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Chapter 19: Problem 35

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 increases during the adiabatic compression process.

Step by step solution

01

Understanding Adiabatic Process

In an adiabatic process, no heat is exchanged with the surroundings. Therefore, the change in internal energy of the gas is equal to the work done on or by the gas. For an ideal gas, this change can be related to temperature changes using the internal energy formula.
02

Initial Parameters Setup

Given: Initial pressure \(P_1 = 1.2\, \text{atm}\), Initial volume \(V_1 = 30 \text{ L}\), and the final volume \(V_2 = \frac{1}{3} \times 30\, \text{L} = 10\, \text{L}\). We need to find the change in internal energy \(\Delta U\).
03

Using Ideal Gas Law

For CO, molar specific heat at constant volume \(C_v = 5/2 R\), where \(R\) is the universal gas constant. The change in internal energy is given by \(\Delta U = nC_v\Delta T\), where \(n = 2\) moles and \(\Delta T\) is the change in temperature.
04

Applying Adiabatic Condition

In adiabatic compression,\(P_1V_1^\gamma = P_2V_2^\gamma\), where \(\gamma = \frac{C_p}{C_v} = \frac{7}{5}\) for a diatomic gas like CO. First solve for final pressure \(P_2\).
05

Use Ideal Gas Equation

Using \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\), and the result from Step 4, solve for \(T_2\). Then determine \(\Delta T = T_2 - T_1\).
06

Calculate Change in Internal Energy

Substitute the values of \(n\), \(C_v\), and \(\Delta T\) into \(\Delta U = nC_v\Delta T\) to calculate the change.
07

Final Interpretation

Since work is done on the gas during adiabatic compression, the internal energy increases. This means the temperature must also increase.

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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 used to relate the pressure, volume, and temperature of an ideal gas. It is expressed by the formula \( PV = nRT \), where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume of the gas
  • \( n \) is the number of moles of the gas
  • \( R \) is the universal gas constant
  • \( T \) is the temperature in Kelvin
This equation helps to understand how different variables interact with each other. In adiabatic processes, it's pivotal for calculating changes in temperature and pressure when volume changes, assuming no heat is exchanged between the system and its surroundings.
When dealing with adiabatic processes, the Ideal Gas Law is combined with adiabatic conditions to solve for unknown variables like final temperature or pressure.
Internal Energy
Internal energy is the total energy contained within a system, arising from the kinetic and potential energy of the molecules. For an ideal gas, the internal energy (\( U \)) depends solely on its temperature, making it easier to manage mathematically.
The change in internal energy can be calculated using the formula:
  • \( \Delta U = nC_v\Delta T \)
  • \( n \) is the number of moles
  • \( C_v \) is the molar specific heat at constant volume
  • \( \Delta T \) is the change in temperature
In adiabatic compression, work is done on the gas, which leads to an increase in internal energy. This is because no heat is lost to the surroundings, and the work results in a temperature increase. Understanding these principles allows us to determine how internal energy changes throughout various thermodynamic processes.
Temperature Change
Temperature change in a thermodynamic process indicates a change in the kinetic energy of gas molecules. During an adiabatic compression, the temperature of the gas increases as work is done on it.
This occurs because the energy added to the system as work cannot be dissipated elsewhere due to the lack of heat transfer with the environment.
To find the change in temperature, the relationship between initial and final states is often manipulated using the Ideal Gas Law and adiabatic condition:
  • \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)
  • Adiabatic condition: \( P_1V_1^\gamma = P_2V_2^\gamma \)
  • Calculate final temperature \( T_2 \) from these equations
  • Find \( \Delta T = T_2 - T_1 \)
This step is crucial in determining how changes in energy affect the overall thermodynamic state of a gas in a closed system.
Carbon Monoxide Gas
Carbon monoxide (CO) is a diatomic molecule often used in chemistry and physics problems due to its simple molecular structure. Its treatment as an ideal gas is valid under certain conditions, especially at moderate temperatures and pressures.
The significance of carbon monoxide in thermodynamic equations lies in its specific heat capacities:
  • Molar specific heat at constant volume, \( C_v = \frac{5}{2} R \)
  • The ratio of specific heats, \( \gamma = \frac{C_p}{C_v} = \frac{7}{5} \)
These properties are essential for solving problems involving heat capacity and for understanding the behavior of CO under different thermodynamic processes.
Recognizing the characteristics of carbon monoxide in an adiabatic process allows for accurate calculations of temperature and energy changes resulting from compression or expansion.

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Most popular questions from this chapter

High-Altitude Research. A large research balloon containing \(2.00 \times 10^{3} \mathrm{m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm \((\text { Fig. } 19.33)\) . Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

On a warm summer day, a large mass of air (armospheric 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 attitude, is called the dry adiabatic lapse rate.)

A gas in a cylinder expands from a volume of 0.110 \(\mathrm{m}^{3}\) to 0.320 \(\mathrm{m}^{3}\) . Heat flows into the gas just rapidly enough to keep the pressure constant at \(1.80 \times 10^{5}\) Pa during the expansion. The total heat added is \(1.15 \times 10^{5} \mathrm{J}\) . (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. it matter whether the gas is ideal? Why or why not?

A cylinder contains 0.0100 mol of helium at \(T=27.0^{\circ} \mathrm{C}\) (a) How much heat is needed to raise the temperature to \(67.0^{\circ} \mathrm{C}\) while keeping the volume constant? Draw a \(p V\) -diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from \(27.0^{\circ} \mathrm{C}\) to \(67.0^{\circ} \mathrm{C}\) ? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? d) If the gas is ideal, what is the change in its internal energy in part (a)? In part \((b) ?\) How do the two answers compare? Why?

An experimenter adds \(970 \mathrm{~J}\) of heat to \(1.75 \mathrm{~mol}\) of an ideal gas to heat it from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C}\) at constant pressure. The gas does \(+223 \mathrm{~J}\) of work during the expansion. (a) Calculate the change in internal energy of the gas. (b) Calculate \(\gamma\) for the gas.
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