/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The volume of a monatomic ideal ... [FREE SOLUTION] | 91Ó°ÊÓ

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The volume of a monatomic ideal gas doubles in an isothermal expansion. By what factor does its pressure change?

Short Answer

Expert verified
The pressure is halved.

Step by step solution

01

Understanding the Problem

The problem involves a monatomic ideal gas undergoing an isothermal expansion. This means that the temperature remains constant. We need to find out how the pressure changes when the volume doubles.
02

Recall the Ideal Gas Law

The ideal gas law states: \[ PV = nRT \]Where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature.
03

Identify Isothermal Conditions

Since the gas is expanding isothermally, \( T \) is constant. Thus, \( nRT \) is constant in the ideal gas law equation during the process.
04

Relate Initial and Final States

The initial state is defined by \( P_1V_1 = nRT \) and the final state by \( P_2V_2 = nRT \). As \( nRT \) remains constant \( P_1V_1 = P_2V_2 \) will hold true.
05

Substitute Given Conditions

We know from the problem that the volume doubles, so \( V_2 = 2V_1 \). Substitute this into the equation: \[ P_1V_1 = P_2(2V_1) \]
06

Solve for Pressure Change

Simplifying the equation for \( P_2 \) we get:\[ P_2 = \frac{P_1V_1}{2V_1} = \frac{P_1}{2} \]This shows that the pressure is halved.

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Key Concepts

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

Isothermal Expansion
In the context of gases, an isothermal process is one where the temperature remains constant throughout. "Iso" means "same," and "thermal" relates to heat, so the words combined refer to a constant temperature process. During isothermal expansion, a gas expands while keeping its temperature unchanged. This type of process is common when considering ideal gases - hypothetical gases that perfectly follow the ideal gas law.

For an ideal gas undergoing isothermal expansion:
  • The energy added into the system is absorbed as work done by the gas, causing it to expand.
  • There is no change in internal energy because the temperature doesn't change.
  • According to the ideal gas equation, the product of pressure and volume remains constant with a constant temperature.
The equation underlying this isothermal change is:
\[ PV = ext{constant} \] This is derived from the ideal gas law, simplifying the expression whenever the temperature and number of moles of the gas remain constant during the change.
Pressure Change in Gases
Pressure change in gases is an important aspect when dealing with physical changes and processes. In our scenario of an isothermal expansion, the pressure changes because of the direct relationship between pressure and volume in the gas law.

Starting with the ideal gas equation \(PV = nRT\):
  • During the isothermal expansion, the volume of the gas doubles while the temperature stays the same.
  • This doubling of volume leads to the halving of pressure to maintain the constant product of pressure and volume.
For example, if the initial pressure (\(P_1\)) is at a certain value with volume (\(V_1\)), and then the volume (\(V_2\)) becomes \(2V_1\) during expansion, the pressure (\(P_2\)) will become \(\frac{P_1}{2}\).

This showcases how the characteristics of the ideal gas are such that when the volume increases in an isothermal expansion, the pressure decreases to reinforce the constant state itemized by the equation.
Monatomic Ideal Gas
Monatomic ideal gases are gases composed of single-atom molecules. Examples include helium, neon, and argon. These gases can be considered under the ideal gas law because they adhere closely to the assumptions inherent in this theoretical model. Understanding these gases helps in analyzing behaviors such as expansion and pressure changes.

For monatomic gases, these properties are noteworthy:
  • They exhibit relatively simple behavior due to the lack of molecular bonds that could affect interactions.
  • Their internal energy can be defined solely by their kinetic energy, which depends only on temperature.
  • When undergoing processes like isothermal expansion, their behavior can be predicted accurately with the ideal gas law.
These fundamental characteristics make it comparatively straightforward to calculate results or predict outcomes when working under idealized conditions, allowing for simplifications that aid in learning and comprehension of basic thermodynamics concepts.

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Most popular questions from this chapter

An engine receives 690 J of heat from a hot reservoir and gives off 430 J of heat to a cold reservoir. What are (a) the work done and (b) the efficiency of this engine?

One mole of an ideal monatomic gas is initially at a temperature of \(263 \mathrm{K}\). (a) Find the final temperature of the gas if \(3280 \mathrm{J}\) of heat are added to it and it does \(722 \mathrm{J}\) of work. (b) Suppose the amount of gas is doubled to two moles. Does the final temperature found in part (a) increase, decrease, or stay the same? Explain.

A refrigerator with a coefficient of performance of 1.75 absorbs \(3.45 \times 10^{4} \mathrm{J}\) of heat from the low-temperature reservoir during each cycle. (a) How much mechanical work is required to operate the refrigerator for a cycle? (b) How much heat does the refrigerator discard to the high-temperature reservoir during each cycle?

Engine A has an efficiency of \(66 \%\). Engine \(B\) absorbs the same amount of heat from the hot reservoir and exhausts twice as much heat to the cold reservoir. (a) Which engine has the greater efficiency? Explain. (b) What is the efficiency of engine B?

Energy from Gasoline Burning a gallon of gasoline releases \(1.19 \times 10^{8} \mathrm{J}\) of internal energy. If a certain car requires \(5.20 \times 10^{5} \mathrm{J}\) of work to drive one mile, (a) how much heat is given off to the atmosphere each mile, assuming the car gets 25.0 miles to the gallon? (b) If the miles per gallon of the car is increased, does the amount of heat released to the atmosphere increase, decrease, or stay the same? Explain.

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