Problem 49 - A Complete Resource Book in Chemistry for JEE Main

Question

When 1 mole gas is heated at constant volume tem perature is raised from 298 to \(308 \mathrm{~K}\). Heat supplied to the gas is \(500 \mathrm{~J}\). Then which of the following state ments is correct? (a) \(\mathrm{Q}=\mathrm{W}=500 \mathrm{~J}, \Delta \mathrm{U}=0\) (b) \(\mathrm{Q}=\Delta \mathrm{U}=500 \mathrm{~J}, \mathrm{~W}=0\) (c) \(\mathrm{Q}=\mathrm{W}=500 \mathrm{~J}, \Delta \mathrm{U}=0\) (d) \(\Delta \mathrm{U}=0, \mathrm{Q}=\mathrm{W}=-500 \mathrm{~J}\)

Step-by-Step Solution

Answer

The correct statement is (b) Q = ΔU = 500 J, W = 0.

Step 1: Understanding the Problem

We have 1 mole of a gas being heated at constant volume from 298 K to 308 K, with 500 J of heat supplied. We need to determine the correct thermodynamic statement regarding heat (Q), work (W), and change in internal energy (ΔU).

Step 2: Recall Thermodynamic Relations

In thermodynamics, for a process at constant volume, the work done, W, is zero because the volume does not change ( W = 0). The first law of thermodynamics is given by Q = ΔU + W. At constant volume, Q = ΔU + 0 or ΔU = Q.

Step 3: Apply the First Law of Thermodynamics

Substitute the given values: Q = 500 J and W = 0. Therefore, ΔU = 500 J. This confirms that the heat supplied is used to increase the internal energy of the gas, with no work done.

Step 4: Identify the Correct Statement

The correct answer is the one where Q = ΔU = 500 J and W = 0, which aligns with the conclusion from Step 3. Thus, answer (b) Q = ΔU = 500 J, W = 0 is correct.

Key Concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental concept in energy dynamics, akin to the principle of conservation of energy. Essentially, it states that energy cannot be created or destroyed, only transferred or transformed from one form to another. In mathematical terms, this law is expressed as:\[ Q = \Delta U + W \]where:

• \( Q \) is the heat added to the system,
• \( \Delta U \) is the change in internal energy of the system,
• \( W \) is the work done by the system.

This formula forms the basis for understanding various thermodynamic processes, including those occurring at constant volume or pressure.
In the context of the exercise, since heat is being added and no work is done (as the volume remains constant), all the heat conversion contributes to changing the internal energy of the gas. This is why the equation simplifies to \( Q = \Delta U \) when the process happens at constant volume.

Constant Volume Process

In a constant volume process, the volume of the substance in question does not change during the heat transfer. This is a key detail because it simplifies the analysis of the thermodynamic process.
For gases, this means that during heating, the piston in the cylinder remains fixed. As a result, no mechanical work (\( W \)) is done on or by the gas, because work depends on volume change and is calculated as:\[ W = P \cdot \Delta V \]where:

• \( W \) is work
• \( P \) is the pressure
• \( \Delta V \) is the change in volume

Since \( \Delta V = 0 \) when volume is constant, it follows that \( W = 0 \). All heat supplied, therefore, converts purely into changing the internal energy of the system. This concept links directly back to the First Law of Thermodynamics, where the equation is simplified to \( \Delta U = Q \). This simplicity makes constant volume processes easier to handle in theoretical and practical applications.

Internal Energy

Internal energy refers to the total energy contained within a thermodynamic system. It encompasses kinetic and potential energy of the molecules within the system. Changes in internal energy are key indicators of energy transformation processes.
During processes such as heating or cooling, internal energy changes reflect the energy balance described by the First Law of Thermodynamics.
In a constant volume process, internal energy changes can be calculated directly from the heat added or removed, as work done is zero. Thus:\[ \Delta U = Q \]This means that any energy added to the system as heat results in a direct increase in internal energy if the volume does not change. Conversely, removing heat decreases internal energy. It is a simple yet powerful principle that allows scientists and engineers to predict and control the behavior of gases under various conditions, enhancing our understanding of material properties and energy efficiencies.