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Particulate Nature of Matter - Gaseous state

Grade 8CBSE

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

🔑Concepts

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In the gaseous state, particles are far apart from each other, meaning there are large intermolecular spaces between them.

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The force of attraction between the particles is negligible, allowing them to move freely in all directions.

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Gaseous particles possess high kinetic energy (KEKE), causing them to move at high speeds.

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Gases have neither a definite shape nor a definite volume; they acquire the shape and volume of the container they are kept in.

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Gases are highly compressible. For example, Compressed Natural Gas (CNGCNG) is used in vehicles, and Liquefied Petroleum Gas (LPGLPG) is used for cooking.

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The rate of diffusion is highest in gases because of the high speed of particles and large spaces between them. This is why the smell of hot food reaches us even from a distance.

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Gaseous particles exert pressure on the walls of the container due to the continuous bombardment of particles against the surface.

📐Formulae

Density (ρ)=Mass (m)Volume (V)\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}

T(K)=T(∘C)+273.15T(K) = T(^\circ C) + 273.15

Pressure (P)=Force (F)Area (A)\text{Pressure } (P) = \frac{\text{Force } (F)}{\text{Area } (A)}

KE=12mv2KE = \frac{1}{2}mv^2

💡Examples

Problem 1:

Convert the temperature of a gas from 35∘C35^\circ C to the Kelvin scale.

Solution:

T(K)=35+273.15=308.15KT(K) = 35 + 273.15 = 308.15 K

Explanation:

To convert Celsius to Kelvin, we add 273.15273.15 to the Celsius value as per the standard temperature conversion formula.

Problem 2:

A sample of gas has a mass of 0.5kg0.5 kg and occupies a volume of 2m32 m^3. Calculate its density.

Solution:

ρ=0.52=0.25kg/m3\rho = \frac{0.5}{2} = 0.25 kg/m^3

Explanation:

Density is defined as mass per unit volume. Using the formula ρ=mV\rho = \frac{m}{V}, we divide the mass by the volume.

Problem 3:

Calculate the pressure exerted by a gas if it applies a force of 500N500 N over an internal container wall area of 2m22 m^2.

Solution:

P=5002=250N/m2 (or Pascal)P = \frac{500}{2} = 250 N/m^2 \text{ (or Pascal)}

Explanation:

Pressure is the force acting perpendicularly per unit area. Using P=FAP = \frac{F}{A}, we find the pressure in PaPa.