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Chapter 15: Problem 32

The work done by one mole of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right)\) in expanding adiabatically is 825 \(\mathrm{J}\) . The initial temperature and volume of the gas are 393 \(\mathrm{K}\) and 0.100 \(\mathrm{m}^{3} .\) Obtain \(\quad\) (a) the final temperature and (b) the final volume of the gas.

Short Answer

Expert verified
(a) Final temperature \( T_f \approx 327.9 \text{ K} \). (b) Final volume \( V_f \approx 0.131 \text{ m}^3 \).

Step by step solution

01

Use the adiabatic process formula

For an adiabatic process, the work done by the gas is given by the formula \( W = \frac{R}{\gamma-1}(T_i - T_f) \), where \( R \) is the ideal gas constant \( 8.314 \text{ J/mol路K} \), \( \gamma = \frac{5}{3} \), \( T_i \) is the initial temperature, and \( T_f \) is the final temperature. Set up the equation substituting in \( W = 825 \text{ J} \), \( T_i = 393 \text{ K} \): \[ 825 = \frac{8.314}{\frac{5}{3} - 1}(393 - T_f) \]
02

Solve for final temperature \(T_f\)

Rearrange the equation from Step 1 to solve for \( T_f \): \[ 825 = \frac{8.314}{\frac{2}{3}}(393 - T_f) \] Calculate \( \frac{8.314}{\frac{2}{3}} = 12.471 \). Then, \[ T_f = 393 - \frac{825}{12.471} \]Calculate to find \[ T_f \approx 327.9 \text{ K} \]
03

Use adiabatic relation for volume and temperature

For adiabatic process, temperature and volume relation is given by \( T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1} \). Substitute given \( \gamma = \frac{5}{3} \), \( T_i = 393 \text{ K} \), \( T_f = 327.9 \text{ K} \), and \( V_i = 0.100 \text{ m}^3 \): \[ 393 \times (0.100)^{\frac{2}{3}} = 327.9 \times V_f^{\frac{2}{3}} \]
04

Solve for final volume \(V_f\)

Rearrange to solve for \( V_f \): \[ V_f^{\frac{2}{3}} = \frac{393 \times (0.100)^{\frac{2}{3}}}{327.9} \]Calculate \( (0.100)^{\frac{2}{3}} \approx 0.04642 \), then compute\[ V_f^{\frac{2}{3}} \approx \frac{393 \times 0.04642}{327.9} \approx 0.0556 \]Raise both sides to power \( \frac{3}{2} \) to solve for \( V_f \):\[ V_f \approx (0.0556)^{\frac{3}{2}} \approx 0.131 \text{ m}^3 \]

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry and physics. It describes the behavior of an ideal gas using the relationship \( PV = nRT \). Here, \( P \) stands for pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
  • The ideal gas constant, \( R \), has a value of 8.314 J/mol路K.
  • The law assumes a gas consists of a large number of small particles that move randomly and collide elastically.
  • This equation is applicable under moderate conditions of temperature and pressure.
Understanding the ideal gas law helps us relate temperature, pressure, and volume, which is crucial in solving thermodynamic problems.
Monatomic Gases
Monatomic gases are gases composed of single atoms. Examples include the noble gases like helium, neon, and argon. They are often used in fundamental studies because of their simplicity compared to more complex polyatomic gases.
The behavior of a monatomic ideal gas under different processes, such as the adiabatic process, can be analyzed precisely:
  • The heat capacity ratio (\( \gamma \)) for a monatomic gas is \( \frac{5}{3} \).
  • This ratio arises from how energy is partitioned among translational degrees of freedom.
In problem-solving, understanding this specific ratio aids in analyzing processes like adiabatic expansions or compressions.
Thermodynamics
Thermodynamics is the study of energy transformations within systems. It provides essential principles to explain how gases behave under various conditions, such as during an adiabatic process.
  • An adiabatic process is one where no heat is transferred to or from the system.
  • In such a process, the work done by or on the gas results in temperature changes.
  • Key laws of thermodynamics guide this: the first law states energy conservation, while the adiabatic condition uses specific heat ratios to relate changes in temperature and volume.
These principles help us understand how an ideal gas's temperature and volume change without external heat addition or loss.
Work Done in Gases
The concept of work in gases relates to the energy transfer that occurs when a gas expands or is compressed. In thermodynamics, the work done by a gas during expansion is significant, especially in adiabatic processes.
  • The work done, \( W \), in an adiabatic process can be expressed using the formula \( W = \frac{R}{\gamma - 1}(T_i - T_f) \).
  • In this case, \( T_i \) and \( T_f \) are the initial and final temperatures.
  • Understanding this helps us track energy flow and predict how state variables, like temperature and volume, will change after the gas does work.
Work calculations within gas systems enable a deeper grasp of energy efficiency and thermodynamic behavior.

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Most popular questions from this chapter

A Carnot heat pump operates between an outdoor temperature of 265 \(\mathrm{K}\) and an indoor temperature of 298 \(\mathrm{K}\) . Find its coefficient of performance.

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.

Five moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 370 to 290 \(\mathrm{K}\) . Determine \(\quad\) (a) the work done (including the algebraic sign) by the gas, and (b) the change in its internal energy.

Carnot engine \(\mathrm{A}\) has an efficiency of \(0.60,\) and Carnot engine \(\mathrm{B}\) has an efficiency of \(0.80 .\) Both engines utilize the same hot reservoir, which has a temperature of 650 \(\mathrm{K}\) and delivers 1200 \(\mathrm{J}\) of heat to each engine. Find the magnitude of the work produced by each engine and the temperatures of the cold reservoirs that they use.

The temperatures indoors and outdoors are 299 and \(312 \mathrm{K},\) respectively. A Carnot air conditioner deposits \(6.12 \times 10^{5} \mathrm{J}\) of heat outdoors. How much heat is removed from the house?
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