Problem 82 - A Complete Resource Book in Chemistry for JEE Main

Question

To calculate the amount of work done in joules during a reversible isothermal expansion of an ideal gas, the volume must be expressed in (a) \(\mathrm{dm}^{3}\) only (b) \(\mathrm{m}^{3}\) only (c) \(\mathrm{cm}^{3}\) only (d) any one of them

Step-by-Step Solution

Answer

(d) any one of them

Step 1: Understand the Formula for Work

In a reversible isothermal expansion of an ideal gas, the work done is given by the formula \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \]where \(W\) is the work done, \(n\) is the number of moles, \(R\) is the ideal gas constant, \(T\) is the temperature, \(V_f\) is the final volume, and \(V_i\) is the initial volume.

Step 2: Consider the Units of Volume

The formula for work involves a ratio of volumes (\(V_f/V_i\)). Since you're taking the natural logarithm of this ratio, the units of volume will cancel each other out. Therefore, the specific unit chosen for volume will not affect the calculation of work as long as both \(V_f\) and \(V_i\) are in the same units.

Step 3: Evaluate the Options

Given the information that any consistent units for \(V_f\) and \(V_i\) will cancel out, the volume can be expressed in any unit system (\(\mathrm{dm}^3\), \(\mathrm{m}^3\), or \(\mathrm{cm}^3\)) as long as both initial and final volumes are measured in the same units. Thus, the correct option is (d) any one of them.

Key Concepts

Ideal Gas

An ideal gas is a theoretical model used to simplify and predict the behavior of gases. In the ideal gas model, certain assumptions are made:

• Molecules are point particles, meaning they have no volume.
• There are no attractive or repulsive forces between molecules.
• All collisions between gas molecules or with the walls of the container are perfectly elastic.

These assumptions allow us to use the ideal gas law, which is expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
An ideal gas exhibits predictable behavior when it undergoes changes like expansion or compression. In particular, during an isothermal process (constant temperature), the equation can help us calculate the work done by or on the gas. The concept of an ideal gas is essential for simplifying complex real-world gas behavior into manageable calculations.

Work Done

When an ideal gas undergoes reversible isothermal expansion, it does work on its surroundings. The work done, \( W \), is calculated using the formula \[ W = nRT \ln\left(\frac{V_f}{V_i}\right) \] Here,

• \( n \) is the number of moles of gas
• \( R \) is the universal gas constant (approximately 8.314 J/(mol·K))
• \( T \) is the absolute temperature in Kelvin
• \( V_f \) and \( V_i \) are the final and initial volumes, respectively

The formula reflects that the work done is proportional to the number of moles and the temperature. The natural logarithm function relates to the change in volume.
Reversible expansion means the process is carried out slowly so the system remains in a state of equilibrium. Calculating work in such conditions gives the maximum possible work done during the process.

Volume Units

Volume is a measure of the space taken up by a gas. When calculating the work done in a reversible isothermal expansion of an ideal gas, the ratio of final volume \( V_f \) to initial volume \( V_i \) appears in the natural logarithm function. Because this is a dimensionless ratio, the actual units of volume cancel each other out.
Therefore, it doesn't matter whether you are using:

• cubic decimeters (\( \mathrm{dm}^3 \))
• cubic meters (\( \mathrm{m}^3 \))
• cubic centimeters (\( \mathrm{cm}^3 \))

As long as both \( V_f \) and \( V_i \) are measured in the same type of unit, you can calculate the work correctly. This flexibility in unit choice makes calculations more convenient depending on the context and the available data.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a mathematical function with a base of \( e \), where \( e \) is an irrational number approximately equal to 2.71828. In the context of thermodynamics, it often comes into play in processes involving expansion or compression of gases.
When we use the formula \( W = nRT \ln\left(\frac{V_f}{V_i}\right) \), the natural logarithm helps illustrate how work is related to the change in volume between the initial and final states. The logarithmic function continuously increases but at a decreasing rate, offering a smooth representation of relative changes, which is why it's valuable in transformations involving gases.
It's important to have a grasp of how to manipulate logarithms for effective problem-solving in chemistry and physics. This mathematical tool is key to understanding exponential growth and decay processes, including those involving gases in thermodynamic systems.