Problem 52 - A Complete Resource Book in Chemistry for JEE Main

Question

1 mole of an ideal gas at \(300 \mathrm{~K}\) is expanded isothermally and reversible from an initial volume of 1 litre to 10 litre. The work in this process is \(\left(\mathrm{R}=2 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(163.7 \mathrm{cal}\) (b) zero (c) \(-1381.8 \mathrm{cal}\) (d) 9 litreatm

Step-by-Step Solution

Answer

The work done is -1381.2 cal, which corresponds to option (c).

Step 1: Understand the Problem

The exercise asks us to calculate the work done when 1 mole of an ideal gas expands isothermally and reversibly from 1 litre to 10 litres at a temperature of 300 K. Given the gas constant, R = 2 cal/mol·K, we need to find the work done during this process.

Step 2: Recall the Work Formula for Isothermal Reversible Expansion

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

Step 3: Substitute the Values

Substitute the given values into the formula. Here, \( n = 1 \), \( R = 2 \) cal/mol·K, \( T = 300 \) K, \( V_i = 1 \) litre, and \( V_f = 10 \) litres. The expression becomes: \[ W = -1 \times 2 \times 300 \times \ln \left(\frac{10}{1}\right) \]

Step 4: Calculate the Natural Logarithm

Calculate the natural logarithm: \( \ln \left(10\right) \approx 2.302 \). This logarithm often arises when dealing with expressions involving powers of 10.

Step 5: Compute the Work Done

Substitute the logarithm value in: \[ W = -2 \times 300 \times 2.302 = -1381.2 \] Therefore, the work done by the gas when it expands isothermally and reversibly is \(-1381.2\) calories.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of ideal gases. It is expressed as:
\[ PV = nRT \]
where:

• \(P\) stands for pressure,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the gas constant (in this scenario, 2 cal/mol·K),
• and \(T\) is the temperature in Kelvin.

This equation helps predict how the gas will behave under different conditions.
It's crucial for solving problems involving gas expansions and compressions.
In isothermal processes, where temperature remains stable, changes in volume and pressure are directly linked.
By knowing the relationship between these variables, we can calculate the work a gas does when it changes volume, as seen in our exercise.

Reversible Process

A reversible process is an ideal type of thermodynamic process that happens infinitely slowly.
This allows the system to remain in a perfect equilibrium at every moment.
In real-world applications, processes are never perfectly reversible, but this concept helps us understand the maximum efficiency a system can achieve.

Isothermal Reversible Expansion:

• During this type of expansion, the temperature of the system remains constant.
• The system exchanges enough heat with its surroundings to maintain this constant temperature.
• The process is so slow that every stage is equilibrium maintained with the environment.

Reversible processes help calculate the maximum work a system can do.
Since the system is in equilibrium at all stages, it requires the least amount of work to return to its initial state.

Work Calculation

Calculating the work done during an isothermal reversible expansion involves using the work formula:
\[ W = -nRT \, \ln \left(\frac{V_f}{V_i}\right) \]
This formula accounts for:

• The number of moles \(n\) participating in the expansion (here, \(n = 1\)),
• The gas constant \(R\) given as 2 cal/mol·K,
• The constant temperature \(T = 300 \) K,
• The initial volume \(V_i = 1 \) litre and the final volume \(V_f = 10 \) litres.

By substituting these values, we calculate the work done by the gas.
It provides insight into how much energy is transferred by the system as it expands.
In this example, the work done is found to be negative, indicating the gas has done work on its surroundings.

Natural Logarithm

The natural logarithm is a mathematical function denoted as \( \ln \).
It is the logarithm to the base \(e\), where \(e\) is approximately 2.718.
This function is essential in various fields, including thermodynamics and kinetics.

Powers of 10 in Logarithms:

• Natural logarithms often appear in equations where ratios are involved.
• In this exercise, we use \( \ln\left(\frac{V_f}{V_i}\right) \) to express how the volume changes.
• Specifically, \( \ln(10) \approx 2.302 \) is used, as we transition from 1 litre to 10 litres.

Understanding how to calculate and apply natural logarithms helps solve equations involving exponential growth or decay.
This knowledge is critical for computing work in thermodynamic processes and further illustrates how minute changes in volume or pressure can lead to significant energy shifts.