Problem 36 - A Complete Resource Book in Chemistry for JEE Main

Question

If two moles of ideal gas at \(540 \mathrm{~K}\) has volume \(44.8 \mathrm{~L}\), then its pressure will be (a) \(1 \mathrm{~atm}\) (b) \(2 \mathrm{~atm}\) (c) \(3 \mathrm{~atm}\) (d) \(4 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The pressure is approximately \(2 \; \mathrm{atm}\), so option (b) is correct.

Step 1: Understand the Problem

We need to find the pressure of an ideal gas given the number of moles, temperature, and volume. The ideal gas law will be used for this calculation.

Step 2: Introduce the Ideal Gas Law

The ideal gas law is given by the formula: \( PV = nRT \). Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant \( (0.0821 \; \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K}) \), and \( T \) is the temperature in Kelvin.

Step 3: Substitute the Known Values

We know \( n = 2 \; \mathrm{moles} \), \( T = 540 \; \mathrm{K} \), \( V = 44.8 \; \mathrm{L} \), and \( R = 0.0821 \; \mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} \cdot \mathrm{K} \). Substitute these values into the ideal gas law equation: \( P \times 44.8 = 2 \times 0.0821 \times 540 \).

Step 4: Calculate the Right Side of the Equation

Calculate \( 2 \times 0.0821 \times 540 = 88.668 \).

Step 5: Solve for Pressure

Now solve the equation \( P \times 44.8 = 88.668 \) for \( P \). Divide both sides by 44.8: \( P = \frac{88.668}{44.8} \approx 1.98 \).

Step 6: Match with the Options

The calculated pressure \( P = 1.98 \approx 2 \; \mathrm{atm} \). Match this with the closest option from the given choices.

Key Concepts

Temperature and pressure relationship

In the context of the ideal gas law, the relationship between temperature and pressure is crucial. As the temperature of a gas increases, its kinetic energy also increases. This results in particles colliding with more force against the walls of the container, leading to an increase in pressure if the volume remains constant. This concept is quantitatively expressed in the ideal gas law formula, which is: \[ PV = nRT \] Here, while the number of moles \( n \) and the volume \( V \) are constant, the temperature \( T \) plays a direct role in determining the pressure \( P \). The higher the temperature, the higher the pressure, assuming volume does not change.

• At a fixed volume, increasing the temperature increases the pressure.
• At a fixed pressure, increasing the temperature would typically require an increase in volume, according to Charles's Law.

Understanding this relationship helps to grasp why changing the temperature affects the pressure of a contained gas.

Ideal gas constant

The ideal gas constant, denoted as \( R \), is a key component of the ideal gas law. It serves as the proportionality constant in the equation \( PV = nRT \). This constant allows the equation to relate pressure, volume, temperature, and the number of moles for an ideal gas. For the units commonly used, \( R \) is expressed as 0.0821 \( \, ext{L} \, ext{atm/mol} \, ext{K} \). This means that for each mole of gas, one atmosphere of pressure, and every Kelvin of temperature, there are 0.0821 liters involved.

• The value of \( R \) depends on the units used, but 0.0821 \( \, ext{L} \, ext{atm/mol} \, ext{K} \) is standard for these units.
• \( R \) ties together the various units and values into a coherent equation, making calculations involving gases possible and straightforward.

Having a consistent value for the ideal gas constant ensures accurate and reliable calculations using the ideal gas law. It is an essential tool for scientists and engineers working with gas-related problems.

Calculation of pressure

To determine the pressure of a gas using the ideal gas law, you first need the number of moles \( n \), the temperature \( T \), the volume \( V \), and the ideal gas constant \( R \). Given these values, you can solve for pressure \( P \) using: \[ P = \frac{nRT}{V} \] In the exercise, we know:

• Number of moles \( n = 2 \)
• Temperature \( T = 540 \, ext{K} \)
• Volume \( V = 44.8 \, ext{L} \)
• The ideal gas constant \( R = 0.0821 \, ext{L atm/mol K} \)

Substituting these values into the formula gives: \[ P = \frac{2 \times 0.0821 \times 540}{44.8} \approx 1.98 \, ext{atm} \] This result illustrates how handling units and constants correctly leads to a precise calculation of gas pressure under specified conditions. It's a simple yet powerful demonstration of the ideal gas law in action.