Problem 155 - A Complete Resource Book in Chemistry for JEE Main

Question

An ideal gas expands in volume from \(1 \times 10^{-3} \mathrm{~m}^{3}\) to 1 \(\times 10^{-2} \mathrm{~m}^{3}\) at \(300 \mathrm{~K}\) against a constant pressure of \(1 \times\) \(10^{5} \mathrm{Nm}^{-2}\). The work done is (a) \(-900 \mathrm{k} \mathrm{J}\) (b) \(-900 \mathrm{~J}\) (c) \(270 \mathrm{~kJ}\) (d) \(940 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The correct answer is (b) -900 J.

Step 1: Understand the Problem

The problem involves an ideal gas expanding against a constant external pressure. We are asked to calculate the work done during this expansion.

Step 2: Recall the Formula for Work Done by Gas

The work done by a gas during expansion or compression at constant pressure is calculated using the formula: \( W = P \Delta V \), where \( W \) is work, \( P \) is pressure, and \( \Delta V \) is the change in volume.

Step 3: Calculate Change in Volume

The initial volume \( V_1 \) is \( 1 \times 10^{-3} \ \mathrm{m}^3 \) and the final volume \( V_2 \) is \( 1 \times 10^{-2} \ \mathrm{m}^3 \). The change in volume \( \Delta V \) is given by \( V_2 - V_1 = (1 \times 10^{-2} - 1 \times 10^{-3}) \ \mathrm{m}^3 = 9 \times 10^{-3} \ \mathrm{m}^3 \).

Step 4: Substitute Values into the Formula

Substitute the values of pressure and change in volume into the formula: \( W = 1 \times 10^5 \ \mathrm{Nm}^{-2} \times 9 \times 10^{-3} \ \mathrm{m}^3 \).

Step 5: Calculate the Work Done

Calculate the work: \( W = 1 \times 10^5 \times 9 \times 10^{-3} = 9 \times 10^2 \ \mathrm{J} = 900 \ \mathrm{J} \). The work done by the gas during expansion is 900 J, but since the gas does work on the surroundings, it is \(- W = -900 \, \mathrm{J}\).

Step 6: Find the Correct Answer Option

The correct option is (b) \(-900 \, \mathrm{J}\) since the work done by the gas expanding against constant pressure is negative.

Key Concepts

Work Done by Gas

During an ideal gas expansion, the work performed by the gas is directed against an external pressure.
This process denotes energy is transferred from the gas to its surroundings. The work done by the gas is calculated by using the formula \[ W = P \Delta V \] where

• \( W \) represents the work done by the gas.
• \( P \) is the constant external pressure.
• \( \Delta V \) is the change in volume of the gas.

When the gas expands, work is performed to push the external pressure outwards, resulting usually in a negative work expression due to the chosen sign convention. It implies the system (gas) is doing work on its surroundings. The calculation from the original exercise shows the work as \[ W = 1 \times 10^5 \, \mathrm{Nm}^{-2} \times 9 \times 10^{-3} \, \mathrm{m}^3 = 900 \, \mathrm{J} \] However, since the work is done by the gas, it's conventionally considered negative: \[ W = -900 \, \mathrm{J} \].

Constant Pressure Process

In a constant pressure process, the pressure exerted against the gas remains the same throughout the whole process of expansion or compression.
This simplifies the equation for work, as the pressure factor becomes a constant, allowing us to focus purely on the volume change.
This scenario is typical when the system is allowed to exchange energy with its surroundings without altering the pressure value it faces. For systems within a fixed environment like a piston with a constant external pressure, the formula reaches a simple product of pressure and volume change, expressing simple thermodynamic exchanges. In academic terms, processes at constant pressure often occur in ideal setups to ease calculations and interpret results easily, making them straightforward for understanding expansions like the one discussed in the original exercise.

Volume Change Calculation

The change in volume, denoted by \( \Delta V \), is an essential part of determining the work done by the gas.
Given initial and final gas volumes, the difference between these provides the \( \Delta V \).
This step involves straightforward subtraction:\[ \Delta V = V_2 - V_1 \] In the original exercise, the initial volume \( V_1 \) was \( 1 \times 10^{-3} \, \mathrm{m}^3 \), and the final volume \( V_2 \) was \( 1 \times 10^{-2} \, \mathrm{m}^3 \). The change was hence calculated as:\[ \Delta V = (1 \times 10^{-2} - 1 \times 10^{-3}) \, \mathrm{m}^3 \] resulting in \( 9 \times 10^{-3} \, \mathrm{m}^3 \).
The volume change serves not only in work calculations but also signifies the extent and type of process occurring, providing insights into system behavior during expansion.