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Chapter 14: Problem 4

Calculate the work done in joules when a mechanical compressor exerting a constant pressure of \(350.0\) kPa compresses an air sample from a volume of \(500.0 \mathrm{~cm}^{3}\) to a volume of \(250.0 \mathrm{~cm}^{3}\).

Short Answer

Expert verified
The work done is -87.5 J.

Step by step solution

01

Understand the Formula for Work Done by a Gas

The work done by a gas when it is compressed or expanded at constant pressure is given by the formula:\[ W = P \cdot \Delta V \]where \( W \) is the work done, \( P \) is the constant pressure, and \( \Delta V \) is the change in volume.
02

Calculate the Change in Volume

First, calculate the change in volume \( \Delta V \). The initial volume is \( 500.0 \mathrm{~cm}^3 \) and the final volume is \( 250.0 \mathrm{~cm}^3 \).\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 250.0 - 500.0 = -250.0 \mathrm{~cm}^3 \]The negative sign indicates that the volume has decreased, which is expected during compression.
03

Convert Pressure to Consistent Units

Pressure is given as \( 350.0 \text{kPa} \). We need to convert this to \( \text{Pa} \) to be consistent with the volume units in \( \text{m}^3 \) (a conversion you'll do in the next step).1 kPa = 1000 Pa,\[ P = 350.0 \times 1000 = 350000 \text{ Pa} \]
04

Convert Volume to Consistent Units

Convert the volume change from \( \text{cm}^3 \) to \( \text{m}^3 \) since 1 m鲁 = 1,000,000 cm鲁:\[ \Delta V = -250.0 \mathrm{~cm}^3 \times \frac{1 ext{ m}^3}{1,000,000 ext{ cm}^3} = -0.00025 ext{ m}^3 \]
05

Calculate the Work Done

Now substitute the values for pressure and change in volume into the formula for work done:\[ W = P \cdot \Delta V = 350000 \times (-0.00025) \]\[ W = -87.5 \text{ J} \]The negative work indicates that energy is supplied to compress the gas.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

work done
In thermodynamics, when a force causes an object to move, work is done. For gases, work can occur when a gas expands or is compressed. It involves transferring energy from one system to another. The formula for calculating work done by a gas at constant pressure is given as:
  • \( W = P \cdot \Delta V \)
Where:
  • \( W \) represents the work done, measured in joules (J).
  • \( P \) is the constant pressure exerted, measured in pascals (Pa).
  • \( \Delta V \) is the change in volume of the gas, measured in cubic meters (m鲁).
In the context of gas compression, when volume decreases, work is typically negative. This indicates that external energy is applied to the gas to compress it. Understanding this concept is key in solving related thermodynamics problems effectively.
pressure-volume work
Pressure-volume work is a specific type of work associated with changes in the volume of a gas. This concept is crucial when studying processes involving gases expanding or compressing under constant pressure.
The idea is simple: if the volume of a gas changes while it remains under a constant pressure, work is performed. Whether it's work done on the gas or by the gas depends on the change:
  • If the volume of the gas decreases (compression), work is done on the gas.
  • If the volume of the gas increases (expansion), the gas does work.
In thermodynamic terms, pressure-volume work is considered when evaluating the energy transfers during these processes. By understanding this, students can better interpret why energy input or output leads to compression or expansion.
gas compression
Gas compression is a basic process in thermodynamics where the volume of a gas is decreased. This is commonly done using a mechanical compressor. During compression, the gas molecules are packed closer together, leading to a reduction in volume.
There are a few key points to keep in mind about gas compression:
  • When gas is compressed, its pressure often increases if temperature stays constant.
  • Compression work is often expressed as a negative value, indicating energy input to the system.
  • Efficient compressors can minimize energy losses, making them important in engineering and industrial applications.
By understanding gas compression, students can better appreciate the nuances of energy changes and work done in real-world systems.

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Most popular questions from this chapter

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