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 \(-1381.8\text{ cal}\). Option (c) is correct.

Step 1: Understanding the Context

We have an ideal gas that is undergoing an isothermal (constant temperature) expansion. The initial volume is given as 1 liter and it expands to 10 liters. The process is reversible, which allows us to use specific formulas for calculations.

Step 2: Formula for Work Done in Isothermal Expansion

In an isothermal and reversible process, the work done by the system can be calculated using the formula: \( W = -nRT \ln \left( \frac{V_f}{V_i} \right) \). Here, \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is the temperature, \(V_f\) is the final volume, and \(V_i\) is the initial volume.

Step 3: Substituting Given Values

We substitute the given values into the formula: \(n = 1\), \(R = 2 \text{ cal mol}^{-1} \text{K}^{-1}\), \(T = 300 \text{ K}\), \(V_i = 1 \text{ L}\), \(V_f = 10 \text{ L}\). Calculating gives:\[W = -1 \times 2 \times 300 \times \ln \left( \frac{10}{1} \right)\]

Step 4: Calculate the Natural Logarithm

Calculate the natural logarithm in the formula: \( \ln(10) \approx 2.302 \).

Step 5: Calculate Work Done

Substitute the value of \( \ln(10) \) into the equation:\[W = -1 \times 2 \times 300 \times 2.302 = -1381.2 \text{ cal}\]Upon approximation, this value matches closest to the option (c), which is \(-1381.8\text{ cal}\).

Step 6: Check Calculation

Double-check the calculations to ensure all steps were followed correctly, confirming the approximation and final answer to align with option (c) \(-1381.8 \text{ cal}\).

Key Concepts

Ideal Gas

An ideal gas is a theoretical gas composed of a set of randomly-moving, non-interacting point particles. It is a simplified model that is often used in thermodynamics to help understand how real gases behave under various conditions.

• **Characteristics of Ideal Gases**: Ideal gases follow the ideal gas law equation: \[ PV = nRT \] where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.


• **Assumptions**: Ideal gases assume that particle collisions are perfectly elastic and that there are no interactions between the particles other than these collisions. They are considered to have zero volume themselves, which means the volume of the gas is just the volume of the container.

Understanding ideal gases allows us to predict and calculate changes in pressure, volume, and temperature accurately, especially under conditions of high temperature and low pressure where real gases behave more ideally. In our exercise, the gas is treated as ideal to simplify the calculations of the work done during its expansion.

Reversible Process

A reversible process is a theoretical concept in thermodynamics where a system changes its state in such a way that the process can be reversed without leaving any net change in both the system and its surroundings.

• **Reversibility in Thermodynamics**: In reality, no process is truly reversible. Reversible processes are ideal and serve as a benchmark for the most efficient processes possible.


• **Characteristics**: Reversible processes are performed infinitely slowly to maintain the system in a constant state of equilibrium. Any change made to the system is so small (infinitesimal) that the system can easily return to its original state by reversing the condition.


• **Energy Efficiency**: Since no energy is wasted in the form of entropy increase in a reversible process, it provides the maximum work output or minimum work input for the process as compared to real, irreversible processes.

In the context of the exercise at hand, the gas expansion was conducted isothermally and reversibly, meaning the calculations for work done assume this infinitely gentle transition between states, maximizing efficiency and allowing for direct calculations using standard thermodynamic equations.

Work Done in Thermodynamics

In thermodynamics, the concept of work revolves around the energy transfer that occurs when an object is moved by a force. Specifically for gases, work is often associated with changes in volume under specific conditions.

• **Calculating Work**: For ideal gases undergoing reversible, isothermal expansions or compressions, the work done \( W \) is derived from the equation: \[ W = -nRT \ln \left( \frac{V_f}{V_i} \right) \] where \( n \) is the number of moles, \( R \) is the gas constant, \( T \) is the absolute temperature, and \( V_i \) and \( V_f \) are the initial and final volumes respectively.


• **Sign Conventions**: The negative sign in the equation reflects the convention that work done by the system (like expansion) is considered negative, while work done on the system (like compression) is positive.


• **Energy Transfer**: Work done during gas expansion is an important concept because it involves energy transfer from the system to the surroundings or vice versa. This energy can cause movement, generate heat, or otherwise change the state of a system.

In our exercise, the work done during the isothermal expansion of one mole of an ideal gas from 1 liter to 10 liters at 300 K is calculated using these principles, resulting in a numeric solution reflecting the specific conditions of the problem.