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Physics: Thermal Physics - Gas Pressure and Particle Motion

Grade 8IB

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

🔑Concepts

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Kinetic Theory of Gases: Gases consist of tiny particles that are in constant, random motion. The particles are far apart relative to their size, and the attractive forces between them are negligible.

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Gas Pressure: Pressure is created when gas particles collide with the internal surfaces of their container. Each collision exerts a small force. The sum of these forces over a specific area is the gas pressure: P=FAP = \frac{F}{A}.

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Temperature and Kinetic Energy: The temperature of a gas is a measure of the average kinetic energy (EkE_k) of its particles. As temperature increases, particles move faster and collide more frequently and forcefully with container walls.

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Boyle's Law: For a fixed mass of gas at a constant temperature, the pressure is inversely proportional to the volume. If volume decreases, the particles are more crowded, leading to more frequent collisions and higher pressure: P∝1VP \propto \frac{1}{V}.

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Pressure-Temperature Relationship (Gay-Lussac's Law): For a fixed mass of gas at a constant volume, the pressure is directly proportional to the absolute temperature (measured in Kelvin). If TT increases, PP increases: P∝TP \propto T.

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Brownian Motion: This refers to the random, erratic motion of larger visible particles (like smoke or pollen) suspended in a fluid, caused by continuous bombardment from invisible, fast-moving atoms or molecules.

📐Formulae

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

P1V1=P2V2P_1 V_1 = P_2 V_2

P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}

T(K)=T(∘C)+273T(K) = T(^{\circ}C) + 273

💡Examples

Problem 1:

A gas cylinder has a volume of 0.10 m30.10 \text{ m}^3 and contains gas at a pressure of 200,000 Pa200,000 \text{ Pa}. If the gas is transferred to a smaller cylinder with a volume of 0.04 m30.04 \text{ m}^3 while the temperature remains constant, calculate the new pressure.

Solution:

Using Boyle's Law: P1V1=P2V2P_1 V_1 = P_2 V_2 200,000×0.10=P2×0.04200,000 \times 0.10 = P_2 \times 0.04 20,000=0.04P220,000 = 0.04 P_2 P2=20,0000.04=500,000 PaP_2 = \frac{20,000}{0.04} = 500,000 \text{ Pa}

Explanation:

Since temperature is constant, we apply the inverse relationship between pressure and volume. Reducing the volume increases the collision frequency, thus increasing pressure.

Problem 2:

Explain why the pressure of a car tire increases after a long drive on a hot road.

Solution:

T↑  ⟹  Ek↑  ⟹  Collision Frequency/Force↑  ⟹  P↑T \uparrow \implies E_k \uparrow \implies \text{Collision Frequency/Force} \uparrow \implies P \uparrow

Explanation:

The friction between the tire and the road, along with the ambient heat, increases the temperature of the air inside the tire. The air particles gain kinetic energy and move faster, hitting the tire walls more often and with greater force, which increases the internal pressure.

Problem 3:

Calculate the pressure exerted by a gas if it applies a force of 1200 N1200 \text{ N} over an area of 3 m23 \text{ m}^2.

Solution:

P=FAP = \frac{F}{A} P=12003=400 PaP = \frac{1200}{3} = 400 \text{ Pa}

Explanation:

The pressure is found by dividing the total force exerted by the area on which the force acts. The unit is Pascals (PaPa), which is equivalent to N/m2N/m^2.