Problem 39 - A Complete Resource Book in Chemistry for JEE Main

Question

If pressure of \(2 \mathrm{~mol}\) of an ideal gas at \(546 \mathrm{~K}\) having volume \(44.8\) litre is (a) \(2 \mathrm{~atm}\) (b) \(3 \mathrm{~atm}\) (c) \(4 \mathrm{~atm}\) (d) \(1 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The correct pressure is (a) 2 atm.

Step 1: Understand the Given Data

We are given the following data for an ideal gas: \( n = 2 \, \text{mol} \), \( V = 44.8 \, \text{liters} \), and \( T = 546 \, \text{K} \). We need to determine the correct pressure \( P \) from the provided options.

Step 2: Use the Ideal Gas Law

The Ideal Gas Law is given by \( PV = nRT \). Here, \( R \) is the ideal gas constant, which is approximately \( 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \).

Step 3: Substitute the Known Values

Insert the known values into the ideal gas law equation: \[ P \times 44.8 = 2 \times 0.0821 \times 546 \]Calculate the right-hand side: \[ 2 \times 0.0821 \times 546 = 89.7888 \].

Step 4: Solve for Pressure \( P \)

Rearrange the equation to solve for \( P \): \[ P = \frac{89.7888}{44.8} \]Calculate \( P \): \[ P \approx 2.005 \approx 2 \] atm.

Key Concepts

Pressure Calculation

The pressure of a gas is an essential aspect of the Ideal Gas Law. It describes the force that the gas exerts on the walls of its container. To find this pressure, you start with the formula from the Ideal Gas Law: \( PV = nRT \). Here, \( P \) represents the pressure we want to calculate.To rearrange for \( P \), the equation becomes:

• \( P = \frac{nRT}{V} \)

This formula explains how the amount of gas, its temperature, and the volume it occupies affect its pressure. Once all other values are known, substituting the values can efficiently determine the pressure of the gas. For the example provided, the calculation yields approximately \( 2 \) atm, indicating the pressure in this specific scenario.

Ideal Gas Constant

The ideal gas constant \( R \) is a crucial part of the Ideal Gas Law equation \( PV = nRT \). It connects pressure, volume, the number of moles, and temperature into a single, unified relationship.The most commonly used value for \( R \) is:

• \( 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \)

In this unit, it helps maintain consistency across different measurements, making it easier to solve equations related to gases. The constant ensures that changes in one variable appropriately reflect in the others. It's a fixed value derived from experimental data, making it universally applicable for problems concerning ideal gases. Using this constant, calculations become more straightforward, allowing for easy manipulation of the equation in various applications.

Volume of Gas

In any gas-related calculation, volume is a critical factor. It refers to the space that the gas occupies. In the Ideal Gas Law, it's designated by \( V \). Volume helps us understand the relationship between the space that the gas takes up and the pressure it exerts.In practice:

• Volume is typically measured in liters (L).
• It can vary due to temperature or pressure changes.

To measure the volume effectively, conditions must be kept constant, such as using a rigid container. In our example, the volume is \( 44.8 \) liters, a significant amount of space for \( 2 \) moles of gas. By maintaining a fixed volume, it becomes easier to explore other properties, such as its pressure or temperature, under controlled circumstances.

Temperature in Kelvin

Temperature, often measured in the Kelvin scale for gas law problems, is another pivotal component of the Ideal Gas Law. The Kelvin scale starts at absolute zero, where there is theoretically no kinetic energy or movement of particles.Key points about temperature in Kelvin:

• It avoids negative numbers, unlike Celsius.
• 0 K corresponds to \(-273.15 ^\circ\)C, the point where particle motion theoretically stops.

Temperature affects the energy and movement of gas particles. A higher temperature means more energetic particles, hence more pressure, assuming volume remains constant. In the given exercise, the temperature is \( 546 \) K. Using Kelvin simplifies the calculations and reflects the direct relationship between temperature and the physical behavior of gases in the equation \( PV=nRT \). By knowing how to handle Kelvin, you manage to maintain the consistency needed for accurate scientific calculations.