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Thermal Physics - Kinetic particle model of matter

Grade 12IGCSEPhysics

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

🔑Concepts

The kinetic particle model states that all matter is made of tiny particles in constant, random motion. In solids, particles are closely packed and vibrate about fixed positions. In liquids, particles are close but can slide over each other. In gases, particles are far apart and move at high speeds in all directions.

Brownian motion is the random movement of microscopic particles (like smoke or pollen) suspended in a fluid, caused by collisions with much smaller, fast-moving molecules of the fluid. This provides evidence for the kinetic theory.

Temperature is proportional to the average kinetic energy (EkE_k) of the particles in a substance. As temperature increases, the average speed of the particles increases.

Gas pressure is exerted when gas particles collide with the walls of a container, exerting a force over an area (P=FAP = \frac{F}{A}). An increase in temperature at constant volume increases the pressure because particles hit the walls more frequently and with greater force.

Boyle's Law states that for a fixed mass of gas at a constant temperature, the pressure PP is inversely proportional to the volume VV. This is represented as P1VP \propto \frac{1}{V} or PV=constantPV = \text{constant}.

Specific Heat Capacity (cc) is the energy required to raise the temperature of 1 kg1\text{ kg} of a substance by 1 C1\text{ }^\circ\text{C} (or 1 K1\text{ K}).

Latent Heat (LL) is the energy required to change the state of 1 kg1\text{ kg} of a substance without a change in temperature. Specific latent heat of fusion (LfL_f) applies to melting/freezing, while specific latent heat of vaporization (LvL_v) applies to boiling/condensing.

Evaporation is a surface phenomenon where the most energetic particles escape the liquid's surface, lowering the average kinetic energy of the remaining particles and resulting in a cooling effect.

📐Formulae

P1V1=P2V2P_1V_1 = P_2V_2

E=mcΔθE = mc\Delta\theta

E=mLfE = mL_f

E=mLvE = mL_v

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

💡Examples

Problem 1:

A gas occupies a volume of 2.0 m32.0\text{ m}^3 at a pressure of 1.0×105 Pa1.0 \times 10^5\text{ Pa}. If the volume is compressed to 0.5 m30.5\text{ m}^3 at a constant temperature, what is the new pressure?

Solution:

Using Boyle's Law: P1V1=P2V2P_1V_1 = P_2V_2. Given P1=1.0×105 PaP_1 = 1.0 \times 10^5\text{ Pa}, V1=2.0 m3V_1 = 2.0\text{ m}^3, and V2=0.5 m3V_2 = 0.5\text{ m}^3. P2=P1V1V2=(1.0×105×2.0)0.5=4.0×105 PaP_2 = \frac{P_1V_1}{V_2} = \frac{(1.0 \times 10^5 \times 2.0)}{0.5} = 4.0 \times 10^5\text{ Pa}

Explanation:

Since the volume decreased by a factor of 4, the pressure must increase by a factor of 4 to maintain the constant PVPV relationship.

Problem 2:

How much energy is required to heat 0.5 kg0.5\text{ kg} of water from 20 C20\text{ }^\circ\text{C} to 100 C100\text{ }^\circ\text{C}? (Specific heat capacity of water c=4200 J/(kg C)c = 4200\text{ J/(kg }^\circ\text{C)}).

Solution:

Using E=mcΔθE = mc\Delta\theta, where m=0.5 kgm = 0.5\text{ kg}, c=4200 J/(kg C)c = 4200\text{ J/(kg }^\circ\text{C)}, and Δθ=10020=80 C\Delta\theta = 100 - 20 = 80\text{ }^\circ\text{C}. E=0.5×4200×80=168,000 J (or 168 kJ)E = 0.5 \times 4200 \times 80 = 168,000\text{ J (or } 168\text{ kJ)}

Explanation:

The formula calculates the total thermal energy transferred based on the mass, the material's specific property, and the change in temperature.

Kinetic particle model of matter - Revision Notes & Key Formulas | IGCSE Grade 12 Physics