Problem 21 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Eight gram each of oxygen and hydrogen at \(27^{\circ} \mathrm{C}\) will have the total kinetic energy in the ratio of \(\ldots \ldots \ldots \ldots .\)

Step-by-Step Solution

Answer

The total kinetic energy of oxygen and hydrogen is in the ratio 1:16.

Step 1: Understanding the Kinetic Energy Formula

The kinetic energy of a gas can be described by the formula \( KE = \frac{3}{2} nRT \), where \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. We need to find the kinetic energy for both oxygen and hydrogen.

Step 2: Converting Temperature to Kelvin

We need to convert the given temperature from Celsius to Kelvin. \( T_{Kelvin} = 27 + 273 = 300 \ K \).

Step 3: Calculating Number of Moles for Each Gas

Find the number of moles for oxygen and hydrogen. Molar mass of oxygen \( O_2 = 32 \ g/mol \) and hydrogen \( H_2 = 2 \ g/mol \).For oxygen: \( n_{O_2} = \frac{8}{32} = 0.25 \ moles \)For hydrogen: \( n_{H_2} = \frac{8}{2} = 4 \ moles \).

Step 4: Calculating Kinetic Energy for Oxygen

Use the kinetic energy formula \( KE_{O_2} = \frac{3}{2} \times 0.25 \times R \times 300 = \frac{3}{2} \times 0.25R \times 300 \).

Step 5: Calculating Kinetic Energy for Hydrogen

Use the kinetic energy formula \( KE_{H_2} = \frac{3}{2} \times 4 \times R \times 300 = \frac{3}{2} \times 4R \times 300 \).

Step 6: Finding the Ratio of Kinetic Energies

The ratio \( \frac{KE_{O_2}}{KE_{H_2}} = \frac{0.25R \times 300}{4R \times 300} = \frac{0.25}{4} = \frac{1}{16} \). Thus, the ratio of the total kinetic energy between oxygen and hydrogen is \( 1:16 \).

Key Concepts

Gas Constant

In the world of chemistry and physics, the gas constant plays a critical role in many equations, especially those concerning gases. The gas constant, often denoted as \( R \), is a constant value that appears in the ideal gas law and is used to relate energy at the molecular level to temperature and moles. The universally accepted value of \( R \) is approximately \( 8.314 \ J/(mol \cdot K) \), which means that for each mole of a gas at a given temperature, this constant helps determine the energy present.
Understanding \( R \) is crucial, as it links the amount of substance (in moles), energy (in joules), and temperature (in Kelvin) together. This connection is what allows us to calculate the kinetic energy of gases, helping to predict their behavior under different conditions.

• The value of \( R \) is derived from experimental observations.
• \( R \) can be used with different units, but \( J/(mol \cdot K) \) is the most common in physical chemistry.

Number of Moles

Moles are a standard unit of measurement in chemistry that represents the amount of a substance. The number of moles is a measure based on Avogadro's number, which is roughly \( 6.022 \times 10^{23} \) entities (atoms, molecules, etc.). Understanding how to calculate the number of moles is vital for composing chemical equations and understanding their outcomes.
To find the number of moles for any substance, you use the formula:

• Number of Moles \( n = \frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}} \)

Let's consider our original example with oxygen and hydrogen:
- **Oxygen**: The molar mass is \( 32 \ g/mol \), and we have \( 8 \ g \). Thus, the moles are \( \frac{8}{32} = 0.25 \) moles.
- **Hydrogen**: With a molar mass of \( 2 \ g/mol \) and a mass of \( 8 \ g \), the moles are \( \frac{8}{2} = 4 \) moles.
This calculation helps us progress to understanding the kinetic energy, as the number of moles \( n \) is directly used in the formula \( KE = \frac{3}{2} nRT \).

Temperature Conversion

When dealing with kinetic energy and other physical properties of gases, temperature is a key factor. However, scientific calculations usually require the temperature to be in Kelvin rather than Celsius. This is because Kelvin begins at absolute zero, the theoretical lowest temperature possible, where all molecular movement ceases.
The conversion from Celsius to Kelvin is straightforward:

• \( T_{Kelvin} = T_{Celsius} + 273.15 \)

In the exercise, the given temperature is \( 27^{\circ} \mathrm{C} \). Using the formula, we convert this to Kelvin:

• \( 27 + 273.15 = 300.15 \ K \).

For ease in calculations, this is often rounded off to \( 300 \ K \). This conversion to Kelvin is crucial because many gas laws, including those involving kinetic energy, are based on absolute thermodynamic temperatures.

Molar Mass

Molar mass is fundamental when calculating moles, which subsequently assist in a wide range of chemical calculations. Molar mass is defined as the mass of one mole of a substance and is expressed in grams per mole \( g/mol \). Each element and compound has a distinct molar mass, which is calculated based on the atomic masses from the periodic table.

• Hydrogen’s molar mass is \( 2 \ g/mol \) for \( H_2 \), because each hydrogen atom has an atomic mass of approximately \( 1 \ g/mol \).
• Oxygen’s molar mass for \( O_2 \) is \( 32 \ g/mol \), as each oxygen atom has an atomic mass of approximately \( 16 \ g/mol \).

In the exercise, knowing the molar masses allowed us to compute the number of moles (\( n \)), which is a component of the kinetic energy formula. Understanding molar mass clarifies how much of a substance you are working with in terms of both its chemical composition and its contribution to properties like kinetic energy.