Problem 35 - A Complete Resource Book in Chemistry for JEE Main

Question

A gas has double the average velocity of \(\mathrm{SO}_{2}\) gas at any temperature. The gas may be (a) \(\mathrm{CO}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{CH}_{4}\) (d) \(\mathrm{O}_{3}\)

Step-by-Step Solution

Answer

The gas is \( \mathrm{CH}_{4} \).

Step 1: Understanding Root Mean Square Velocity

The root mean square velocity of a gas is given by the formula \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas. If a gas has double the velocity of another, then its \( v_{rms} \) is twice that of this other gas.

Step 2: Calculate the RMS Velocity Ratio

For two gases, if the gas in question has double the root mean square velocity of \( \mathrm{SO}_{2} \), then:\[ v_{rms, \text{gas}} = 2 \times v_{rms, \mathrm{SO}_2} \]Substituting in the formula, this becomes:\[ \sqrt{\frac{3RT}{M_{\text{gas}}}} = 2 \times \sqrt{\frac{3RT}{M_{\mathrm{SO}_2}}} \]

Step 3: Simplify the Equation

We can square both sides of the equation to eliminate the square root:\[ \frac{3RT}{M_{\text{gas}}} = 4 \times \frac{3RT}{M_{\mathrm{SO}_2}} \]Cancel \( 3RT \) from both sides:\[ \frac{1}{M_{\text{gas}}} = \frac{4}{M_{\mathrm{SO}_2}} \]Therefore, \( M_{\text{gas}} = \frac{1}{4} \times M_{\mathrm{SO}_2} \).

Step 4: Calculate the Molar Mass of Sulfur Dioxide

The molar mass of \( \mathrm{SO}_{2} \) is calculated as follows:Sulfur (S): \( 32 \) g/molOxygen (O): \( 16 \) g/mol\( M_{\mathrm{SO}_2} = 32 + (2 \times 16) = 64 \) g/mol.

Step 5: Solve for the Molar Mass of the Gas

Since \( M_{\text{gas}} = \frac{1}{4} \times 64 \),\( M_{\text{gas}} = 16 \) g/mol.

Step 6: Identify the Gas

Compare the calculated molar mass (16 g/mol) with the provided gas options:(a) \( \mathrm{CO}_{2} \): \( 44 \) g/mol(b) \( \mathrm{C}_{2} \mathrm{H}_{4} \): \( 28 \) g/mol(c) \( \mathrm{CH}_{4} \): \( 16 \) g/mol(d) \( \mathrm{O}_{3} \): \( 48 \) g/molThe gas with a molar mass of 16 g/mol is \( \mathrm{CH}_{4} \).

Key Concepts

Gas Molar Mass

The molar mass of a gas is an essential property that helps us understand how a gas behaves under different conditions. It represents the mass of one mole of any given gas. Knowing the molar mass is crucial for various calculations, especially when dealing with gas laws and their applications.
When you know the molar mass (M) of a gas, you can relate it to its density, its behavior during reactions, and its velocity at a given temperature.
For instance, if we consider different gases, the molecular weight directly influences key characteristics such as:

• Molecular speeds: Lighter gases tend to move faster.
• Diffusion rates: Gases with lower molar masses diffuse more quickly than heavier ones.
• Pressure exerted by gases: This is always related to the mass and speed of gas particles.

Thus, understanding the molar mass is not only about knowing the composition but also about predicting how a gas will interact with its environment.

Velocity Comparison

When comparing the velocities of gases, it's crucial to understand the concept of root mean square (RMS) velocity. RMS velocity helps us measure the average speed of gas molecules in a sample. For different gases, despite having the same temperature, the RMS velocities will vary because of differences in their molar masses.
The formula for the root mean square velocity is:\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]where:

• \(v_{rms}\) is the root mean square velocity.
• \(R\) is the gas constant.
• \(T\) is the absolute temperature.
• \(M\) is the molar mass of the gas.

With this formula, you can demonstrate that lighter gases move faster. For example, if a gas must have double the velocity of \(\mathrm{SO}_2\), its molar mass will be a fraction of \(\mathrm{SO}_2\)'s molar mass, as the relation \(v_{rms, \text{gas}} = 2 \times v_{rms, \mathrm{SO}_2}\) states.
This inverse relationship between molar mass and velocity is a fundamental aspect of kinetic molecular theory.

Chemical Calculations

Chemical calculations involving gases are often centered around various formulas that interlink gas properties with measurable variables like pressure, volume, and temperature. Understanding these calculations requires familiarity with a few key concepts.
First and foremost, determine the given values, such as the molar mass of gases. This value plays a significant role in further calculations, as seen in this exercise where the molar mass of sulphur dioxide (\(\mathrm{SO}_2\)) was calculated as 64 g/mol.
The steps in such calculations often include:

• Identifying known and unknown variables from a problem statement.
• Using appropriate gas laws and equations, such as the ideal gas law or RMS velocity equation, based on the scenario.
• Solving equations step-by-step to find the unknown value, performed using algebraic manipulation.
• Comparing results with known properties of substances to identify the gas in question or predict behavior.

In this specific exercise, following the logical steps of setting up equations from given conditions and solving algebraically helped us identify \(\mathrm{CH}_4\) as the gas with the desired velocity at a given temperature. By mastering these techniques, you can solve various computational problems in chemistry with confidence.