Problem 173 - A Complete Resource Book in Chemistry for JEE Main

Question

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of \(10 \mathrm{dm}^{3}\) to a volume of \(100 \mathrm{dm}^{3}\) at \(27^{\circ} \mathrm{C}\) is: (a) \(35.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (b) \(38.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (c) \(45.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) (d) \(23.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

Step-by-Step Solution

Answer

The entropy change is (b) 38.3 J mol⁻¹ K⁻¹.

Step 1: Understand the Problem

We need to find the change in entropy for the isothermal reversible expansion of an ideal gas. We are given the initial and final volumes, the temperature, and the number of moles of gas.

Step 2: Use the Entropy Change Formula for Gases

For an isothermal process, the entropy change \( \Delta S \) of an ideal gas can be given by the formula:\[\Delta S = nR \ln \left( \frac{V_f}{V_i} \right)\]where \( n \) is the number of moles, \( R \) is the ideal gas constant \( 8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \), \( V_f \) is the final volume, and \( V_i \) is the initial volume.

Step 3: Substitute the Known Values

Substitute the values into the formula: \( n = 2 \), \( V_f = 100 \, \text{dm}^3 \), and \( V_i = 10 \, \text{dm}^3 \).\[\Delta S = 2 \times 8.314 \times \ln \left( \frac{100}{10} \right)\]

Step 4: Calculate the Natural Logarithm

Calculate the natural logarithm part: \( \ln \left( \frac{100}{10} \right) = \ln(10) \). Use the approximation \( \ln(10) \approx 2.302 \).

Step 5: Solve for the Entropy Change

Now solve:\[\Delta S = 2 \times 8.314 \times 2.302\]\[\Delta S = 2 \times 19.137828\]\[\Delta S = 38.275656 \, \text{J} \, \text{K}^{-1}\]Round to one decimal place: \( \Delta S \approx 38.3 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \).

Step 6: Verify with Given Options

Compare the calculated entropy change with the options given. It matches option (b) \( 38.3 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \).

Key Concepts

Ideal Gas

An ideal gas is a theoretical concept that simplifies the behavior of gases. It's based on several assumptions:

• The gas consists of a large number of small particles (atoms or molecules) that are in constant random motion.
• The particles are considered point particles with no volume.
• They do not exert any forces on each other, except during elastic collisions.
• These collisions with each other and with the walls of the container are perfectly elastic, meaning there is no energy loss.

These assumptions allow us to use simpler mathematical models to study the gas's behavior under various conditions.
For example, the ideal gas law, expressed as \( PV = nRT \), relates the pressure \( P \), volume \( V \), number of moles \( n \), and temperature \( T \) of an ideal gas through the gas constant \( R \).
This concept is extremely useful in thermodynamics and chemistry for simplifying and solving problems that involve gases.

Isothermal Process

An isothermal process is a type of thermodynamic process in which the temperature remains constant. This implies that any heat added to the system is used to do work or any work done on the system changes the internal energy while the temperature stays unchanged.

In the context of an ideal gas undergoing an isothermal process, the internal energy remains constant because the temperature does not change. Given the ideal gas law, any change in volume must be accompanied by a change in pressure to maintain a constant temperature.
For practical applications:

• Isothermal processes can be seen in the slow compression or expansion of gases where temperature is kept constant deliberately.
• They often require a mechanism to either absorb or release heat to the surroundings.

This type of process is particularly significant in understanding and calculating entropy changes, as temperature is a crucial factor in these calculations.

Entropy Formula

Entropy is a measure of the disorder or randomness in a system, and it plays a crucial role in the second law of thermodynamics.For an ideal gas undergoing an isothermal process, the change in entropy \( \Delta S \) can be calculated using the formula:

\[ \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \] Here:

• \( n \) represents the number of moles of the gas.
• \( R \) is the ideal gas constant \( 8.314 \, \text{J} \, \text{mol}^{-1} \, \text{K}^{-1} \).
• \( V_f \) and \( V_i \) are the final and initial volumes, respectively.

This formula originates from the fundamental changes in heat when a gas expands isothermally and reversibly.
It provides insight into the energy dispersal within the system as the gas changes from one state to another, reflecting the increase in disorder when volume increases.

Reversible Expansion

Reversible expansion is a key concept in thermodynamics, especially concerning ideal gases and entropy change. It describes a process that occurs in such a way that the system can be returned to its original state by an infinitesimally small change without any net effect on the system or its surroundings.

In theory, reversible processes occur infinitely slowly to maintain equilibrium and minimize energy losses through friction or dissipation.
With ideal gases:

• Reversible expansion implies that the gas expands slowly enough that the pressure difference between the system and its surroundings is always zero.
• This allows for maximum work to be done by the gas because it's expanding against maximum opposing pressure continuously.

This kind of process is essential when calculating entropy changes because it results in a more accurate measure of energy distribution within the system. In our exercise, the reversible expansion of the ideal gas is what allows us to calculate the exact change in entropy using the entropy formula.