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Pressure, Winds, Storms, and Cyclones - Pressure Exerted by Air

Grade 8CBSE

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

🔑Concepts

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Air around us exerts pressure in all directions. This is demonstrated when a tin can filled with boiling water and sealed collapses when cold water is poured over it, because the air pressure inside becomes lower than the pressure outside.

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Atmospheric pressure is the force per unit area exerted against a surface by the weight of the air above that surface. It is measured using the formula P=FAP = \frac{F}{A}.

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High-speed winds are always accompanied by reduced air pressure. This is why roofs of huts or warehouses can be blown off during high-speed winds if they are weak; the fast wind creates low pressure above the roof, while the pressure inside remains higher.

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Air expands on heating and occupies more space. When the same volume of air occupies more space, it becomes less dense and therefore lighter. This is why warm air rises up and creates a low-pressure area.

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

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Air pressure decreases with an increase in altitude because the column of air above becomes shorter and less dense as we move higher into the atmosphere.

📐Formulae

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

1 Pascal (Pa)=1 N/m21 \text{ Pascal (Pa)} = 1 \text{ N/m}^2

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

💡Examples

Problem 1:

A force of 200 N200 \text{ N} is applied over an area of 5 m25 \text{ m}^2 by the wind hitting a wall. Calculate the pressure exerted by the wind.

Solution:

P=200 N5 m2=40 PaP = \frac{200 \text{ N}}{5 \text{ m}^2} = 40 \text{ Pa}

Explanation:

Pressure is defined as force per unit area. By dividing the total force (200 N200 \text{ N}) by the area (5 m25 \text{ m}^2), we find that the wind exerts a pressure of 4040 Pascals.

Problem 2:

During a cyclone, the atmospheric pressure in a coastal town dropped from a normal of 101325 Pa101325 \text{ Pa} to 98500 Pa98500 \text{ Pa}. Calculate the drop in pressure.

Solution:

101325−985002825\begin{array}{r} 101325 \\ -98500 \\ \hline 2825 \end{array} The pressure drop is 2825 Pa2825 \text{ Pa}.

Explanation:

To find the change in pressure, we perform vertical subtraction of the storm pressure from the standard atmospheric pressure.

Problem 3:

Why does a balloon expand when air is blown into it?

Solution:

Air molecules inside the balloon collide with the inner walls, exerting pressure (PP). As more air is added, the number of collisions increases, increasing the internal pressure until the balloon expands.

Explanation:

This demonstrates that air exerts pressure in all directions on the container it occupies.