krit.club logo

Particulate Nature of Matter - What Decides Different States of Matter?

Grade 8CBSE

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

🔑Concepts

•

Matter is made up of tiny particles which are in continuous motion. The state of matter (Solid, Liquid, or Gas) is decided by the competition between two factors: Intermolecular Force of Attraction and Kinetic Energy of particles.

•

Intermolecular Force of Attraction (FattractionF_{attraction}): This force tends to keep the particles together. In solids, this force is strongest, while in gases, it is weakest.

•

Intermolecular Space (SS): This is the empty space between particles. It is minimum in solids (SsolidS_{solid}), moderate in liquids (SliquidS_{liquid}), and maximum in gases (SgasS_{gas}).

•

Kinetic Energy (KEKE): Particles possess energy due to their motion. KEKE is directly proportional to the temperature (TT). As temperature increases, KEKE increases, allowing particles to overcome attractive forces.

•

Solids: Particles are tightly packed because FattractionF_{attraction} is very high and KEKE is low. Particles only vibrate about their fixed positions.

•

Liquids: Particles have enough KEKE to move around each other but not enough to escape the bulk. FattractionF_{attraction} is intermediate.

•

Gases: Particles have very high KEKE and negligible FattractionF_{attraction}, allowing them to move randomly in all directions and occupy all available space.

•

Phase Change: By changing temperature or pressure, we can change the state. Increasing temperature increases KEKE, which leads to melting or vaporization.

📐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

KE∝T (where T is Absolute Temperature in Kelvin)KE \propto T \text{ (where } T \text{ is Absolute Temperature in Kelvin)}

Force of Attraction∝1Intermolecular Space\text{Force of Attraction} \propto \frac{1}{\text{Intermolecular Space}}

💡Examples

Problem 1:

A sample of gas is at 27∘C27^{\circ}C. Convert this temperature into the Kelvin scale.

Solution:

T(K)=27+273.15=300.15KT(K) = 27 + 273.15 = 300.15 K

Explanation:

To convert Celsius to Kelvin, we add 273.15273.15 to the Celsius value as per the SI unit standards.

Problem 2:

Compare the density of a substance in its solid and gaseous states if the mass remains constant at 100 g100\text{ g} but the volume increases from 20 cm320\text{ cm}^3 (solid) to 8000 cm38000\text{ cm}^3 (gas).

Solution:

Density of solid: ρsolid=10020=5 g/cm3\rho_{solid} = \frac{100}{20} = 5\text{ g/cm}^3 Density of gas: ρgas=1008000=0.0125 g/cm3\rho_{gas} = \frac{100}{8000} = 0.0125\text{ g/cm}^3

Explanation:

Since particles in gases are far apart (large intermolecular space), the volume is much higher, leading to a significantly lower density compared to solids.

Problem 3:

A substance undergoes a temperature change. If the initial temperature was 300K300 K and it was cooled by 50K50 K, what is the final temperature in Celsius?

Solution:

300−50250\begin{array}{r} 300 \\ - 50 \\ \hline 250 \end{array} Final Temperature in K=250KK = 250 K. In Celsius: t(∘C)=250−273.15=−23.15∘Ct(^{\circ}C) = 250 - 273.15 = -23.15^{\circ}C

Explanation:

We first find the final Kelvin temperature using subtraction and then convert it back to Celsius using the formula t(∘C)=T(K)−273.15t(^{\circ}C) = T(K) - 273.15.