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Matter in Our Surroundings - Can Matter Change its State?

Grade 9CBSE

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

🔑Concepts

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Matter can change its state from solid to liquid and from liquid to gas by changing temperature or pressure conditions.

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The temperature at which a solid melts to become a liquid at atmospheric pressure is called its Melting Point. For ice, this is 273.15 K273.15 \text{ K} or 0∘C0^\circ\text{C}.

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Latent Heat of Fusion is the amount of heat energy required to change 1 kg1 \text{ kg} of a solid into a liquid at its melting point at atmospheric pressure without a change in temperature.

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The Boiling Point is the temperature at which a liquid starts boiling at atmospheric pressure. It is a bulk phenomenon. For water, it is 373.15 K373.15 \text{ K} (100∘C100^\circ\text{C}).

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Latent Heat of Vaporization is the heat energy required to change 1 kg1 \text{ kg} of a liquid to gas at atmospheric pressure at its boiling point.

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Sublimation is the change of state directly from solid to gas without changing into liquid state (e.g., Ammonium Chloride NH4ClNH_4Cl or Camphor). Deposition is the direct change from gas to solid.

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Applying pressure and reducing temperature can liquefy gases. Solid carbon dioxide (CO2CO_2) is called 'Dry Ice' because it converts directly into gaseous state on decrease of pressure to 1 atmosphere1 \text{ atmosphere} without coming into liquid state.

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Evaporation is a surface phenomenon where a liquid changes into vapor at any temperature below its boiling point. Factors affecting evaporation include surface area, temperature, humidity, and wind speed.

📐Formulae

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

Q=m⋅Lf (where Lf is Latent Heat of Fusion)Q = m \cdot L_f \text{ (where } L_f \text{ is Latent Heat of Fusion)} Rose

Q=m⋅Lv (where Lv is Latent Heat of Vaporization)Q = m \cdot L_v \text{ (where } L_v \text{ is Latent Heat of Vaporization)}

💡Examples

Problem 1:

Convert the temperature of 25∘C25^\circ\text{C} to the Kelvin scale.

Solution:

T(K)=25+273=298 KT(K) = 25 + 273 = 298 \text{ K}

Explanation:

To convert Celsius to Kelvin, we use the relation T(K)=T(∘C)+273T(K) = T(^\circ\text{C}) + 273.

Problem 2:

Why does ice at 273 K273 \text{ K} cause more cooling than water at the same temperature?

Solution:

Ice at 273 K273 \text{ K} absorbs more heat (LfL_f) from the surroundings to overcome the forces of attraction.

Explanation:

Water at 273 K273 \text{ K} possesses additional energy in the form of Latent Heat of Fusion (LfL_f) compared to ice at the same temperature. Therefore, ice is more effective in cooling as it absorbs this extra energy from the medium.

Problem 3:

Suggest a method to liquefy atmospheric gases.

Solution:

By applying high pressure and reducing the temperature (T↓T \downarrow, P↑P \uparrow).

Explanation:

Increasing pressure brings the gas particles closer together, and decreasing temperature slows down their kinetic energy, allowing intermolecular forces to bind them into a liquid state.