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A quantity of an ideal gas undergoes an isother-mal expansion at \(20^{\circ} \mathrm{C}\) and does \(3.0 \times 10^{3} \mathrm{~J}\) of work on its surroundings in the process. (a) Will the entropy of the gas (1) increase, (2) remain the same, or (3) decrease? Explain. (b) What is the change in the entropy of the gas?

Short Answer

Expert verified
(a) The entropy will increase. (b) The change in entropy is approximately 10.24 J/K.

Step by step solution

01

Analyze the Process

This is an isothermal expansion, which means the temperature of the gas remains constant. During an isothermal expansion of an ideal gas, the volume increases, and hence the entropy of the gas increases as well.
02

Identify the Formula for Entropy Change

For an isothermal process in a gas, the change in entropy \( \Delta S \) can be calculated using the formula:\[ \Delta S = \frac{Q}{T} \]where \( Q \) is the heat transferred to the gas and \( T \) is the temperature in Kelvin.
03

Calculate Work Done

Given that the work done \( W \) by the gas on the surroundings is \( 3.0 \times 10^{3} \) J, for an isothermal process of an ideal gas, \( Q = W \) because the internal energy does not change. Therefore, \( Q = 3.0 \times 10^{3} \) J.
04

Convert Temperature to Kelvin

The given temperature is \( 20^{\circ} \mathrm{C} \). To convert this to Kelvin, use the conversion formula:\[ T = 20 + 273 = 293 \text{ K} \]
05

Calculate Change in Entropy

Substitute \( Q = 3.0 \times 10^{3} \) J and \( T = 293 \) K into the entropy change formula:\[ \Delta S = \frac{3.0 \times 10^{3}}{293} \approx 10.24 \text{ J/K} \]

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Key Concepts

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

Understanding Entropy
Entropy is a fundamental concept in thermodynamics that quantifies the degree of disorder or randomness in a system.
In more straightforward terms, it measures how spread out or chaotic energy is in a system.
When an ideal gas expands isothermally (at constant temperature), its entropy increases because the gas molecules have more space to occupy and hence can be arranged in more ways.
This is because:
  • The number of microstates, or possible configurations, accessible to the gas increases.
  • More energy paths lead to greater disorder.
Consequently, in an isothermal process, where temperature remains constant but volume increases, the system absorbs heat, causing an increase in entropy.
For example, if an ideal gas does work on its surroundings by expanding isothermally, as in the original exercise, it indicates that heat flows into the system, further increasing entropy.
Isothermal Processes Explained
An isothermal process is a type of thermodynamic process where the temperature remains constant.
This means any heat added to the system is used to do work, without altering the internal energy.
Here’s why isothermal processes are significant:
  • They often occur in slow processes where thermal equilibrium is maintained.
  • Since temperature is constant, the internal energy of an ideal gas remains unchanged.
In isothermal expansion, like in the given exercise:
  • The gas expands, doing work on the surroundings.
  • Heat is added to the gas to keep the temperature constant.
  • The entire heat energy is transferred to the work done, typically measured by the equation \( Q = W \).
Understanding this concept can explain why the given expansion does not change internal energy and why the heat supplied results directly in the work done, leading to an increase in entropy due to energy redistribution.
Ideal Gas Law Fundamentals
The Ideal Gas Law is a cornerstone in understanding how gases behave under various conditions.
Its formula is expressed as:
  • \( PV = nRT \)
  • Where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature in Kelvin.
In the context of the problem, this law helps explain how, despite temperature being constant during an isothermal expansion, volume changes interact with pressure changes:
  • As the gas expands, volume \( V \) increases.
  • To maintain the product \( PV \) at constant \( nRT \), pressure \( P \) must decrease.
  • In essence, while one property increases, another must decrease to keep the equation balanced.
This principle is vital for solving many thermodynamic problems, as it provides a clear relationship between different properties of the gas, such as in our original exercise with isothermal processes, ensuring that even as volume changes, the other variables adjust appropriately to maintain equilibrium.

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Most popular questions from this chapter

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