Problem 18 - A Complete Resource Book in Chemistry for JEE Main

Question

The volume occupied by \(9.0 \mathrm{~g}\) of nitrogen gas at \(300 \mathrm{~K}\) and \(750 \mathrm{mmHg}\) pressure is (a) \(5.854\) (b) \(6.7432\) (c) \(8.8462\) (d) \(8.0225\)

Step-by-Step Solution

Answer

The volume of nitrogen gas is approximately \(8.0225\, \text{L}\), option (d).

Step 1: Identify Given Information

We have the mass of nitrogen gas as \(9.0 \, \text{g}\), temperature \(T = 300 \, \text{K}\), and pressure \(P = 750 \, \text{mmHg}\). The molecular weight of nitrogen \( \text{N}_2 \) is approximately \(28 \, \text{g/mol}\). We'll use the Ideal Gas Law, \( PV = nRT \), to find the volume \(V\).

Step 2: Convert Pressure to atm

Convert the pressure from \( \text{mmHg} \) to \( \text{atm} \):\[ 1 \, \text{atm} = 760 \, \text{mmHg} \]. Thus, \( 750 \, \text{mmHg} = \frac{750}{760} \, \text{atm} \approx 0.9868 \, \text{atm} \).

Step 3: Calculate Number of Moles

Calculate the number of moles \(n\) of nitrogen using its mass and molecular weight: \[ n = \frac{9.0 \, \text{g}}{28 \, \text{g/mol}} \approx 0.3214 \, \text{mol} \].

Step 4: Apply Ideal Gas Law

Use the Ideal Gas Law formula \( PV = nRT \) to solve for \(V\). Here, \(R = 0.0821 \, \text{L} \, \text{atm/K} \, \text{mol} \). Substitute values: \[ V = \frac{nRT}{P} = \frac{0.3214 \, \text{mol} \times 0.0821 \, \text{L} \, \text{atm/K} \, \text{mol} \times 300 \, \text{K}}{0.9868 \, \text{atm}} \].

Step 5: Calculate the Volume

Perform the calculation to find \(V\):\[ V = \frac{0.3214 \times 0.0821 \times 300}{0.9868} \approx 8.0225 \, \text{L} \]. Thus, the volume occupied by the nitrogen gas is approximately \(8.0225\, \text{L}\).

Key Concepts

Moles Calculation

Calculating the number of moles is a key part of applying the Ideal Gas Law. First, it's important to understand that a mole is a unit in chemistry that measures the amount of a substance. It corresponds to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities, like atoms, molecules, or ions.
To find the number of moles of a particular gas, you use its mass and molecular weight. The formula is:

• \( n = \frac{\text{mass}}{\text{molar mass}} \)

For example, if you have 9 grams of nitrogen gas \(\text{N}_2\), and the molar mass of \(\text{N}_2\) is about 28 g/mol, the number of moles \(n\) is calculated as follows:

• \( n = \frac{9.0\, \text{g}}{28\, \text{g/mol}} \approx 0.3214\, \text{mol} \)

This calculation is crucial because it tells us how many moles of nitrogen we have, which we then use to determine other properties like volume and pressure under different conditions.

Pressure Conversion

Understanding pressure units is essential for applying the Ideal Gas Law, especially since pressure needs to be in consistent units for these calculations. In many textbook problems, pressures are often given in millimeters of mercury (mmHg).
To align with the Ideal Gas Law, however, you must convert this to atmospheres (atm), because the Gas Constant \(R\) is typically given with pressure in atm. The conversion factor is:

• \(1\, \text{atm} = 760\, \text{mmHg}\)

So, if you're given a pressure of 750 mmHg, convert it like this:

• \(750\, \text{mmHg} = \frac{750}{760}\, \text{atm} \approx 0.9868\, \text{atm}\)

This step is necessary because using incorrect units would result in an incorrect volume calculation when applying the Ideal Gas Law.

Gas Volume Calculation

Calculating the volume gas occupies is a common application of the Ideal Gas Law. \(PV = nRT\) is the equation we use, where:

• \(P\) is pressure
• \(V\) is volume
• \(n\) is the number of moles
• \(R\) is the universal gas constant (\(0.0821 \text{L atm/K mol}\))
• \(T\) is temperature in Kelvin

To find the volume \(V\), rearrange the formula:

• \( V = \frac{nRT}{P} \)

Using the number of moles (0.3214 mol), \(R\) the gas constant, temperature (300 K), and pressure in atm (0.9868 atm), the calculation becomes:

• \( V = \frac{0.3214 \times 0.0821 \times 300}{0.9868} \approx 8.0225\, \text{L}\)

This formula helps us find the volume of the nitrogen gas under these specific conditions, illustrating how different factors like temperature, pressure, and moles of gas interact in the Ideal Gas Law.