Problem 18 - Pearson Physics

Question

A gas with a constant pressure of \(270 \mathrm{kPa}\) does \(36,000 \mathrm{~J}\) of work as it expands. What was the change in volume of the gas?

Step-by-Step Solution

Answer

The change in volume is approximately \(0.1333 \mathrm{~m}^3\).

Step 1: Identify the formula

To find the change in volume of the gas when a known amount of work is done at constant pressure, we use the formula: \[ W = P \Delta V \]where:- \( W \) is the work done by the gas, given as \( 36,000 \mathrm{~J} \).- \( P \) is the constant pressure, given as \( 270 \mathrm{kPa} \).- \( \Delta V \) is the change in volume, which is what we need to find.

Step 2: Convert pressure units

Given that the pressure is in kPa, we convert it to Pa (Pascals) since work is given in Joules (1 J = 1 N·m). This ensures consistent SI units:\[270 \mathrm{kPa} = 270,000 \mathrm{~Pa}\]

Step 3: Rearrange the formula

Rearrange the formula to solve for the change in volume \( \Delta V \):\[ \Delta V = \frac{W}{P} \]

Step 4: Substitute known values

Substitute \( W = 36,000 \mathrm{~J} \) and \( P = 270,000 \mathrm{~Pa} \) into the rearranged formula:\[ \Delta V = \frac{36,000}{270,000} \]

Step 5: Calculate the change in volume

Perform the division to find \( \Delta V \):\[ \Delta V = 0.1333 \mathrm{~m}^3 \]

Key Concepts

Work Done

In thermodynamics, when we talk about **work done** by a gas, we refer to the energy transferred when the gas expands or contracts against an external pressure. Work done is a crucial part of many thermodynamic processes, as it quantifies the energy output or input during these transformations.

To calculate the work done by a gas, you can use the formula:

• \( W = P \Delta V \)
• "\( W \)" is the work done, usually measured in joules (J).
• "\( P \)" is the pressure exerted on the gas, in pascals (Pa).
• "\( \Delta V \)" is the change in the volume of the gas.

The work is positive if the gas does work on the surroundings (like pushing a piston), and negative if work is done on the gas. This concept helps us understand energy exchanges in engines and other systems where gas expansion occurs.

Pressure

**Pressure** is a measure of the force exerted by a substance per unit area. In gases, pressure results from the constant bombardment of gas molecules against the walls of its container.

When we deal with thermodynamics, pressure is often in units of pascals (Pa), where 1 pascal equals 1 newton per square meter. Sometimes, kilopascals (kPa) are used for convenience, where 1 kPa equals 1,000 pascals. In the exercise, we had:

• The gas pressure was given as 270 kPa, which we converted to 270,000 Pa for calculations.

Understanding pressure is essential because it affects how gases expand and contract. An increase in pressure can lead to a decrease in volume, and vice versa, which is a central theme in gas laws.

Volume Change

Volume change, represented as \( \Delta V \), is the difference between the final and initial volumes of a gas. It's crucial for determining how much a gas has expanded or contracted during a process. In our exercise, we found the change in volume using:

• \( \Delta V = \frac{W}{P} \)

Here, with a constant pressure and known work done, solving for \( \Delta V \) shows us how much space the gas occupies after expansion. Knowing volume change helps in applications like calculating capacities in engines, designing balloons, and understanding natural phenomena.

Gas Expansion

**Gas expansion** occurs when a gas increases in volume, usually due to added heat or a decrease in pressure. This expansion is a vital principle in thermodynamics, linked with temperature and pressure changes.

During expansion:

• The energy stored in the gas is used to push out against the pressure, doing work in the process.
• This work can be harnessed as mechanical energy, like in the cylinders of an engine.

In the exercise, the gas expanded while maintaining constant pressure, resulting in work done measurable as energy (joules). Understanding gas expansion explains operations in heat engines, atmospheric sciences, and even in simple combustion engines.