States of Matter 11 Class Notes - Arjuna JEE 2025 PDF
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2025
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Nikhil Saini Sir
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These notes present a lecture on states of matter, likely for JEE 2025 preparation. Topics covered include eudiometry, collision theory, and virial equations. Example problems and calculations are included, focusing on the different methods and equations related to these topics.
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# 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 | |---|---| | CO<sub>2</sub>, SO<sub>2</sub> (acidi...
# 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 | |---|---| | CO<sub>2</sub>, SO<sub>2</sub> (acidic gas) | Alkali = NaOH, KOH | | O<sub>2</sub> | Alkaline Pyrogallol | | O<sub>3</sub> | Turpentine Oil | | CO, C<sub>2</sub>H<sub>2</sub>, NH<sub>3</sub> | Ammonical Cu<sub>2</sub>Cl<sub>2</sub>, CuSO<sub>4</sub> solution in acidic solution | | N<sub>2</sub> | Heated Mg or sometimes inert | | NO | FeSO<sub>4</sub> solution | | H<sub>2</sub>O | Neglected (Liquid) | ### Eudiometry example * A mixture of 100 ml of O<sub>2</sub>, CO<sub>2</sub> and Ne is treated with NaOH, resulting in a decrease of 30 ml. * This indicates that the volume of CO<sub>2</sub> was 30 ml. * The remaining 70 ml of O<sub>2</sub> and Ne is then treated with alkaline pyrogallol, resulting in a decrease of 50 ml. * This indicates that the volume of O<sub>2</sub> was 50 ml and the remaining 20 ml is Ne. ## Eudiometry (After Mole Concept) - 1 CxHy + (x + y/4) O2 -> xCO<sub>2</sub> + (y/2) H<sub>2</sub>O - 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) O<sub>2</sub>-> xCO<sub>2</sub> + (y/2) H<sub>2</sub>O - 16 ml -> 48 ml -> 48 ml - 1 CxHy -> xCO<sub>2</sub> - 16 ml -> 48 ml - x = 3 - 1 mol CxHy -> (y/2) mol H<sub>2</sub>O - 16 ml -> 48 ml - y = 6 - **C<sub>3</sub>H<sub>6</sub>** ## 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 z<sup>th</sup> virial coefficient Z = 1 + B/V + C/V<sup>2</sup> + D/V<sup>3</sup> + ----- * **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]<sup>-1</sup> - a/VRT * b/V << 1 * (1 - x)<sup>-1</sup> = (1 + x + x<sup>2</sup> + x<sup>3</sup> + -----) * x << 1 Z = [1 + b/V + b<sup>2</sup>/V<sup>2</sup> + b<sup>3</sup>/V<sup>3</sup> + -----] - a/VRT Z = 1 + (b - a/RT)/V + b<sup>2</sup>/V<sup>2</sup> + b<sup>3</sup>/V<sup>3</sup> + ----- * 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 * T<sub>B</sub> = a/Rb ## Boyle's Temperature * Temperature at which real gas behaves as ideal gas. * T<sub>B</sub> = a/Rb ## Inversion Temperature * Temperature at which real gas on expansion neither shows cooling nor heating is called inversion temperature (T<sub>i</sub>) * T<sub>I</sub> = 2a/Rb = 2T<sub>B</sub> ## Equation of Law of Corresponding State: * T<sub>C</sub>: 8a/27Rb * P<sub>C</sub>: a/27b<sup>2</sup> * V<sub>C</sub>: 3b * T/T<sub>C</sub> = θ * P/P<sub>C</sub> = Π * V/V<sub>C</sub> = Φ * Reduced Temperature * Reduced Pressure * Reduced Volume *(P + a/V<sup>2</sup>)(V - b) = RT* * Put T = θT<sub>C</sub>, P = ΠP<sub>C</sub>, V = ΦV<sub>C</sub> [(ΠP<sub>C</sub> + a/Φ<sup>2</sup>V<sub>C</sub><sup>2</sup>)][(ΦV<sub>C</sub> - b)] = RθT<sub>C</sub> [(Πa/27b<sup>2</sup> + a/9b<sup>2</sup>)][(3Φb - b)] = Rθ8a/27Rb [Π/27b<sup>2</sup> + 1/9b<sup>2</sup>][(3Φb - b)] = 8θ/27Rb [(Π + 3/Φ<sup>2</sup>)[(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 = (πσ<sup>2</sup>)" * "Volume swept by particle in 1 sec = (πσ<sup>2</sup>)(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 -> (πσ<sup>2</sup>)V<sub>a</sub><sub>v</sub> x N* * No. of collisions suffered by particle in 1 sec -> 2[(πσ<sup>2</sup>)V<sub>a</sub><sub>v</sub> x N*] = Collision Number ## Collision Number * Number of collisions suffered by one molecule in 1 second. * Z<sub>1</sub> = √2(πσ<sup>2</sup>)(V<sub>a</sub><sub>v</sub>)(N*) * σ - Radius of gas molecule * V<sub>a</sub><sub>v</sub> - 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 "Z<sub>1</sub>" * 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: * Z<sub>1</sub> = √2(πσ<sup>2</sup>)(V<sub>a</sub><sub>v</sub>)N* * 1C -> 1C * 2C -> 1C (x) * "collision frequency" * Z<sub>1</sub> x N*/2 = [√2(πσ<sup>2</sup>)V<sub>a</sub><sub>v</sub> x N*N*/2 = Z<sub>1</sub><sup>2</sup>/2 ## Collision Frequency continued * Total number of collisions suffered by all molecules per unit time per unit volume. * Z<sub>2</sub> = √2(πσ<sup>2</sup>)(V<sub>a</sub><sub>v</sub>)N*<sup>2</sup>/2 = πσ<sup>2</sup>V<sub>a</sub><sub>v</sub><sup>2</sup>N*<sup>2</sup>/√2 * (Z<sub>1</sub> x N*)/2 ## Factors affecting collision frequency * Z<sub>2</sub> = √2πσ<sup>2</sup>V<sub>a</sub><sub>v</sub>N*<sup>2</sup>/2 * σ - Radius of gas molecule * Dependent on size * Dependent on "b" * For a given gas * Z<sub>2</sub> ∝ V<sub>a</sub><sub>v</sub>N*<sup>2</sup> ## Factors affecting collision frequency continued * V<sub>a</sub><sub>v</sub> = √8RT/πM * V<sub>a</sub><sub>v</sub> ∝ √T * Z<sub>2</sub> ∝ T*<sup>3/2</sup>(P/V)<sup>2</sup> * Z<sub>2</sub> ∝ P<sup>2</sup>/T<sup>3/2</sup> * Constant Pressure * Z<sub>2</sub> ∝ 1/T<sup>3/2</sup> ## Factors affecting collision frequency continued * PV = nRT * V = RT/nP * Z<sub>2</sub> ∝ T*<sup>1/2</sup>(1/V)<sup>2</sup> * Z<sub>2</sub> ∝ 1/V<sup>2</sup> * Z<sub>2</sub> ∝ P<sup>2</sup>/T<sup>2</sup> * V ∝ 1/P * 1/V ∝ P/T ## Factors affecting collision frequency continued * At constant pressure * Z<sub>2</sub> ∝ 1/T<sup>3/2</sup> * Z<sub>2</sub> ∝ 1/T<sup>3/2</sup> * At constant volume * Z<sub>2</sub> ∝ (T)<sup>3/2</sup> * **At constant volume, Z<sub>2</sub> increases with increasing temperature.** ## Mean Free Path * 1 sec = Distance -> V<sub>a</sub><sub>v</sub> * λ = l<sub>1</sub> + l<sub>2</sub> + l<sub>3</sub>/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/Z<sub>1</sub> * λ = V<sub>a</sub><sub>v</sub>/Z<sub>1</sub> * λ = 1/√2πσ<sup>2</sup>N* ## Factors affecting mean free path * λ = 1/√2πσ<sup>2</sup>N* * Dependent on size (b) * λ ∝ 1/b * λ<sub>1</sub> -> ## Factors affecting mean free path continued * λ = 1/√2πσ<sup>2</sup>N* * 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 * Z<sub>1</sub> = √2(πσ<sup>2</sup>)V<sub>a</sub><sub>v</sub> x N* * Z<sub>2</sub> = √2(πσ<sup>2</sup>)(V<sub>a</sub><sub>v</sub>) x N*<sup>2</sup>/2 * Z<sub>2</sub> ∝ P<sup>2</sup>/T<sup>3/2</sup> * Z<sub>2</sub> ∝ 1/V<sup>2</sup> * λ = 1/√2πσ<sup>2</sup>N* * λ ∝ 1/b * λ ∝ (T/P) * λ ∝ V * σ - Radius of molecule * V<sub>a</sub><sub>v</sub> - 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 C<sub>a</sub><sub>v</sub>) * C<sub>a</sub><sub>v</sub> = √8RT/πM * C<sub>a</sub><sub>v</sub> ∝ √T * T -> 4T * P -> 2P * DC -> x(DC) * DC ∝ (T<sup>1/2</sup> x T) * DC ∝ (T<sup>3/2</sup>/P) * DC' ∝ ((4T)<sup>3/2</sup>/2P) * DC' ∝ 8T<sup>3/2</sup>/2P * DC' ∝ 4T<sup>3/2</sup>/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.
