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States of Matter - Kinetic particle theory

Grade 12A LevelChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Kinetic Particle Theory states that all matter is composed of tiny particles (atoms, molecules, or ions) that are in constant, random motion.

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In a solid, particles are packed closely in a regular lattice, vibrating about fixed positions with strong intermolecular forces of attraction.

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In a liquid, particles are close together but arranged irregularly; they have enough kinetic energy to slide over each other, allowing liquids to flow.

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In a gas, particles are far apart and move rapidly in random directions; intermolecular forces are negligible except during collisions.

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The average kinetic energy of particles is directly proportional to the absolute temperature in Kelvin (TT). As TT increases, the velocity (vv) of the particles increases.

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Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration. The rate of diffusion is inversely proportional to the square root of the relative molecular mass (MrM_r).

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Brownian Motion refers to the random movement of visible particles (like pollen or smoke) suspended in a fluid, caused by collisions with invisible, fast-moving atoms or molecules.

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An Ideal Gas obeys the gas laws under all conditions of temperature and pressure, assuming particles have zero volume and no intermolecular forces.

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Phase changes occur at constant temperature: Melting (Solid→LiquidSolid \rightarrow Liquid), Boiling (Liquid→GasLiquid \rightarrow Gas), Condensation (Gas→LiquidGas \rightarrow Liquid), and Freezing (Liquid→SolidLiquid \rightarrow Solid).

📐Formulae

T(K)=θ(∘C)+273.15T(K) = \theta(^{\circ}C) + 273.15

PV=nRTPV = nRT

P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

KEavg=32kTKE_{avg} = \frac{3}{2}kT

Rate of Diffusion∝1Mr\text{Rate of Diffusion} \propto \frac{1}{\sqrt{M_r}}

P=FAP = \frac{F}{A}

💡Examples

Problem 1:

Compare the rates of diffusion of ammonia (NH3NH_3) and hydrogen chloride (HClHCl) gases at the same temperature and pressure.

Solution:

Using Graham's Law: RateNH3RateHCl=Mr(HCl)Mr(NH3)\frac{\text{Rate}_{NH_3}}{\text{Rate}_{HCl}} = \sqrt{\frac{M_{r(HCl)}}{M_{r(NH_3)}}}. Given Mr(NH3)≈17M_{r(NH_3)} \approx 17 and Mr(HCl)≈36.5M_{r(HCl)} \approx 36.5, the ratio is 36.517≈1.46\sqrt{\frac{36.5}{17}} \approx 1.46.

Explanation:

Because NH3NH_3 has a lower relative molecular mass than HClHCl, its particles move faster on average, resulting in a higher rate of diffusion by a factor of approximately 1.461.46.

Problem 2:

A sample of gas occupies 2.0 dm32.0\text{ dm}^3 at 300 K300\text{ K} and 100 kPa100\text{ kPa}. Calculate the volume it will occupy if the temperature is increased to 600 K600\text{ K} at constant pressure.

Solution:

Using Charles's Law: V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}. Rearranging for V2V_2: V2=V1×T2T1=2.0×600300=4.0 dm3V_2 = V_1 \times \frac{T_2}{T_1} = 2.0 \times \frac{600}{300} = 4.0\text{ dm}^3.

Explanation:

At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. Doubling the temperature results in doubling the volume.

Problem 3:

Determine the number of moles of an ideal gas present in a 5.0 L5.0\text{ L} container at 2.0 atm2.0\text{ atm} and 27∘C27^{\circ}C. (Use R=0.0821 L⋅atm/mol⋅KR = 0.0821\text{ L}\cdot\text{atm}/\text{mol}\cdot\text{K})

Solution:

First, convert temperature to Kelvin: T=27+273=300 KT = 27 + 273 = 300\text{ K}. Use PV=nRT⇒n=PVRTPV = nRT \Rightarrow n = \frac{PV}{RT}. n=2.0×5.00.0821×300≈0.406 moln = \frac{2.0 \times 5.0}{0.0821 \times 300} \approx 0.406\text{ mol}.

Explanation:

The Ideal Gas Law relates pressure, volume, temperature, and moles. All units must be consistent, necessitating the conversion of Celsius to Kelvin.