Suppose that we have \(1 \mathrm{~mol}\) of a monatomic gas \(B_{1}\), whose nuclei are in their lowest energy state, and \(1 \mathrm{~mol}\) of a monatomic gas \(B_{2}\) consisting of exactly the same atoms as \(B_{1}\), except that the nuclei are in an excited state whose energy \(\epsilon\) is much larger than \(k T\) and whose lifetime is much larger than the time for diffusion to occur. Both gases are at the same pressure and are maintained at the same temperature \(T\) by a heat reservoir. (a) Immediately after the nuclei of gas \(B_{2}\) have been excited, diffusion takes place. Calculate the entropy change of the universe. (b) After the nuclei of gas \(B_{2}\) have been excited, a time much larger than the lifetime of the excited state is allowed to elapse, and then diffusion takes place. Calculate the entropy change of the universe. (c) Show that the answer to part \((b)\) is larger than that to part \((a)\). (This problem is due to M. J. Klein.)

Step by step solution

01

Understand the Problem

We need to calculate the entropy change of the universe for a two-part process involving two gases, both initially with non-interacting nuclei: one with nuclei in the lowest energy state, the other with nuclei excited to a state with energy \(\epsilon\) which is much larger than \(kT\). The gases are at the same pressure and temperature.

02

Define Entropy Change for Initial Excitation State - (a)

Consider the entropy change when the diffusion occurs immediately after the nuclei of gas \(B_{2}\) have been excited. For gas \(B_{2}\), the energy per atom is \(\epsilon\). Entropy change for \(B_{2}\) can be calculated using the relation: \[ \Delta S = nR \ln\left( \frac{V_f}{V_i} \right) \] where \(n\) is the number of moles, \(R\) is the gas constant, and \(V_f\) and \(V_i\) are the final and initial volumes respectively.

03

Calculate Entropy Change for Diffusion in Initial Stage of Excitation - (a)

For \(1\) mol of gas diffusing into another without changing temperature or pressure, the change in entropy for \(B_2\) is zero as it’s a similar gas structure. However, consider the Boltzmann factor for the excited state. For system \(B_{1}\) and \(B_{2}\) considered isolated under equilibrium, conclude the most probable distribution condition yielding the net entropy increase.

04

Define Long-Term Relaxed State and Entropy - (b)

After allowing the excited nuclei to relax over a period, the energy returns to the equilibrium state. The system’s entropy change now involves both gases re-achieving a more stabilized thermodynamic equilibrium.

05

Entropy Change for Relaxed Equilibrium State - (b)

Entropy will involve conventional mixing terms since energy states in \(B_{2}\) have lowered back essentially. Calculate the change in the distribution of the system and associated heat exchanges reclaiming equilibrium state.

06

Compare Entropy Changes in Scenarios - (c)

To show that part (b)'s entropy change is larger than that in part (a), evaluate both scenarios through thermodynamic added entropy calculations. Highlight the additional entropy contribution arriving from excited states stabilization processes.

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Key Concepts

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

Monatomic Gas

A monatomic gas consists of individual atoms, each acting as an independent particle. These gases are the simplest form of gases and perfect for studying basic thermodynamic behaviors.
Some examples of monatomic gases include noble gases like Helium (He), Neon (Ne), and Argon (Ar).
Monatomic gases have specific properties:

• Each atom moves independently.
• They have three degrees of freedom (motion in three spatial directions).
• Their internal energy is strictly kinetic.

These characteristics make monatomic gases excellent subjects for exploring fundamental thermodynamics concepts.

Excited State

When atoms absorb energy, they can move to a higher energy level called the excited state. This state is temporary and usually unstable compared to the ground state.
Key points about excited states:

• Energy required for excitation is much larger than the thermal energy, described by \(\epsilon \gg kT\).
• Atoms will eventually return to the ground state by releasing energy, usually as photons.
• The excited state lifetime affects how atoms behave before they return to the ground state.

Understanding the excited state is crucial because it changes the thermodynamic properties of the gas, influencing entropy and energy calculations.

Heat Reservoir

A heat reservoir is a large system that can absorb or supply heat without changing its temperature. It maintains thermodynamic systems at a constant temperature.
In our exercise<, the gases are kept at the same temperature by a heat reservoir. The presence of the heat reservoir ensures:

• Constant temperature for both gases despite other changes.
• Facilitation of equilibrium processes.
• No significant temperature change of the reservoir itself due to its large size.

Heat reservoirs are essential in thermodynamic problems as they simplify calculations by providing a stable thermal environment.

Thermodynamic Equilibrium

Thermodynamic equilibrium is when a system’s properties do not change over time. All parts of the system are at the same temperature, and there is no net flow of energy or matter.
For the gases in the exercise, achieving equilibrium means:

• Both gases reach a uniform temperature and pressure.
• Energy distribution among molecules stabilizes.
• Entropy, the measure of disorder, reaches its maximum allowable value.

Reaching thermodynamic equilibrium after diffusion is key to understanding the calculated changes in entropy and energy for the system. Understanding equilibrium helps in predicting the final state and behavior of the gases involved.