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What is the average mechanical energy of the atoms of an ideal monatomic gas at \(300 \mathrm{K}\) ?

Short Answer

Expert verified
The total average mechanical energy of the atoms of an ideal monatomic gas at \(300 \ \mathrm{K}\) is approximately \(6.21 \times 10^{-21} \ \mathrm{J}\).

Step by step solution

01

Identify given and required information

: We have the following given information: - The gas is an ideal monatomic gas - The temperature is \(T = 300 \ \mathrm{K}\) We need to find the average mechanical energy of the atoms in this gas.
02

Use the equipartition theorem to find the average energy of each degree of freedom

: The equipartition theorem states that the average energy of each degree of freedom is \(\frac{1}{2}kT\), where \(k\) is the Boltzmann constant and \(T\) is the temperature in kelvin. In this case, the temperature is \(T = 300 \ \mathrm{K}\), so the average energy of each degree of freedom is: \[ \frac{1}{2}kT = \frac{1}{2}(1.38 \times 10^{-23}\, \mathrm{J/K})(300\, \mathrm{K}) \]
03

Calculate the average energy of each degree of freedom

: Now, we just need to calculate the value using the given temperature: \[ \frac{1}{2}kT = \frac{1}{2}(1.38 \times 10^{-23}\, \mathrm{J/K})(300\, \mathrm{K}) = 2.07 \times 10^{-21}\, \mathrm{J} \] So the average energy of each degree of freedom is approximately \(2.07 \times 10^{-21} \ \mathrm{J}\).
04

Calculate the total average mechanical energy

: Since there are 3 translational degrees of freedom for a monatomic gas, we can find the total average mechanical energy by multiplying the average energy of each degree of freedom by the number of degrees of freedom: \[ \mathrm{Total\ average\ mechanical\ energy} = 2.07 \times 10^{-21}\, \mathrm{J} \times 3 \]
05

Calculate the total average mechanical energy

: Now, we just need to multiply the numbers to find the total average mechanical energy: \[ \mathrm{Total\ average\ mechanical\ energy} = 2.07 \times 10^{-21}\, \mathrm{J} \times 3 = 6.21 \times 10^{-21}\, \mathrm{J} \] So the total average mechanical energy of the atoms of an ideal monatomic gas at 300 K is approximately \(6.21 \times 10^{-21} \ \mathrm{J}\).

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Key Concepts

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

Equipartition Theorem
When we talk about the energy inside a system, such as a gas, we often refer to thermal energy. This energy is shared among the microscopic constituents of the system (like atoms and molecules) through their motion and vibrations. Now, the equipartition theorem is a powerful principle in statistical mechanics that helps us understand this distribution of energy.

The theorem states that, for a system in thermal equilibrium at temperature T, the energy is equally partitioned among its degrees of freedom. Each degree of freedom contributes \frac{1}{2}kT\ to the total energy, where \(k\) is the Boltzmann constant and \(T\) is the temperature in kelvins. It applies to any classical (non-quantum) degree of freedom that appears quadratically in the energy (like the x, y, or z components of kinetic energy for a gas particle).

In simple terms, if you think of a gas as a crowd of atoms bouncing around, the equipartition theorem tells us that on average, each direction of bounce (up-down, left-right, back-forth) has the same amount of energy given by \(\frac{1}{2}kT\).
Monatomic Gas
Gases can be complex, consisting of molecules with many atoms, or they can be simple, made of individual atoms, such as helium or neon. These single-atom gases are called monatomic gases. In the realm of physics, monatomic gases serve as an ideal model because they have the simplest form of motion - translational.

Since a monatomic gas has no chemical bonds or complex structures like molecules do, it won't have rotational or vibrational degrees of freedom that more complex gases have. Instead, a monatomic gas has three translational degrees of freedom, corresponding to motion in three-dimensional space: moving left-right (x-axis), up-down (y-axis), and back-forth (z-axis). When we're using the equipartition theorem to calculate the energy in a monatomic gas, we only consider these three degrees of freedom.
Degrees of Freedom
The concept of degrees of freedom is crucial in understanding the movement and energy distribution in systems at the microscopic level. A degree of freedom refers to an independent mode of motion that contributes to the energy of a system. In a physical sense, this often translates to the number of ways a particle, or a complex of particles, can move or vibrate in space.

In the context of an ideal monatomic gas, as we've mentioned, there are only three translational degrees of freedom. This simplicity leads to straightforward calculations when applying the equipartition theorem, as you only have to consider these translational movements. Degrees of freedom are not just limited to translation; in more complex molecules, they also include rotations and vibrations. But for our monatomic gas, these don't come into play, making it an idealized system to study thermal energy distribution.
Boltzmann Constant
The Boltzmann constant (\(k\)) is a fundamental physical constant that bridges the macroscopic and microscopic worlds. It appears in various contexts in physics, but when it comes to understanding the average mechanical energy of atoms in a gas, it plays a key role as part of the equipartition theorem.

With a value of approximately \(1.38 \times 10^{-23} \, \mathrm{J/K}\), the Boltzmann constant provides the conversion factor between temperature in kelvins and the kinetic energy possessed by the particles in a given system. It's named after Ludwig Boltzmann, a pioneer in the field of statistical mechanics, which is the branch of physics that deals with the behavior of systems with a large number of particles.

So, the Boltzmann constant helps us understand how something as abstract as temperature relates to the very concrete energies of particles whizzing around in a gas. It's the reason we can calculate that each atomic degree of freedom has an average energy of \(\frac{1}{2}kT\) at a given temperature.

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Most popular questions from this chapter

What is the internal energy of 6.00 mol of an ideal monatomic gas at \(200^{\circ} \mathrm{C}\) ?

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A monatomic ideal gas undergoes a quasi-static process that is described by the function \(p(V)=p_{1}+3\left(V-V_{1}\right),\) where the starting state is \(\left(p_{1}, V_{1}\right)\) and the final state \(\left(p_{2}, V_{2}\right) .\) Assume the system consists of \(\mathrm{n}\) moles of the gas in a container that can exchange heat with the environment and whose volume can change freely. (a) Evaluate the work done by the gas during the change in the state. (b) Find the change in internal energy of the gas. (c) Find the heat input to the gas during the change. (d) What are initial and final temperatures?

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