Problem 134 - A Complete Resource Book in Chemistry for JEE Main

Question

Pressure in a mixture of \(4 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) confined in a container of 1 litre capacity at \(0{ }^{\circ} \mathrm{C}\) is (a) \(25.2 \mathrm{~atm}\) (b) \(35.6 \mathrm{~atm}\) (c) \(15.4 \mathrm{~atm}\) (d) \(48.2 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The pressure of the gas mixture is 25.2 atm, answer (a).

Step 1: Find Moles of Each Gas

First, calculate the number of moles of each gas. The molar mass of \(\mathrm{O}_2\) is \(32\ \mathrm{g/mol}\), and the molar mass of \(\mathrm{H}_2\) is \(2\ \mathrm{g/mol}\).\[\text{Moles of } \mathrm{O}_2 = \frac{4}{32} = 0.125 \text{ moles}\]\[\text{Moles of } \mathrm{H}_2 = \frac{2}{2} = 1.0 \text{ mole}\]

Step 2: Calculate Total Moles in the Mixture

Add the moles of each gas together to find the total moles in the mixture.\[\text{Total moles} = 0.125 + 1.0 = 1.125 \text{ moles}\]

Step 3: Use the Ideal Gas Law to Find Pressure

To find the pressure exerted by the gas mixture, use the ideal gas law \(PV = nRT\). Rearrange to solve for pressure \(P\):\[\P = \frac{nRT}{V}\]Where:- \(n = 1.125\ \text{moles}\)- \(R = 0.0821\ \text{L·atm/mol·K}\)- \(T = 273\ \text{K}\) (since \(0^{\circ}C = 273\ \text{K}\))- \(V = 1\ \text{L}\)Plug the values into the formula:\[P = \frac{(1.125)(0.0821)(273)}{1} = 25.2 \text{ atm}\]

Step 4: Select the Correct Answer

Compare the value calculated with the given options to select the correct answer, which is \(25.2 \text{ atm}\).

Key Concepts

Pressure Calculation

Understanding how to calculate pressure is essential in applying the ideal gas law. In this problem, we need to determine how much pressure is exerted by a gas mixture composed of oxygen and hydrogen.
The ideal gas law formula is:\[ PV = nRT \]Here:

• \( P \) is the pressure, which is what we want to find.
• \( V \) is the volume of the container, given as \( 1 \text{ L} \).
• \( n \) is the total number of moles of gas in the mixture.
• \( R \) is the ideal gas constant, \( 0.0821 \text{ L·atm/mol·K} \).
• \( T \) is the temperature in Kelvin, which is \( 273 \text{ K} \) when converted from \( 0 ^{\circ} \text{C} \).

To find the pressure \( P \), we rearrange the formula:\[ P = \frac{nRT}{V}\]By plugging in the values:\[ P = \frac{1.125 \times 0.0821 \times 273}{1} = 25.2 \text{ atm}\]This calculation shows that the pressure exerted by the gas mixture is \( 25.2 \text{ atm} \). This physical understanding of pressure helps us predict how gas behavior changes with conditions.

Moles of Gases

Calculating the moles of gases helps us understand the amount of substance present in a gas. The number of moles is crucial because it connects directly to the gas's mass and volume. For this exercise, the problem provides us with the masses of oxygen and hydrogen, requiring us to convert these masses into moles using their molar masses.
The molar mass of a substance is the mass of one mole of that substance. Here are the molar masses used:

• For oxygen (\( \mathrm{O}_2 \)), the molar mass is \( 32 \text{ g/mol} \).
• For hydrogen (\( \mathrm{H}_2 \)), the molar mass is \( 2 \text{ g/mol} \).

Using the formula:\[ \text{Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}}\]- For \( \mathrm{O}_2 \):\[\frac{4}{32} = 0.125 \text{ moles}\]- For \( \mathrm{H}_2 \):\[\frac{2}{2} = 1.0 \text{ mole}\]These calculations help us determine how many moles of each gas are present in the mixture. Knowing the total moles in the space helps us in further pressure calculations and understanding chemical reactions when these gases are mixed.

Gas Mixture

When dealing with gas mixtures, it's important to understand that each component gas contributes to the total pressure essentially independent of each other. This is based on Dalton's Law of Partial Pressures.
To understand the mixture concept in this exercise, recognize:

• Each gas contributes to the total pressure based on its amount in moles and occupies the entire volume of the container.
• The individual pressures from each gas add up to the total pressure of the gas mixture.

In our example, we first calculated the moles of both \( \mathrm{O}_2 \) and \( \mathrm{H}_2 \), achieving a sum of 1.125 moles. These total moles play a central role in determining the pressure exerted by the gas mixture, as seen in the ideal gas law principle.
Thus, when we assessed the gas mixture, the key takeaway is that both gases freely mix and the combined effect of their individual pressures results in a total observed pressure, which we found to be \( 25.2 \text{ atm} \). This understanding of gas mixtures is crucial in predicting how different gases behave together in various chemical and physical processes.