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Pressure, Winds, Storms, and Cyclones - Formation of Wind

Grade 8CBSE

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

🔑Concepts

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Air exerts pressure on all objects and surfaces. This pressure is caused by the weight of the air molecules in the atmosphere acting on a unit area: P=FAP = \frac{F}{A}.

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Air expands on heating and becomes lighter (less dense). When air is heated, its volume (VV) increases, leading to a decrease in density (ρ\rho) because ρ=mV\rho = \frac{m}{V}. Consequently, warm air rises.

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The movement of air from a high-pressure region to a low-pressure region is called Wind. The speed of the wind depends on the magnitude of the pressure difference between the two regions.

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High-speed winds are accompanied by reduced air pressure. This phenomenon explains why roofs of huts can be blown off during high-velocity winds; the pressure above the roof decreases, while the pressure inside the house remains higher, creating an upward lift.

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Uneven heating of the Earth is the primary cause of wind. The Equator (around 0∘0^\circ latitude) receives more solar energy than the Poles (90∘90^\circ latitude). Warm air at the Equator rises, creating a low-pressure area, while cool air from the 0∘0^\circ to 30∘30^\circ latitude belt moves towards the Equator.

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Land and Sea Breezes: During the day, land heats up faster than water. Warm air over land rises, and cool air from the sea moves in (Sea Breeze). At night, land cools faster than water, and cool air moves from the land to the sea (Land Breeze).

📐Formulae

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

ρ=mV\rho = \frac{m}{V}

V∝T (at constant pressure)V \propto T \text{ (at constant pressure)}

ΔP∝Wind Speed\Delta P \propto \text{Wind Speed}

💡Examples

Problem 1:

Calculate the pressure exerted by a wind blowing with a force of 800 N800\text{ N} on a wall with an area of 20 m220\text{ m}^2.

Solution:

Given: Force (FF) = 800 N800\text{ N}, Area (AA) = 20 m220\text{ m}^2. Using the formula P=FAP = \frac{F}{A}, we get: 800÷2040\begin{array}{r} 800 \div 20 \\ \hline 40 \end{array} So, P=40 PaP = 40\text{ Pa}.

Explanation:

Pressure is the force acting per unit area. By dividing the total force of the wind by the surface area of the wall, we find the pressure in Pascals (Pa\text{Pa}).

Problem 2:

Why does a balloon expand when placed in the sun?

Solution:

As the air inside the balloon is heated by the sun, its temperature (TT) increases. According to the principle of expansion, the volume (VV) of the air increases (V∝TV \propto T).

Explanation:

Heating causes the air molecules to move faster and spread out, increasing the internal pressure against the balloon's walls, causing it to stretch and expand.

Problem 3:

Explain the direction of wind flow between a region at 1015 mb1015\text{ mb} (millibars) and a region at 990 mb990\text{ mb}.

Solution:

Wind flows from high pressure to low pressure. 1015 (High Pressure)−990 (Low Pressure)25 (Pressure Gradient)\begin{array}{r} 1015 \text{ (High Pressure)} \\ - 990 \text{ (Low Pressure)} \\ \hline 25 \text{ (Pressure Gradient)} \end{array} The wind will flow from the 1015 mb1015\text{ mb} region toward the 990 mb990\text{ mb} region.

Explanation:

The pressure difference (gradient) of 25 mb25\text{ mb} creates a force that pushes air molecules toward the area with fewer molecules (lower pressure).

Formation of Wind Class 8 Notes & Examples | CBSE Science