Problem 48 - A Complete Resource Book in Chemistry for JEE Main

Question

2 mole of an ideal gas at \(27^{\circ} \mathrm{C}\) temperature is expanded reversibly from \(2 \mathrm{~L}\) to \(20 \mathrm{~L}\). Find entropy change in cal. \((\mathrm{R}=2 \mathrm{cal} / \mathrm{mol} \mathrm{K})\) (a) \(92.1\) (b) 0 (c) 4 (d) \(9.2\)

Step-by-Step Solution

Answer

The entropy change is 9.2 cal, corresponding to option (d).

Step 1: Understand the Problem

We need to find the change in entropy for a reversible expansion of an ideal gas. We are given the initial and final volume, the temperature, and the gas constant for calculation.

Step 2: Use the Entropy Change Formula for Ideal Gas

The formula for the entropy change when an ideal gas expands is given by: \[ \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \] where \( n \) is the number of moles, \( R \) is the gas constant, \( V_f \) is the final volume, and \( V_i \) is the initial volume.

Step 3: Substitute Values in the Formula

Given that \( n = 2 \) moles, \( R = 2 \) cal/mol·K, \( V_f = 20 \) L, and \( V_i = 2 \) L, substitute these values into the entropy change formula: \[ \Delta S = 2 \times 2 \ln \left( \frac{20}{2} \right) \] \[ \Delta S = 4 \ln(10) \]

Step 4: Calculate the Natural Logarithm

Calculate the natural logarithm of 10, which is approximately 2.302. Substitute this value back into the equation: \[ \Delta S = 4 \times 2.302 \]

Step 5: Compute the Final Entropy Change

Perform the multiplication to find \( \Delta S \): \[ \Delta S = 4 \times 2.302 = 9.208 \] Round this to one decimal place to match the answer choices, giving us \( 9.2 \).

Key Concepts

Understanding Ideal Gas

An *ideal gas* is a theoretical concept that represents a gas where the molecules are point particles interacting only through elastic collisions with the walls of their container.

Ideal gases obey the ideal gas law, which is formulated as:

• Pressure (\( P \)) is directly proportional to Temperature (\( T \)) and Volume (\( V \))
• The constant of proportionality is \( nR \), where \( n \) represents the number of moles and \( R \) is the universal gas constant

The ideal gas equation is expressed as:\[PV = nRT\]Ideal gases are useful for modeling behavior in real gases under many standard conditions, as deviations are mostly negligible.

However, at very low temperatures or extremely high pressures, real gases diverge from ideal behavior, requiring corrections like the Van der Waals equation for accurate predictions.

Reversible Expansion Process

A *reversible process* is a conceptual sequence of states through which a system can be retraced exactly back to its initial state, without causing any changes in the surroundings.

In the context of thermodynamics and ideal gases, a reversible expansion occurs when the gas expands incrementally, with the system being in a quasi-equilibrium state at each step. This means that every small change in external pressure is matched by an integrated change in the gas's volume.

Why is reversibility important? Because it ensures maximum efficiency in physical processes:

• It absorbs maximum work from the surroundings during compression.
• It utilizes minimum energy from the system during expansion.

Reversible processes are a standard in theoretical studies because they optimize conditions for calculating changes such as entropy.

Entropy Change Formula

Entropy, a measure of disorder or randomness, can increase or decrease based on system changes. For an ideal gas undergoing a reversible process, the formula to calculate the entropy change (\( \Delta S \)) is:\[\Delta S = nR \ln \left( \frac{V_f}{V_i} \right)\]Here's a step-by-step understanding of the variables:

• \( n \): Number of moles of the gas.
• \( R \): The gas constant, which is specific to the units of energy and temperature in which it's defined.
• \( V_f \): The final volume of the gas after expansion.
• \( V_i \): The initial volume of the gas before expansion.

This formula arises from integrating an infinitesimal change in entropy \( dS \) which corresponds to a small volume change \( dV \), efficiently capturing the shift as a function of \( \ln \left( \frac{V_f}{V_i} \right) \). The natural logarithm reflects how the entropy relates to multiplicative changes, offering insight into system disorder.

Temperature Effect on Entropy

Temperature heavily influences the entropy of a system. At higher temperatures, gases tend to possess higher entropy due to increased molecular motion.

Entropy's temperature dependence is crucial because:

• At constant pressure, heating the gas raises the entropy as molecules spread further.
• Conversely, cooling reduces entropy by constricting molecular motion into a more ordered state.

When calculating entropy changes during expansion at a constant temperature (isothermal process), the focus shifts from temperature increases to changes in volume and pressure:\[\Delta S = nR \ln \left( \frac{V_f}{V_i} \right)\]This equation shows how entropy variation occurs without temperature change.

Understanding these temperature effects allows us to predict how a system will respond to various state changes, emphasizing the balance between entropy contribution and thermal energy.