Problem 59 - A Complete Resource Book in Chemistry for JEE Main

Question

Ratio of \(\mathrm{C}_{\mathrm{p}}\) and \(\mathrm{C}_{\mathrm{v}}\) of a gas ' \(\mathrm{X}^{\prime}\) is \(1.4\). The number of atoms of the gas ' \(\mathrm{X}\) ' present in \(11.2\) litres of it at NTP will be (a) \(6.02 \times 10^{21}\) (b) \(60.2 \times 10^{23}\) (c) \(6.02 \times 10^{23}\) (d) \(1.02 \times 10^{23}\)

Step-by-Step Solution

Answer

The number of atoms is \(6.02 \times 10^{23} \), so the answer is (c).

Step 1: Use Ratio of Cp and Cv to Determine Atomicity

The ratio of specific heats, given by \( \gamma = \frac{C_p}{C_v} \), is 1.4. For a monoatomic gas, \( \gamma = 1.67 \), for a diatomic gas, \( \gamma = 1.4 \), and for a triatomic gas with linear shape, \( \gamma \) is close to 1.33. Since \( \gamma \) for our gas 'X' is 1.4, it indicates that the gas 'X' is diatomic.

Step 2: Apply Ideal Gas Law to Calculate Moles

At NTP (Normal Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Given that the volume of gas 'X' is 11.2 liters, calculate the number of moles using the relation: \[ \text{Number of moles} = \frac{\text{Volume of gas at NTP}}{22.4} = \frac{11.2}{22.4} = 0.5 \text{ moles} \]

Step 3: Calculate Number of Molecules

Avogadro's number states that one mole of any substance contains \( 6.02 \times 10^{23} \) molecules. Therefore, 0.5 moles of gas 'X' contain: \[ 0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23} \text{ molecules} \]

Step 4: Determine Number of Atoms

Since gas 'X' is diatomic, each molecule comprises 2 atoms. The total number of atoms in 3.01 \( \times 10^{23} \) molecules is: \[ 3.01 \times 10^{23} \times 2 = 6.02 \times 10^{23} \text{ atoms} \]

Key Concepts

Specific Heat Ratio

The specific heat ratio, often denoted by \( \gamma \), is an important concept when studying gases. It is represented by the ratio of two specific heats: the specific heat at constant pressure \( C_p \) and the specific heat at constant volume \( C_v \). In formula terms, \( \gamma = \frac{C_p}{C_v} \). This ratio provides insight into the type of gas we are dealing with, because different gases will have distinct specific heat ratios.

• For a **monoatomic gas**, \( \gamma \) is approximately 1.67.
• If the gas is **diatomic** in nature, \( \gamma \) tends to be around 1.4.
• A **triatomic linear gas** typically has a \( \gamma \) value close to 1.33.

In the exercise, the gas in question has a \( \gamma \) value of 1.4. This directly indicates that the gas 'X' is diatomic. Understanding this helps us predict and calculate further properties of the gas.

Atomicity Determination

Atomicity refers to the number of atoms that compose a molecule of a substance. It is a useful way to describe the molecular composition of a gas. With the specific heat ratio, we have already determined that the gas 'X' is diatomic, meaning each molecule of gas 'X' consists of two atoms.

• **Monoatomic** gases have an atomicity of 1.
• **Diatomic** gases have an atomicity of 2, suggesting two atoms per molecule, such as oxygen \( O_2 \).
• **Triatomic** gases, like carbon dioxide \( CO_2 \), have an atomicity of 3.

Knowing the atomicity is crucial as it affects how many atoms exist in a given quantity of gas. For instance, if 3.01 \( \times 10^{23} \) molecules of a diatomic gas are present, they equate to twice that number in atoms, because each molecule contains two atoms. This amounts to 6.02 \( \times 10^{23} \) atoms in total, as seen in our calculation.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry and physics. It is the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. Avogadro's number is expressed as \( 6.02 \times 10^{23} \). This immense number helps us translate between the macroscopic scale, which we interact with in labs, and the microscopic scale, at the atomic level.

• This concept allows us to determine how many molecules or atoms are present in a given number of moles.
• For instance, one mole contains exactly \( 6.02 \times 10^{23} \) molecules.
• When you have half a mole, like in the exercise, you simply multiply by this number and discover that you have approximately 3.01 \( \times 10^{23} \) molecules.

Applying Avogadro's number reveals the total number of individual atoms within a gas sample when combined with the gas's atomicity. In our example, since gas 'X' is diatomic, the number of atoms is double the number of molecules, leading us to our final calculation of 6.02 \( \times 10^{23} \) atoms.