1 - a/VRT * b/V << 1 * (1 - x)-1 = (1 + x + x2 + x3 + -----) * x << 1 Z = [1 + b/V + b2/V2 + b3/V3 + -----] - a/VRT Z = 1 + (b - a/RT)/V + b2/V2 + b3/V3 + ----- * Equating * b - a/RT = B * Under normal conditions * Z = 1 + B/V + ----- Z = 1 + [(b - a/RT)/V] * Real gas behaves ideally * Z = 1 b - a/RT = 0 b = a/RT * Boyle's Temperature * TB = a/Rb ## Boyle's Temperature * Temperature at which real gas behaves as ideal gas. * TB = a/Rb ## Inversion Temperature * Temperature at which real gas on expansion neither shows cooling nor heating is called inversion temperature (Ti) * TI = 2a/Rb = 2TB ## Equation of Law of Corresponding State: * TC: 8a/27Rb * PC: a/27b2 * VC: 3b * T/TC = θ * P/PC = Π * V/VC = Φ * Reduced Temperature * Reduced Pressure * Reduced Volume *(P + a/V2)(V - b) = RT* * Put T = θTC, P = ΠPC, V = ΦVC [(ΠPC + a/Φ2VC2)][(ΦVC - b)] = RθTC [(Πa/27b2 + a/9b2)][(3Φb - b)] = Rθ8a/27Rb [Π/27b2 + 1/9b2][(3Φb - b)] = 8θ/27Rb [(Π + 3/Φ2)[(3Φ - 1)] = 8θ * (Equation of State) * Π, Φ find * θ find ## Law of Corresponding State: * Two gases having the same reduced pressure and reduced volume would also have the same reduced temperature ## Collision Theory * Gas particles are considered as solid spheres. * N* = (N/V) = No. of particles present per unit volume ## Collision Theory diagram * A circle is drawn with a line through the center. The line represents the average velocity (Vav) of the gas particles. * The circle is labeled with * "Gas particles - spherical", * σ = "Radius of gas molecule", * Vav = "Average Velocity" * Arrows point away from the circle showing the direction of movement of the gas particle. * The following text is also present: * T = (1 sec) * "Distance covered by particle in 1 sec" * "Area of cross-section = (πσ2)" * "Volume swept by particle in 1 sec = (πσ2)(Vav)" * "N* / V = Number Density" ## Collision Theory diagram continuation * Two circles are drawn. * The circles represent gas molecules. * The first circle is labeled "Head-on Collision" and shows the particles travelling directly towards one another. * The second is labeled "Grazing collision" and shows the particles passing close to one another without colliding directly. * Arrows point away from the center of each circle indicating the direction of movement of each particle. ## Collision Theory continued * No. of collisions suffered by particle in 1 sec -> (πσ2)Vav x N* * No. of collisions suffered by particle in 1 sec -> 2[(πσ2)Vav x N*] = Collision Number ## Collision Number * Number of collisions suffered by one molecule in 1 second. * Z1 = √2(πσ2)(Vav)(N*) * σ - Radius of gas molecule * Vav - Average Velocity * N* - No. of collision per unit volume ## Collision Frequency * Default - Binary collision * Ternary collision * We will assume binary collisions only ## Collision frequency diagram * A circle with a black dot in the center is drawn inside a rectangle. The dot is traveling to the right and is captioned with "Z1" * Arrows surrounding the rectangle, point inwards, and are labeled with "N*" * The caption "Total no. of collisions in 1 sec" is written below the rectangle. * The following text is present: * Z1 = √2(πσ2)(Vav)N* * 1C -> 1C * 2C -> 1C (x) * "collision frequency" * Z1 x N*/2 = [√2(πσ2)Vav x N*N*/2 = Z12/2 ## Collision Frequency continued * Total number of collisions suffered by all molecules per unit time per unit volume. * Z2 = √2(πσ2)(Vav)N*2/2 = πσ2Vav2N*2/√2 * (Z1 x N*)/2 ## Factors affecting collision frequency * Z2 = √2πσ2VavN*2/2 * σ - Radius of gas molecule * Dependent on size * Dependent on "b" * For a given gas * Z2 ∝ VavN*2 ## Factors affecting collision frequency continued * Vav = √8RT/πM * Vav ∝ √T * Z2 ∝ T*3/2(P/V)2 * Z2 ∝ P2/T3/2 * Constant Pressure * Z2 ∝ 1/T3/2 ## Factors affecting collision frequency continued * PV = nRT * V = RT/nP * Z2 ∝ T*1/2(1/V)2 * Z2 ∝ 1/V2 * Z2 ∝ P2/T2 * V ∝ 1/P * 1/V ∝ P/T ## Factors affecting collision frequency continued * At constant pressure * Z2 ∝ 1/T3/2 * Z2 ∝ 1/T3/2 * At constant volume * Z2 ∝ (T)3/2 * **At constant volume, Z2 increases with increasing temperature.** ## Mean Free Path * 1 sec = Distance -> Vav * λ = l1 + l2 + l3/3 ## Mean Free Path continued * Average distance travelled by gas molecules between 2 successive collisions. * λ = Distance travelled by particle in 1 sec/No. of collisions per second by particle in 1 sec * λ = Distance travelled by particle in 1 sec/Z1 * λ = Vav/Z1 * λ = 1/√2πσ2N* ## Factors affecting mean free path * λ = 1/√2πσ2N* * Dependent on size (b) * λ ∝ 1/b * λ1 -> ## Factors affecting mean free path continued * λ = 1/√2πσ2N* * For a given gas * λ ∝ 1/N* * λ ∝ 1/P * λ ∝ (T/P) * λ ∝ V ## Factors affecting mean free path continued * PV = nRT * PNA/RT = (N/V) = N* ## Summary * Z1 = √2(πσ2)Vav x N* * Z2 = √2(πσ2)(Vav) x N*2/2 * Z2 ∝ P2/T3/2 * Z2 ∝ 1/V2 * λ = 1/√2πσ2N* * λ ∝ 1/b * λ ∝ (T/P) * λ ∝ V * σ - Radius of molecule * Vav - Average velocity * N* - no. of molecules per unit volume ## #Q. * 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. * What is the value of x? * DC α (λ x Cav) * Cav = √8RT/πM * Cav ∝ √T * T -> 4T * P -> 2P * DC -> x(DC) * DC ∝ (T1/2 x T) * DC ∝ (T3/2/P) * DC' ∝ ((4T)3/2/2P) * DC' ∝ 8T3/2/2P * DC' ∝ 4T3/2/P * DC' ∝ 4DC * x = 4 ## Thank You * A silhouette illustration of Arjuna shooting his bow and arrow is shown against an orange background. * The text "THANK YOU" is written in bold letters below the illustration. * The "P" logo from the PhysicsWallah website is present in the top right corner.

States of Matter 11 Class Notes - Arjuna JEE 2025 PDF

Document Details

Tags

Related

Summary

Full Transcript

Upgrade to continue