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Chapter 19: Problem 61

A monatomic ideal gas expands slowly to twice its original volume, doing 300 \(\mathrm{J}\) of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is (a) isothermal; (b) adiabatic; (c) isobaric.

Short Answer

Expert verified
(a) Heat added: 300 J, Change in internal energy: 0 J. (b) Heat added: 0 J, Change in internal energy: -300 J. (c) Heat added: 500 J, Change in internal energy: 200 J.

Step by step solution

01

Identify Given Data and Process Type

We start by noting that the ideal gas is monatomic and performs 300 J of work, expanding to twice its original volume. We will solve this problem for three different processes: (a) isothermal, (b) adiabatic, and (c) isobaric.
02

Understand Isothermal Process (a)

In an isothermal process, the temperature remains constant. For an ideal gas, the change in internal energy is related to the change in temperature: \( \Delta U = 0 \). According to the first law of thermodynamics \( \Delta U = Q - W \), hence, \( Q = W \). Therefore, the heat added is \( Q = 300 \text{ J} \).
03

Analyze Adiabatic Process (b)

For an adiabatic process, there is no heat transfer into or out of the system, so \( Q = 0 \). The work done by the gas is equal to the change in internal energy: \( \Delta U = -W \). Thus, the change in internal energy is \( \Delta U = -300 \text{ J} \).
04

Examine Isobaric Process (c)

In an isobaric process, the pressure remains constant while the gas expands. Using the first law of thermodynamics \( \Delta U = Q - W \), and noting that the specific heat at constant pressure for a monatomic ideal gas is given by \( C_p = \frac{5}{2}R \), we need to calculate \( \Delta U \) and \( Q \). Using \( \Delta U = \frac{3}{2}nR\Delta T \) and \( W = p \Delta V \). Since \( p \Delta V = 300 \text{ J} \), and to find \( Q \), we use: \( Q = \Delta U + W \) which requires solving \( \Delta U = \frac{2}{3}W \). Therefore, \( \Delta U = 0.4 * 300 \text{ J} = 200 \text{ J} \), thus \( Q = 200 \text{ J} + 300 \text{ J} = 500 \text{ J} \).

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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 gas composed of a vast number of randomly moving particles that are not subject to intermolecular forces. The concept of ideal gases lays the groundwork for understanding various gas processes and simplifies the analysis by making certain assumptions:
  • Particles have negligible volume compared to the container's volume.
  • The particles move in straight lines and only change direction when they collide with the walls of the container.
  • All collisions are perfectly elastic, meaning there is no energy loss during collisions.
  • Gas behavior can be described by the ideal gas law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature.
While no gas is truly "ideal," many real gases behave like ideal gases under standard conditions, making this model useful in predictions and calculations. Understanding an ideal gas is essential for studying different thermodynamic processes.
Isothermal Process
An isothermal process is a thermodynamic process where the temperature of a system remains constant. For an ideal gas undergoing an isothermal change, the internal energy does not change because internal energy depends on temperature:
  • According to the first law of thermodynamics, the change in internal energy \( \Delta U = Q - W \), where \( Q \) is the heat added to the system and \( W \) is the work done by the system.
  • In an isothermal process for an ideal gas, since \( \Delta U = 0 \), it follows that \( Q = W \). This means all the work done by the gas is compensated by heat flowing into the gas.
For example, if an ideal gas does 300 J of work while expanding isothermally, 300 J of heat must be added to the gas to keep the temperature constant. Isothermal processes are typically slow to allow for heat exchange and can often be represented on pressure-volume graphs as hyperbolic curves.
Adiabatic Process
An adiabatic process is characterized by the complete absence of heat exchange between the system and its surroundings. This implies that all the energy expended as work comes from the interior of the gas:
  • The first law of thermodynamics again tells us \( \Delta U = Q - W \), but for an adiabatic process, \( Q = 0 \), simplifying to \( \Delta U = -W \).
  • This means if a gas does work on its surroundings, it loses an amount of internal energy equal to the work done. Conversely, if work is done on the gas, its internal energy increases.
In our exercise's case, the gas performs 300 J of work, meaning \( \Delta U = -300 \text{ J} \). Adiabatic processes can occur quickly, such as in gas compressions or expansions, like in piston engines, where there isn't enough time for heat exchange.
Isobaric Process
An isobaric process occurs at a constant pressure. This is typical in scenarios where gases expand or contract within a piston that can move freely to adjust to pressure changes:
  • When studying an isobaric process of an ideal gas, we use the first law again: \( \Delta U = Q - W \).
  • For a monatomic ideal gas, the relationship with changes in internal energy is given by \( \Delta U = \frac{3}{2}nR\Delta T \).
  • The work done by the gas is \( W = P \Delta V \).
In the step-by-step exercise, this is expressed as the gas doing 300 J of work. Knowing that \( \Delta U = 0.4 \cdot 300 = 200 \text{ J} \), the heat \( Q \) added to the system includes both the work done and the increase in internal energy, leading to \( Q = 500 \text{ J} \). Isobaric processes are useful to describe many natural phenomena, such as atmospheric processes in meteorology, and can often be analyzed in heat engines where expansion or compression occurs at constant pressure.

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Most popular questions from this chapter

A cylinder with a piston contains 0.150 mol of mitrogen at \(1.80 \times 10^{5} \mathrm{Pa}\) and 300 \(\mathrm{K}\) . The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a \(p V\) -diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

Two moles of carbon monoxide (CO) start at a pressure of 1.2 atm and a volume of 30 liters. The gas is then compressed adiabatically to \(\frac{1}{3}\) this volume. Assume that the gas may be treated as ideal. What is the change in the internal energy of the gas? Does the internal energy increase or decrease? Does the temperature of the gas increase or decrease during this process? Explain.

Two moles of an ideal gas are heated at constant pressure from \(T=27^{\circ} \mathrm{C}\) to \(T=107^{\circ} \mathrm{C}\) (a) Draw a \(p V\) -diagram for this process. (b) Calculate the work done by the gas.

Doughnuts: Breakfast of Champions! A typical doughnut contains \(2.0 \mathrm{~g}\) of protein. \(17.0 \mathrm{~g}\) of carbohydrates, and \(7.0 \mathrm{~g}\) of fat. The average food energy values of these substances are \(4.0 \mathrm{kcal} / \mathrm{g}\) for protein and carbohydrates and \(9.0 \mathrm{kcal} / \mathrm{g}\) for fat. (a) During hcavy exercise, an average person uscs cnergy at a rate of 510 keal \(/\) h. How long would you have to exercise to "work off"one doughnut? (b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be \(60 \mathrm{~kg}_{4}\) and express your answer in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\)

On a warm summer day, a large mass of air (atmospheric pressure \(1.01 \times 10^{5} \mathrm{Pa}\) ) is heated by the ground to a temperature of \(26.0^{\circ} \mathrm{C}\) and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why? Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only \(0.850 \times 10^{5} \mathrm{Pa}\) . Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cool- ing for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of altitude, is called the dry adiabatic lapse rate.)
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