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Our Home: Earth, a Unique Life Sustaining Planet - What Allows Life to Be Sustained on Earth?

Grade 8CBSE

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

🔑Concepts

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The Goldilocks Zone: Earth is located at an ideal distance from the Sun, known as the 'Habitable Zone', where temperatures allow water to exist in liquid form (H2OH_2O).

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Atmospheric Composition: Earth's atmosphere contains a unique mix of gases including Nitrogen (78%78\%), Oxygen (21%21\%), and Carbon Dioxide (0.03%0.03\%), which are essential for life and maintaining temperature.

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The Greenhouse Effect: Greenhouse gases like CO2CO_2 and CH4CH_4 trap solar heat, keeping the Earth's average temperature at approximately 15∘C15^\circ C. Without this, the temperature would be about −18∘C-18^\circ C.

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Protection by Ozone Layer: The stratospheric ozone layer (O3O_3) absorbs harmful ultraviolet (UVUV) radiation from the sun, protecting living organisms from DNA damage.

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The Role of Liquid Water: Water acts as a universal solvent, facilitating biochemical reactions and regulating global temperatures through its high specific heat capacity.

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Magnetic Field: Earth's core generates a magnetic field that deflects the solar wind, preventing the atmosphere from being stripped away.

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Photosynthesis and Respiration: Plants convert solar energy into chemical energy (C6H12O6C_6H_{12}O_6), providing the base of the food chain and replenishing Oxygen (O2O_2).

📐Formulae

6CO2+6H2O→Sunlight/ChlorophyllC6H12O6+6O26CO_2 + 6H_2O \xrightarrow{\text{Sunlight/Chlorophyll}} C_6H_{12}O_6 + 6O_2

C6H12O6+6O2→6CO2+6H2O+Energy (ATP)C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy (ATP)}

TK=TC+273.15T_K = T_C + 273.15

Oxygen Percentage=Volume of O2Total Volume of Air×100\text{Oxygen Percentage} = \frac{\text{Volume of } O_2}{\text{Total Volume of Air}} \times 100

💡Examples

Problem 1:

If the average temperature of a specific region on Earth is 25∘C25^\circ C, what is this temperature in the Kelvin scale?

Solution:

Using the formula TK=TC+273.15T_K = T_C + 273.15, we substitute TC=25T_C = 25: TK=25+273.15=298.15 KT_K = 25 + 273.15 = 298.15\text{ K}

Explanation:

Scientific measurements of planetary temperatures are often expressed in Kelvin (KK) to provide an absolute scale for thermodynamic calculations.

Problem 2:

A sample of air has a total volume of 500 mL500\text{ mL}. If the volume of Oxygen in this sample is 105 mL105\text{ mL}, calculate the percentage of Oxygen. Does this match Earth's standard atmospheric composition?

Solution:

Percentage of O2=(105500)×100=21%O_2 = \left( \frac{105}{500} \right) \times 100 = 21\%.

Explanation:

The result is 21%21\%, which perfectly matches the standard percentage of Oxygen in Earth's atmosphere required to sustain aerobic life.

Problem 3:

Explain the chemical importance of the Ozone layer formation from Oxygen molecules.

Solution:

3O2→UV rays2O33O_2 \xrightarrow{UV \text{ rays}} 2O_3

Explanation:

High-energy UVUV radiation breaks O2O_2 molecules into atomic oxygen, which then reacts with other O2O_2 molecules to form Ozone (O3O_3). This layer is crucial for blocking lethal radiation.