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The temperature of \(2.00 \mathrm{~mol}\) of an ideal monatomic gas is raised \(15.0 \mathrm{~K}\) in an adiabatic process. What are (a) the work \(W\) done by the gas, (b) the energy transferred as heat \(Q,(\mathrm{c})\) the change \(\Delta E_{\text {int }}\) in internal energy of the gas, and (d) the change \(\Delta K\) in the average kinetic energy per atom?

Short Answer

Expert verified
(a) Work done, \( W = 374.13 \) J; (b) Heat transferred, \( Q = 0 \); (c) \( \Delta E_{\text{int}} = 374.13 \) J; (d) \( \Delta K = 3.10 \times 10^{-22} \) J/atom.

Step by step solution

01

Understanding the Adiabatic Process

In an adiabatic process, the system does not exchange heat with its surroundings. Therefore, the heat transferred, \( Q \), is zero. It is important to note that this implies all changes in energy are due to work done or changes in internal energy.
02

Determine the Change in Internal Energy

For an ideal gas, the change in internal energy \( \Delta E_{\text{int}} \) is related to the number of moles \( n \), the molar heat capacity at constant volume \( C_V \), and the temperature change \( \Delta T \). For a monatomic ideal gas, \( C_V = \frac{3}{2} R \), where \( R \) is the ideal gas constant (8.314 J/mol·K). Thus, \( \Delta E_{\text{int}} = n C_V \Delta T = 2.00 \times \frac{3}{2} \times 8.314 \times 15.0 \). Calculate this value.
03

Calculate Work Done by the Gas

Since the process is adiabatic, the work done by the gas, \( W \), is equal to the change in internal energy, as \( Q = 0 \) (no heat exchange). Thus, \( W = \Delta E_{\text{int}} \). Use the result from Step 2 to obtain \( W \).
04

Average Kinetic Energy Change Per Atom

The change in average kinetic energy per atom can be determined from the relation with the temperature change. For a monatomic ideal gas, the average kinetic energy per molecule is given by \( \frac{3}{2} k T \), where \( k \) is Boltzmann's constant (\(1.38 \times 10^{-23} \) J/K). Therefore, the change in average kinetic energy \( \Delta K \) is \( \frac{3}{2} k \Delta T \). Substitute \( \Delta T = 15.0 \) to find \( \Delta K \).

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Key Concepts

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

Adiabatic Process
An adiabatic process is a thermodynamic operation where a system does not exchange heat with its surroundings. This means that, during the process, the heat transfer, denoted as \( Q \), is zero. In simpler terms, the system is insulated from its environment. Any change that occurs in the thermodynamic quantities is due to work done by or on the system, or a change in the system's internal energy.

Adiabatic processes are common in natural phenomena, such as the compression or expansion of gas in the atmosphere. They are characterized by rapid changes where heat doesn't have the time to transfer. In calculations, these processes help simplify the equations of thermodynamics since no heat transfer terms are involved.
  • If a gas expands adiabatically, it does work and cools down.
  • If compressed, it heats up as no heat escapes.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermodynamics. It relates the pressure, volume, and temperature of an ideal gas with the formula \( PV = nRT \), where:
  • \( P \) is the pressure.
  • \( V \) is the volume.
  • \( n \) is the number of moles of gas.
  • \( R \) is the ideal gas constant (8.314 J/mol·K).
  • \( T \) is the temperature in Kelvin.
Ideal gases are hypothetical gases composed of many randomly moving point particles that interact only via elastic collisions. In an adiabatic process, although heat transfer is zero, the Ideal Gas Law can still apply to determine changes in state variables (pressure, volume, and temperature) due to work done on or by the gas.

Real gases approximate ideal behavior under many conditions, but ideally, this law helps us explore fundamental concepts in thermodynamics.
Kinetic Theory
Kinetic Theory provides a molecular perspective on thermodynamics. It explains macroscopic properties like temperature and pressure by looking at the microscopic details of molecular motion. According to this theory, the temperature of a gas is a measure of the average kinetic energy of its molecules.
  • For an ideal monatomic gas, the average kinetic energy per molecule is given by \( \frac{3}{2} kT \), where \( k \) is Boltzmann's constant (\(1.38 \times 10^{-23} \) J/K).
  • The kinetic energy determines the speed and direction of molecules, influencing pressure and temperature.
In the context of an adiabatic process, the change in kinetic energy is directly related to the change in temperature because no heat is exchanged. Every increase in internal energy corresponds to more vigorous molecular motion, translating to an increase in kinetic energy.
Internal Energy Change
Internal energy is the total energy contained within a system. For an ideal gas, changes in internal energy \( \Delta E_{\text{int}} \) depend solely on changes in temperature. This is because internal energy is related to the motion and potential of molecules, and temperature is a direct translation of kinetic energy for ideal gases.
For monatomic gases, which lack rotational or vibrational modes, the change in internal energy is calculated as:
\[ \Delta E_{\text{int}} = nC_V\Delta T \] where:
  • \( n \) is the number of moles.
  • \( C_V \) is the molar heat capacity at constant volume (\( \frac{3}{2}R \) for monatomic gases).
  • \( \Delta T \) is the change in temperature.
In an adiabatic process, any work done, \( W \), by or on the system directly changes the internal energy. Thus, \( \Delta E_{\text{int}} = W \), since \( Q = 0 \).

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Most popular questions from this chapter

An ideal diatomic gas, with rotation but no oscillation, undergoes an adiabatic compression. Its initial pressure and volume are \(1.20 \mathrm{~atm}\) and \(0.200 \mathrm{~m}^{3} .\) Its final pressure is \(2.40 \mathrm{~atm} .\) How much work is done by the gas?

A sample of ideal gas expands from an initial pressure and volume of \(32 \mathrm{~atm}\) and \(1.0 \mathrm{~L}\) to a final volume of \(4.0 \mathrm{~L}\). The initial temperature is \(300 \mathrm{~K}\). If the gas is monatomic and the expansion isothermal, what are the (a) final pressure \(p_{f}\), (b) final temperature \(T_{f}\), and (c) work \(W\) done by the gas? If the gas is monatomic and the expansion adiabatic, what are (d) \(p_{f}\), (e) \(T_{f}\), and (f) \(W ?\) If the gas is diatomic and the expansion adiabatic, what are \((\mathrm{g}) p_{f},(\mathrm{~h})\) \(T_{f}\), and (i) \(W ?\)

During a compression at a constant pressure of \(250 \mathrm{~Pa}\), the volume of an ideal gas decreases from \(0.80 \mathrm{~m}^{3}\) to \(0.20 \mathrm{~m}^{3}\). The initial temperature is \(360 \mathrm{~K}\), and the gas loses \(210 \mathrm{~J}\) as heat. What are (a) the change in the internal energy of the gas and (b) the final temperature of the gas?

In an industrial process the volume of \(25.0 \mathrm{~mol}\) of a monatomic ideal gas is reduced at a uniform rate from \(0.616 \mathrm{~m}^{3}\) to \(0.308 \mathrm{~m}^{3}\) in \(2.00 \mathrm{~h}\) while its temperature is increased at a uniform rate from \(27.0^{\circ} \mathrm{C}\) to \(450^{\circ} \mathrm{C}\). Throughout the process, the gas passes through thermodynamic equilibrium states. What are (a) the cumulative work done on the gas, (b) the cumulative energy absorbed by the gas as heat, and (c) the molar specific heat for the process? (Hint: To evaluate the integral for the work, you might use $$\int \frac{a+b x}{A+B x} d x=\frac{b x}{B}+\frac{a B-b A}{B^{2}} \ln (A+B x)$$ an indefinite integral.) Suppose the process is replaced with a twostep process that reaches the same final state. In step 1 , the gas volume is reduced at constant temperature, and in step 2 the temperature is increased at constant volume. For this process, what are (d) the cumulative work done on the gas, (e) the cumulative energy absorbed by the gas as heat, and (f) the molar specific heat for the process?

At what frequency would the wavelength of sound in air be equal to the mean free path of oxygen molecules at \(1.0 \mathrm{~atm}\) pressure and \(0.00^{\circ} \mathrm{C} ?\) The molecular diameter is \(3.0 \times 10^{-8} \mathrm{~cm}\).

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