Problem 10 - A Complete Resource Book in Chemistry for JEE Main

Question

If a gas absorbs \(200 \mathrm{~J}\) of heat and expands by 500 \(\mathrm{cm}^{3}\) against a constant pressure of \(2 \times 10^{\mathrm{s}} \mathrm{Nm}^{-2}\), then change in internal energy is (a) \(-200 \mathrm{~J}\) (b) \(-100 \mathrm{~J}\) (c) \(+100 \mathrm{~J}\) (d) \(+300 \mathrm{~J}\)

Step-by-Step Solution

Answer

The change in internal energy is +100 J, corresponding to option (c).

Step 1: Understand the Given Values

The problem states that a gas absorbs heat, which can be denoted as \( Q = +200 \, \text{J} \). The gas also expands by \( 500 \, \text{cm}^3 \) against a constant pressure of \( P = 2 \times 10^5 \, \text{N/m}^2 \). We are asked to find the change in internal energy.

Step 2: Convert Volume to Proper Units

The volume change is given in \( \text{cm}^3 \), so we need to convert it to \( \text{m}^3 \). We know that \( 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \). Thus, \( \Delta V = 500 \, \text{cm}^3 = 500 \times 10^{-6} \, \text{m}^3 = 5 \times 10^{-4} \, \text{m}^3 \).

Step 3: Calculate Work Done by the Gas

The work done by the gas during expansion is calculated using the formula \( W = P \Delta V \). Substituting the given values, \( W = 2 \times 10^5 \, \text{N/m}^2 \times 5 \times 10^{-4} \, \text{m}^3 = 100 \, \text{J} \).

Step 4: Apply the First Law of Thermodynamics

According to the first law of thermodynamics, the change in internal energy \( \Delta U \) is given by \( \Delta U = Q - W \). Substituting the calculated work and the given heat, \( \Delta U = 200 \, \text{J} - 100 \, \text{J} = 100 \, \text{J} \).

Step 5: Determine the Correct Answer

The change in internal energy is \( 100 \, \text{J} \), which corresponds to option (c) \(+100 \, \text{J} \).

Key Concepts

First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle in physics and a crucial part of understanding energy changes in a system. It states that the total energy of a closed system is constant. Simply put, energy cannot be created or destroyed but can only be transformed from one form to another. This is often represented by the formula:\[ \Delta U = Q - W \]where \( \Delta U \) represents the change in internal energy of the system, \( Q \) is the heat absorbed by the system, and \( W \) is the work done by the system. In simpler terms, the change in the internal energy of a system equals the heat added to the system minus the work done by the system.

• If \( Q > W \), the internal energy of the system increases.
• If \( Q < W \), the internal energy decreases.
• If \( Q = W \), the internal energy remains unchanged.

This law is a cornerstone in thermodynamics because it bridges the concepts of heat transfer with energy conservation in mechanical systems.

Internal Energy Change

Internal energy is the total energy contained within a system, encompassing all kinetic and potential energies of the molecules or atoms. The change in this energy, \( \Delta U \), tells us how much the energy level has shifted due to heat exchange and work done. In the exercise, the gas underwent a process where it absorbed heat and performed work, leading to a change in internal energy.
Using the First Law of Thermodynamics, we can calculate this change by subtracting the work done (\( W \)) from the heat absorbed (\( Q \)). - In this situation, the gas absorbed \(200 \, \text{J}\) of heat and performed \(100 \, \text{J}\) of work, leading to: \[ \Delta U = 200 \, \text{J} - 100 \, \text{J} = 100 \, \text{J} \] A positive \( \Delta U \) indicates an increase in internal energy, meaning the system's energy state has risen after the process.
- This reflects how the gas used its absorbed heat to do work while still increasing its energy state.

Work Done by Gas

In thermodynamics, work is carried out when a force is applied over a distance. For a gas, this typically involves expansion or contraction under pressure. The work done by a gas (\[W\]) during its expansion or compression can be expressed as:\[ W = P \Delta V \]Where \( P \) is the pressure the gas exerts and \( \Delta V \) is the change in volume. Work is positive when a gas expands (as it pushes against external pressure) and negative when it is compressed. In the exercise provided:
- The gas expanded by \(500 \text{ cm}^3\) or \(5 \times 10^{-4} \, \text{m}^3\).- Under a constant pressure of \(2 \times 10^5 \, \text{N/m}^2\), the work done was calculated as:\[ W = 2 \times 10^5 \, \text{N/m}^2 \times 5 \times 10^{-4} \, \text{m}^3 = 100 \, \text{J}. \]This work done shows how energy can be used by the gas to overcome an external force or to change its volume. Understanding this helps highlight the relationship between volume change and energy required or released in thermodynamic processes.