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The pressure of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\) doubles during an adiabatic compression. What is the ratio of the final volume to the initial volume?

Short Answer

Expert verified
The ratio of the final volume to the initial volume is approximately 0.660.

Step by step solution

01

Understanding Adiabatic Process Equation

For an adiabatic process, the relation between pressure \( P \), volume \( V \), and the specific heat ratio \( \gamma \) is given by \( PV^{\gamma} = \text{constant} \). In this case, \( \gamma = \frac{5}{3} \).
02

Applying the Pressure Change

Since the pressure doubles, we have \( P_2 = 2P_1 \). We need to find the volume ratio \( \frac{V_2}{V_1} \).
03

Using the Adiabatic Condition

From the relation \( PV^{\gamma} = \text{constant} \), we have \( P_1V_1^{\gamma} = P_2V_2^{\gamma} \). Substitute \( P_2 = 2P_1 \) to get \( P_1V_1^{\gamma} = 2P_1V_2^{\gamma} \).
04

Simplifying the Equation

Cancel \( P_1 \) from both sides of the equation: \( V_1^{\gamma} = 2V_2^{\gamma} \).
05

Isolating Volume Ratio

Rearrange the equation: \( \left(\frac{V_2}{V_1}\right)^{\gamma} = \frac{1}{2} \). Substitute \( \gamma = \frac{5}{3} \) to get \( \left(\frac{V_2}{V_1}\right)^{\frac{5}{3}} = \frac{1}{2} \).
06

Solving for Volume Ratio

Take both sides to the power of \( \frac{3}{5} \): \( \frac{V_2}{V_1} = \left(\frac{1}{2}\right)^{\frac{3}{5}} \). Calculate this to find the ratio, \( \frac{V_2}{V_1} \approx 0.660 \).

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical concept in physics and chemistry. It describes a gas composed of many randomly moving point particles that do not interact except when they collide elastically. The ideal gas law is expressed as \( PV = nRT \), where:
  • \( P \) is the pressure,
  • \( V \) is the volume,
  • \( n \) is the number of moles,
  • \( R \) is the gas constant,
  • and \( T \) is the temperature in Kelvin.

Ideal gases help us understand and predict the behavior of real gases under different conditions. However, in reality, no gas perfectly follows the ideal gas law. Deviations occur at high pressures and low temperatures. Despite this, the ideal gas concept is incredibly useful for simplifying calculations and gaining insights into thermodynamics.
Specific Heat Ratio
The specific heat ratio, often denoted by \( \gamma \), is a crucial factor in thermodynamics. It represents the ratio of specific heats at constant pressure \( C_p \) and constant volume \( C_v \). Mathematically, \( \gamma = \frac{C_p}{C_v} \). For a monatomic ideal gas, this ratio is generally \( \frac{5}{3} \).
The specific heat ratio provides vital information about the energy required to change the temperature of a gas under constant volume and pressure conditions. This ratio plays a key role in adiabatic processes, where no heat is exchanged with the surroundings.
  • If \( \gamma \) is higher, the gas will experience greater temperature changes during compression or expansion.
  • Larger \( \gamma \) values indicate stronger molecular interactions.
Understanding \( \gamma \) helps us predict gas behavior in various thermodynamic transformations. It is especially valuable when analyzing adiabatic processes.
Volume Ratio
In the context of the exercise, the volume ratio \( \frac{V_2}{V_1} \) represents the change in a gas's volume before and after an adiabatic process.
When the pressure of a gas changes adiabatically, its volume changes in a specific way according to the relation \( PV^{\gamma} = \text{constant} \), where \( \gamma \) is the specific heat ratio.
  • In adiabatic compression, as pressure increases, volume decreases.
  • The volume ratio can help determine whether the final volume is larger or smaller than the initial volume.
This calculation is vital because it indicates how much a gas has been compressed or expanded without transferring heat. The exercise shows how the specific heat ratio \( \gamma = \frac{5}{3} \) plays into finding this volume ratio after a pressure change.
Thermodynamics
Thermodynamics is the science of energy and its transformations. It encompasses several key concepts, including temperature, energy transfer, and the laws governing these processes.
The primary laws of thermodynamics include:
  • The Zeroth Law, which defines temperature.
  • The First Law, equivalent to the conservation of energy.
  • The Second Law, which introduces the concept of entropy.
  • The Third Law, which concerns absolute zero temperature.

In the context of gases and their behavior, thermodynamics provides frameworks for understanding processes like adiabatic transformations.
Adiabatic processes are those where no heat is transferred to or from the system. Instead, the system's internal energy changes and affects temperature and pressure.
Thermodynamic principles allow us to analyze how gases interact under different conditions, contributing to applications in engines, heating systems, and atmospheric science.

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Most popular questions from this chapter

A Carnot engine uses hot and cold reservoirs that have temperatures of 1684 and 842 \(\mathrm{K}\) , respectively. The input heat for this engine is \(\left|Q_{\mathrm{H}}\right|\) The work delivered by the engine is used to operate a Carnot heat pump. The pump removes heat from the \(842-\mathrm{K}\) reservoir and puts it into a hot reservoir at a temperature \(T^{\prime}\) . The amount of heat removed from the \(842-\mathrm{K}\) reservoir is also \(\left|Q_{\mathrm{H}}\right| .\) Find the temperature \(T^{\prime}\)

Heat \(Q\) flows spontaneously from a reservoir at 394 \(\mathrm{K}\) into a reservoir at 298 \(\mathrm{K}\) . Because of the spontaneous flow, 2800 \(\mathrm{J}\) of energy is rendered unavailable for work when a Carnot engine operates between the reservoir at 298 \(\mathrm{K}\) and a reservoir at 248 \(\mathrm{K}\) . Find \(Q\) .

Suppose that the gasoline in a car engine burns at \(631^{\circ} \mathrm{C}\) , while the exhaust temperature (the temperature of the cold reservoir) is \(139^{\circ} \mathrm{C}\) and the outdoor temperature is \(27^{\circ} \mathrm{C}\) . Assume that the engine can be treated as a Carnot engine (a gross oversimplification). In an attempt to increase mileage performance, an inventor builds a second engine that functions between the exhaust and outdoor temperatures and uses the exhaust heat to produce additional work. Assume that the inventor's engine can also be treated as a Carnot engine. Determine the ratio of the total work produced by both engines to that produced by the first engine alone.

Multiple-Concept Example 6 deals with the same concepts as this problem does. What is the efficiency of a heat engine that uses an input heat of \(5.6 \times 10^{4} \mathrm{J}\) and rejects \(1.8 \times 10^{4} \mathrm{J}\) of heat?

A monatomic ideal gas expands at constant pressure. (a) What percentage of the heat being supplied to the gas is used to increase the internal energy of the gas? (b) What percentage is used for doing the work of expansion?

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