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

Short Answer

Expert verified
(a) Work done by the gas is 4988.4 J. (b) Change in internal energy is -4988.4 J.

Step by step solution

01

Understand the Problem

In an adiabatic expansion, a gas expands without heat exchange with its surroundings. For an ideal gas, changes in internal energy are related to temperature changes. We need to find the work done by the gas and the change in its internal energy given the temperature drop from 370 K to 290 K.
02

Use the Adiabatic Process Formula

For an adiabatic process in an ideal monatomic gas, the relationship between temperature and volume is given by \( TV^{\gamma-1} = \text{constant} \), where \( \gamma = \frac{5}{3} \) for a monatomic gas. However, to find work done, we need to use the First Law of Thermodynamics: \( \Delta U = Q - W \). Since \( Q = 0 \), \( \Delta U = -W \).
03

Calculate Internal Energy Change

The change in internal energy \( \Delta U \) for an ideal monatomic gas is given by the formula \( \Delta U = \frac{3}{2} nR \Delta T \), where \( n = 5 \) moles and \( \Delta T = 290 - 370 = -80 \text{ K} \). Thus, \( \Delta U = \frac{3}{2} \times 5 \times R \times (-80) \).
04

Substitute and Calculate

Substitute the values into the internal energy change formula: \( \Delta U = \frac{3}{2} \times 5 \times 8.314 \times (-80) \). Calculate \( \Delta U \). This results in \( \Delta U = -4988.4 \text{ J} \).
05

Calculate Work Done by the Gas

Since \( Q = 0 \) in an adiabatic process and using the relationship \( \Delta U = -W \), we have \( W = - \Delta U \). Therefore, \( W = 4988.4 \text{ J} \).
06

Note on Work and Sign

In thermodynamics, positive work done by the gas means the gas is doing work on the surroundings. Since the process is expansion, and work is done by the gas, \( W = 4988.4 \text{ J} \) is positive.

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental concept in physics that describes the conservation of energy. It states that the change in the internal energy of a system is equal to the amount of heat added to the system minus the work done by the system on its surroundings. In simpler terms, it can be written as: \[ \Delta U = Q - W \] Where:
  • \( \Delta U \) is the change in internal energy
  • \( Q \) is the heat added to the system
  • \( W \) is the work done by the system

In an adiabatic process, like the one described in the exercise, there is no heat exchange with the surroundings which makes \( Q = 0 \). This simplifies the equation to \( \Delta U = -W \). Hence, any change in internal energy is a result of the work done by or on the system.
Understanding this law, especially in adiabatic processes, is critical as it helps describe how energy transformations occur without external heat transfer.
Monatomic Ideal Gas
A monatomic ideal gas is a theoretical gas composed of single-atom gas particles, and it assumes that the particles do not interact with each other except via elastic collisions. The ideal gas law, which applies to monatomic gases, simplifies how we understand and predict their behavior.
This ideal scenario follows the equations: - **Equation of State**: \[ PV = nRT \]
  • Where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant (8.314 J/mol·K), and \( T \) is temperature.
- **Degrees of Freedom**: A monatomic gas has three degrees of translational freedom, making its specific heat capacities quite specific.These equations form the basis for exploring changes in an ideal gas's state, particularly in cases of temperature fluctuation or volume alterations, such as in an adiabatic expansion.
Internal Energy Change
The change in internal energy \( \Delta U \) of a gas is directly related to its temperature change and is crucial for understanding energy dynamics. In the context of a monatomic ideal gas, this change is calculated using the formula:\[ \Delta U = \frac{3}{2} nR \Delta T \]Where:
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant
  • \( \Delta T \) is the change in temperature
Here, the factor \( \frac{3}{2} \) comes from the three translational degrees of freedom in a monatomic gas. It reflects how the energy is distributed among these degrees of freedom, affecting kinetic energy and, by extension, temperature.In the example, this calculation considers a specific drop in temperature. Therefore, identifying how internal energy changes is vital for determining associated energetic phenomena.
Work Done by Gas
In thermodynamic processes, particularly notable in adiabatic expansions, the concept of work done by the gas plays a key role. Work \( W \) involves energy transfer that results when a gas expands or compresses against a force. For gases, this can be calculated by integrating the pressure with respect to volume.In adiabatic conditions where no heat is exchanged, the work done by the gas is directly responsible for changes in internal energy. Specifically, using the First Law of Thermodynamics, the relationship is: \[ W = - \Delta U \]Thus, in this exercise, calculating work becomes straightforward once the energy change is known, which was determined through temperature variations. The work done is positive if the gas expands, meaning it does work on its surroundings. Hence, achieving clarity on such terms provides comprehensive insights into the mechanics of gas behaviors under various thermodynamic processes.

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Most popular questions from this chapter

Suppose 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 concepts that are important in this problem. In doing \(16600 \mathrm{~J}\) of work, an engine rejects \(9700 \mathrm{~J}\) of heat. What is the efficiency of the engine?

Even at rest, the human body generates heat. The heat arises because of the body's metabolism - that is, the chemical reactions that are always occurring in the body to generate energy. In rooms designed for use by large groups, adequate ventilation or air conditioning must be provided to remove this heat. Consider a classroom containing 200 students. Assume that the metabolic rate of generating heat is \(130 \mathrm{~W}\) for each student and that the heat accumulates during a fifty-minute lecture. In addition, assume that the air has a molar specific heat of \(C_{V}=\frac{5}{2} R\) and that the room (volume \(=1200 \mathrm{~m}^{3}\), initial pressure \(=1.01 \times 10^{5} \mathrm{~Pa}\), and intial temperature \(=21^{\circ} \mathrm{C}\) ) is sealed shut. If all the heat generated by the students were absorbed by the air, by how much would the air temperature rise during a lecture?

A system undergoes a two-step process. In the first step, the internal energy of the system increases by \(228 \mathrm{~J}\) when \(166 \mathrm{~J}\) of work is done on the system. In the second step, the internal energy of the system increases by \(115 \mathrm{~J}\) when \(177 \mathrm{~J}\) of work is done on the system. For the overall process, find the heat. What type of process is the overall process? Explain.

A heat pump removes \(2090 \mathrm{~J}\) of heat from the outdoors and delivers \(3140 \mathrm{~J}\) of heat to the inside of a house. (a) How much work does the heat pump need? (b) What is the coefficient of performance of the heat pump?

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