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Chapter 18: Problem 24

During an adiabatic process, the temperature of 3.92 moles of a monatomic ideal gas drops from \(485^{\circ} \mathrm{C}\) to \(205^{\circ} \mathrm{C}\). For this gas, find (a) the work it does, (b) the heat it exchanges with its surroundings, and (c) the change in its internal energy.

Short Answer

Expert verified
Work done: 13721.376 J; Heat exchanged: 0; Change in internal energy: -13721.376 J.

Step by step solution

01

Convert Temperatures to Kelvin

First, convert the temperatures from degrees Celsius to Kelvin so that you can use them in calculations involving the ideal gas law and thermodynamic equations. Convert the initial temperature: \( T_1 = 485 + 273.15 = 758.15 \text{ K} \) and the final temperature: \( T_2 = 205 + 273.15 = 478.15 \text{ K} \).
02

Use the First Law of Thermodynamics

During an adiabatic process, there is no heat exchange between the system and its surroundings, which means \( Q = 0 \). The first law states \( \Delta U = Q - W \), and since \( Q = 0 \), \( \Delta U = -W \).
03

Determine Change in Internal Energy

The change in internal energy for a monatomic ideal gas is given by \( \Delta U = \frac{3}{2} nR\Delta T \), where \( n = 3.92 \) moles and \( \Delta T = T_2 - T_1 = 478.15 - 758.15 = -280 \) K. Substituting the values, \( \Delta U = \frac{3}{2} \times 3.92 \times 8.314 \times (-280) \). Calculate \( \Delta U \).
04

Calculate Change in Internal Energy

Calculate \( \Delta U \) using the expression from Step 3: \[ \Delta U = \frac{3}{2} \times 3.92 \times 8.314 \times (-280) = -13721.376 \text{ J} \]. This change indicates that the internal energy has decreased.
05

Determine Work Done by the Gas

Since \( \Delta U = -W \), the work done by the gas is \( W = -(-13721.376) = 13721.376 \text{ J} \).
06

Determine Heat Exchanged

During an adiabatic process, as established, there is no heat exchanged with the surroundings, so \( Q = 0 \).
07

Summarize Results

Finally, summarize the results: (a) The work done by the gas is \( 13721.376 \text{ J} \). (b) The heat exchanged \( Q = 0 \). (c) The change in internal energy \( \Delta U = -13721.376 \text{ J} \).

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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 principle that forms the basis of understanding energy transformations in thermodynamics. It can be simply stated as "energy cannot be created or destroyed, only transformed from one form to another." In mathematical terms, this law is written as \( \Delta U = Q - W \), where:
  • \( \Delta U \) represents the change in internal energy of the system,
  • \( Q \) symbolizes the heat exchanged with the surroundings,
  • \( W \) signifies the work done by or on the system.
In an adiabatic process, no heat is exchanged with the surroundings, thus \( Q = 0 \). This simplifies the equation to \( \Delta U = -W \), indicating that any work done by the system results in a change in the internal energy. Therefore, the First Law establishes a direct relationship between the work done by the gas and its internal energy change during an adiabatic process.
Internal Energy of Ideal Gas
The internal energy of an ideal gas is a measure of the total kinetic energy of the gas particles. For a monatomic ideal gas, this can be represented as \( U = \frac{3}{2} nRT \), where:
  • \( n \) is the number of moles of the gas,
  • \( R \) is the universal gas constant, approximately \( 8.314 \text{ J/mol路K} \),
  • \( T \) is the temperature in Kelvin.
During an adiabatic process, the internal energy changes due to work done, rather than heat exchange. The change in internal energy is given by \( \Delta U = \frac{3}{2} nR \Delta T \), where \( \Delta T \) is the change in temperature. For the provided exercise, where the temperature drops, the calculation yields a negative \( \Delta U \), indicating a reduction in the internal energy of the gas.
Work Done by Gas
In thermodynamics, work done by a gas during a process is an expression of energy transfer resulting from volume changes under pressure. For an adiabatic process, the work done is directly related to the change in the gas's internal energy. Given that \( \Delta U = -W \) in adiabatic conditions, any decrease in internal energy corresponds to positive work done by the gas.The work can be calculated when you know the change in internal energy: \( W = - \Delta U \). In our exercise, we found the work done by the gas to be \( 13721.376 \text{ J} \). This positive value indicates that the gas has done work on its surroundings as it expanded, employing the energy lost in its internal structure.

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Most popular questions from this chapter

A cylinder with a movable piston holds 2.75 mol of argon at a constant temperature of \(295 \mathrm{K}\). As the gas is compressed isothermally, its pressure increases from 101 kPa to 121 kPa. Find \((a)\) the final volume of the gas, \((b)\) the work done by the gas, and \((c)\) the heat added to the gas.

If a heat engine does 2700 J of work with an efficiency of \(0.18,\) find \((a)\) the heat taken in from the hot reservoir and (b) the heat given off to the cold reservoir. \((c)\) If the efficiency of the engine is increased, do your answers to parts (a) and (b) increase, decrease, or stay the same? Explain.

A Carnot engine operates between a hot reservoir at the Kelvin temperature \(T_{h}\) and a cold reservoir at the Kelvin temperature \(T_{C}\) (a) If both temperatures are doubled, does the efficiency of the engine increase, decrease, or stay the same? Explain. (b) If both temperatures are increased by \(50 \mathrm{K}\), does the efficiency of the engine increase, decrease, or stay the same? Explain.

An ideal gas doubles its volume in one of three different ways: (i) at constant pressure; (ii) at constant temperature; (iii) adiabatically. Explain your answers to each of the following questions: (a) In which expansion does the gas do the most work? (b) In which expansion does the gas do the least work? (c) Which expansion results in the highest final temperature? (d) Which expansion results in the lowest final temperature?

Suppose the temperature of the hot reservoir is increased by \(16 \mathrm{K}\), from \(576 \mathrm{K}\) to \(592 \mathrm{K}\), and that the temperature of the cold reservoir is also increased by \(16 \mathrm{K}\), from \(305 \mathrm{K}\) to \(321 \mathrm{K}\). (a) Is the new efficiency greater than, less than, or equal to 0.470? Explain. (b) What is the new efficiency? (c) What is the change in entropy of the hot reservoir when \(1050 \mathrm{J}\) of heat is drawn from it? (d) What is the change in entropy of the cold reservoir?
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