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Chapter 19: Problem 31

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.

Short Answer

Expert verified
The internal energy and temperature of the gas increase during the adiabatic compression.

Step by step solution

01

Understand the Adiabatic Compression

In an adiabatic compression, no heat is exchanged with the surroundings. For an ideal gas, the change in internal energy (\[\Delta U\]) is equal to the work done on the gas, since \(\Delta U = q + w\), where \(q = 0\) in adiabatic processes. Thus, the change in internal energy equals the work done on the gas.
02

Apply the First Law of Thermodynamics

In an adiabatic process, \(q = 0\), so the first law of thermodynamics simplifies to \(\Delta U = w\). Since work is done on the gas during compression, the internal energy will increase.
03

Use the Adiabatic Process Relationship

For an adiabatic process, the relation between initial and final states of an ideal gas is given by \(P_1 V_1^\gamma = P_2 V_2^\gamma\), where \(\gamma = \frac{C_p}{C_v}\). For diatomic gases like CO, \(\gamma \approx 1.4\).
04

Calculate Final Pressure Using Adiabatic Relation

Given \(V_2 = \frac{1}{3} V_1\), substitute into the adiabatic relation: \[ P_1 V_1^\gamma = P_2 (\frac{1}{3} V_1)^\gamma \]Solve for \(P_2\):\[ P_2 = P_1 \left(3\right)^\gamma \]\(P_2 = 1.2 \text{ atm} \times 3^{1.4}\).
05

Calculate Change in Temperature

Use the relation for temperatures in adiabatic processes: \(T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}\). Since \(V_2 = \frac{1}{3}V_1\), solve for \(T_2\):\[ T_2 = T_1 \left(3\right)^{\gamma-1} \]. Elementary manipulation shows \(T_2 > T_1\), indicating an increase in temperature.
06

Conclusion about Internal Energy and Temperature Change

Since there is work done on the gas during compression, as indicated by the increasing temperature, the internal energy of the gas increases. Therefore, temperature and internal energy both increase during adiabatic compression of an ideal gas.

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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 principle for understanding gas behaviors in different conditions. It is represented by the equation \( PV = nRT \), where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume it occupies
  • \( n \) is the number of moles of the gas
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
This equation helps correlate the pressure, volume, and temperature of a gas under a given set of conditions. It assumes the gas behaves ideally, which means the gas particles do not attract or repel each other, and occupy no volume. In the context of the adiabatic process, the Ideal Gas Law is indirectly used to understand changes in these variables as the gas volume changes.
By knowing how these factors interact, we can predict how a gas will behave if it is compressed or expanded, such as in an adiabatic process.
Internal Energy
Internal energy is the total energy contained within a system, derived from the motion and interactions of the molecules within the gas. In ideal gases, internal energy is primarily dependent on the temperature of the gas and the number of particles it contains. Internal energy can be calculated using the equation:\[ \Delta U = nC_v\Delta T \]
Where \( \Delta U \) is the change in internal energy, \( n \) is the number of moles, \( C_v \) is the molar specific heat capacity at constant volume, and \( \Delta T \) is the change in temperature.
During an adiabatic process, since no heat is exchanged with the surroundings, any change in the internal energy of an ideal gas is due to the work done on or by the gas. Therefore, in an adiabatic compression, the work done on the gas increases its internal energy and temperature.
First Law of Thermodynamics
The First Law of Thermodynamics is a version of the law of conservation of energy, tailored for thermodynamic systems and processes. It is typically stated as:\[\Delta U = q + w\]Where:
  • \( \Delta U \) is the change in internal energy of the system
  • \( q \) is the heat added to the system
  • \( w \) is the work done on the system
For adiabatic processes, this law simplifies because no heat exchange occurs with the surroundings \((q = 0)\). Hence, it reduces to \( \Delta U = w \), meaning the entire change in internal energy stems from the work done on the gas.
In the context of the given problem, understanding this law helps explain why the internal energy and temperature of the gas increase during adiabatic compression. The work done compressing the gas increases its internal energy, illustrating a fundamental principle of energy transformation.

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Most popular questions from this chapter

CALC During the time 0.305 mol of an ideal gas under- goes an isothermal compression at \(22.0^{\circ} \mathrm{C}, 468 \mathrm{J}\) of work is done on it by the surroundings. (a) If the final pressure is 1.76 atm, what was the initial pressure? (b) Sketch a pV-diagram for the process.

CALC The temperature of 0.150 mol of an ideal gas is held constant at \(77.0^{\circ} \mathrm{C}\) while its volume is reduced to 25.0\(\%\) of its initial volume. The initial pressure of the gas is 1.25 atm. (a) Determine the work done by the gas. (b) What is the change in its internal energy? (c) Does the gas exchange heat with its surroundings? If so, how much? Does the gas absorb or liberate heat?

On a warm summer day, a large mass of air (atmospheric pressure \(1.01 \times 10^{5}\) 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 \(0.850 \times 10^{5}\) Pa. Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cooling for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of altitude, is called the dry adiabatic lapse rate.)

A gas undergoes two processes. In the first, the volume remains constant at 0.200 \(\mathrm{m}^{3}\) and the pressure increases from \(2.00 \times 10^{5}\) Pa to \(5.00 \times 10^{5}\) Pa. The second process is a compression to a volume of 0.120 \(\mathrm{m}^{3}\) at a constant pressure of \(5.00 \times 10^{5} \mathrm{Pa}\) (a) In a \(p V\) -diagram, show both processes. (b) Find the total work done by the gas during both processes.

An air pump has a cylinder 0.250 \(\mathrm{m}\) long with a mov- able piston. The pump is used to compress air from the atmosphere (at absolute pressure \(1.01 \times 10^{5}\) Pa) into a very large tank at \(4.20 \times 10^{5}\) Pa gauge pressure. (For air, \(C_{V}=20.8 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) ) (a) The piston begins the compression stroke at the open end of the cylinder. How far down the length of the cylinder has the piston moved when air first begins to flow from the cylinder into the tank? Assume that the compression is adiabatic. (b) If the air is taken into the pump at \(27.0^{\circ} \mathrm{C}\) , what is the temperature of the compressed air? (c) How much work does the pump do in putting 20.0 mol of air into the tank?
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