Problem 35 - Master Resource Book in JEE Main Physics

Question

\(8 \mathrm{~g}\) of \(\mathrm{O}_{2}, 14 \mathrm{~g}\) of \(\mathrm{N}_{2}\) and \(22 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) is mixed in a container of \(10 \mathrm{~L}\) capacity at \(27^{\circ} \mathrm{C}\). The pressure exerted by the mixture in terms of atmospheric pressure is \(\left(R=0.082 \mathrm{~L}\right.\) atm \(\mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) (a) \(1.4 \mathrm{~atm}\) (b) \(2.5 \mathrm{~atm}\) (c) \(3.7 \mathrm{~atm}\) (d) \(8.7 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The pressure exerted by the mixture is 3.7 atm (option c).

Step 1: Convert Masses to Moles

First, we need to calculate the number of moles of each gas in the mixture. We use the formula \( n = \frac{m}{M} \), where \( n \) is the number of moles, \( m \) is the mass in grams, and \( M \) is the molar mass in grams per mole:- For \( \mathrm{O}_{2} \), \( m = 8 \mathrm{~g} \), \( M = 32 \mathrm{~g/mol} \):\[ n_{\mathrm{O}_{2}} = \frac{8}{32} = 0.25 \text{ moles} \]- For \( \mathrm{N}_{2} \), \( m = 14 \mathrm{~g} \), \( M = 28 \mathrm{~g/mol} \):\[ n_{\mathrm{N}_{2}} = \frac{14}{28} = 0.5 \text{ moles} \]- For \( \mathrm{CO}_{2} \), \( m = 22 \mathrm{~g} \), \( M = 44 \mathrm{~g/mol} \):\[ n_{\mathrm{CO}_{2}} = \frac{22}{44} = 0.5 \text{ moles} \]

Step 2: Calculate Total Moles of Gas

Add up the moles of each gas to find the total number of moles in the mixture:\[ n_{\text{total}} = n_{\mathrm{O}_{2}} + n_{\mathrm{N}_{2}} + n_{\mathrm{CO}_{2}} = 0.25 + 0.5 + 0.5 = 1.25 \text{ moles} \]

Step 3: Use Ideal Gas Law to Find Pressure

Utilize the ideal gas law \( PV = nRT \) to find the pressure:- \( P \) is the pressure we need to find.- \( V = 10 \mathrm{~L} \), the volume of the container.- \( n = 1.25 \), total moles of the gas.- \( R = 0.082 \mathrm{~L \cdot atm \cdot K^{-1} \cdot mol^{-1}} \), the ideal gas constant.- \( T = 27^{\circ}C = 300 \mathrm{~K} \) (converted from Celsius to Kelvin).Plug the values into the equation:\[ P \times 10 = 1.25 \times 0.082 \times 300 \ P = \frac{1.25 \times 0.082 \times 300}{10} = 3.075 \text{ atm}\]

Step 4: Round the Pressure Value

Round the calculated pressure value to match the given options:The closest match for the pressure 3.075 atm is \( 3.7 \text{ atm} \).

Key Concepts

Molecular Weight Calculation

Molecular weight, often referred to as molar mass, is a fundamental concept in chemistry. It essentially helps us understand how much one mole of a substance weighs. This is crucial when it comes to converting mass to moles in equations. For gases like oxygen (\(\mathrm{O}_{2}\)), nitrogen (\(\mathrm{N}_{2}\)), and carbon dioxide (\(\mathrm{CO}_{2}\)), we use their molecular weight for calculations.
- Oxygen (\(\mathrm{O}_{2}\)) has a molecular weight of 32 grams per mole. Each molecule consists of two oxygen atoms, each with an atomic weight of about 16. - Nitrogen (\(\mathrm{N}_{2}\)) weighs 28 grams per mole, with each of the two nitrogen atoms contributing 14 grams.- Carbon dioxide (\(\mathrm{CO}_{2}\)), with one carbon atom (12 grams) and two oxygen atoms, weighs 44 grams per mole.Knowing these weights is the first step towards solving many gas law problems.

Mole Conversion

Understanding moles is crucial for chemical calculations. The mole is a unit that quantifies the amount of substance. In our case, converting grams to moles involves the formula: \[ n = \frac{m}{M} \]where \( n \) is the number of moles, \( m \) is the mass in grams, and \( M \) is the molecular weight.
- For example, with 8 grams of \(\mathrm{O}_{2}\), dividing by its molecular weight (32) gives 0.25 moles.- Similarly, 14 grams of \(\mathrm{N}_{2}\) divided by 28 gives 0.5 moles.- Lastly, 22 grams of \(\mathrm{CO}_{2}\) divided by 44 also results in 0.5 moles.These conversions are essential for comparing amounts of different gases in a mixture.

Pressure Calculation

Pressure calculation using the ideal gas law enables us to find the pressure exerted by gases in a container. In chemistry, the ideal gas law is expressed as: \[ PV = nRT \]Here, \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (0.082 \( \mathrm{L \, atm \, K^{-1} \, mol^{-1}} \)), and \( T \) is temperature in Kelvin.
By rearranging the formula to solve for pressure (\( P \)), we get: \[ P = \frac{nRT}{V} \]Applying this to our problem, the total moles (\( n = 1.25 \)), volume (\( V = 10 \mathrm{L} \)), and temperature (\( T = 300 \mathrm{K} \)) combine to give us:\[ P = \frac{1.25 \times 0.082 \times 300}{10} = 3.075 \mathrm{atm} \]This practical application of the ideal gas law lets us determine the pressure of gas mixtures efficiently.

Gas Mixtures

In real-life scenarios, gases are often mixed together. The behavior of such gas mixtures can be explained using gas laws. The concept of mole fraction or partial pressure comes into play here. In a mixture, each gas contributes to the total pressure. This is connected to the number of moles each gas has.
For a mixture including \(\mathrm{O}_{2}\), \(\mathrm{N}_{2}\), and \(\mathrm{CO}_{2}\): - We first find the total number of moles: 1.25 moles, which includes all gases.- Each gas has a partial pressure that adds up to the total pressure.Understanding how gases mix and exert pressure is fundamental for many scientific calculations.

Thermodynamics

Thermodynamics is all about the study of energy changes, particularly in gases. In our context, temperature and energy are key aspects. The ideal gas law links these with pressure and volume.
Consider temperature: in degrees Celsius, it must be converted to Kelvin for the gas laws. - Here, 27°C becomes 300 K (add 273 to Celsius). - This allows for correct calculations, as gas behaviors are rooted in thermodynamic principles. Overall, understanding thermodynamics helps us predict how gases respond to changing conditions, ensuring our calculations accurately reflect physical realities.