Problem 83 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(0.75\) mole of an ideal gas is expanded isothermally at \(27^{\circ} \mathrm{C}\) from 15 litres to 25 litres, then work done by the gas during this process is \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(-1054.2 \mathrm{~J}\) (b) \(-896.4 \mathrm{~J}\) (c) \(-954.2 \mathrm{~J}\) (d) \(-1254.3 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done is -954.2 J, option (c).

Step 1: Understanding the Problem

We need to calculate the work done by an ideal gas that expands isothermally from an initial volume of 15 liters to a final volume of 25 liters at 27°C. The amount of gas is 0.75 moles, and the universal gas constant R is given as 8.314 J K⁻¹ mol⁻¹.

Step 2: Formula for Isothermal Expansion Work

The work done by an ideal gas during an isothermal expansion can be calculated using the formula: \( W = -nRT\ln\left(\frac{V_f}{V_i}\right) \), where \( n \) is the number of moles, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( V_f \) is the final volume, and \( V_i \) is the initial volume.

Step 3: Convert Temperature to Kelvin

The temperature is given in Celsius (27°C). Convert it to Kelvin using the formula: \( T(K) = T(°C) + 273.15 \). Thus: \( T = 27 + 273.15 = 300.15 \) K.

Step 4: Substitute Values into the Formula

Substitute the given values into the formula. We have \( n = 0.75 \), \( R = 8.314 \), \( T = 300.15 \), \( V_f = 25 \), and \( V_i = 15 \). Calculate: \( W = -0.75 \times 8.314 \times 300.15 \times \ln\left(\frac{25}{15}\right) \).

Step 5: Calculate the Natural Logarithm

Calculate the natural logarithm of the volume ratio: \( \ln\left(\frac{25}{15}\right) = \ln(1.6667) = 0.5108 \).

Step 6: Calculate the Work Done

Finally, substitute all the calculated values into the work done formula: \( W = -0.75 \times 8.314 \times 300.15 \times 0.5108 = -954.2 \) J. Therefore, the work done by the gas is -954.2 J.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that describes the behavior of ideal gases. It combines several gas laws such as Boyle's, Charles's, and Avogadro's laws into a single equation. The formula is written as \( PV = nRT \), where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume.
• \( n \) is the number of moles.
• \( R \) is the universal gas constant (8.314 J K⁻¹ mol⁻¹).
• \( T \) is the temperature in Kelvin.

This equation helps in understanding how gases change under different conditions of pressure, volume, and temperature. In isothermal processes, where temperature remains constant, the relationship simplifies to \( PV = \text{constant} \), proving particularly useful in calculating work done on or by the gas.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat, work, and forms of energy. It explains how energy is transferred between physical systems and how it influences the matter. Some key principles of thermodynamics include:

• The conservation of energy, known as the First Law.
• The concept of entropy and its relation to the Second Law.
• The distinction in energy changes in various thermodynamic processes like isothermal, adiabatic, and isobaric.

In an isothermal expansion, energy in the form of heat is absorbed by the gas, which does work on its surroundings without a change in temperature. This is a perfect application of these laws, illustrating the conversion and conservation of energy.

Work Done in Thermodynamic Processes

In thermodynamics, work done by or on the system is a crucial aspect. For an isothermal expansion of an ideal gas, the work done can be described by the formula: \[ W = -nRT\ln\left(\frac{V_f}{V_i}\right) \] Where:

• \( W \) is the work done by the gas.
• \( n \) is the number of moles.
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin.
• \( V_f \) and \( V_i \) are the final and initial volumes, respectively.

The negative sign indicates that the work is done by the system. During an expansion, the gas does work on the surroundings, indicating energy transfer outward. Understanding this process helps in analyzing how different conditions and parameters affect energy exchange.

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. In mathematics, it is extensively used along with exponential functions. In thermodynamics, particularly in the calculation for work done during isothermal expansions, the natural logarithm helps quantify exponential changes in volume or pressure ratios. For instance, in calculating work done: \[ W = -nRT\ln\left(\frac{V_f}{V_i}\right) \] \( \ln\left(\frac{V_f}{V_i}\right) \) provides a measure of how the volumes change. This calculation is crucial as it translates physical change into a form that can be used to compute the energy involved in the expansion.