Question

Suppose an atom of volume \(V_{\mathrm{A}}\) is inside a container of volume \(V\). The atom can occupy any position within this volume. For this simple model, the number of states available to the atom is given by \(V / V_{A}\). Now suppose the same atom is inside a container of volume \(2 V .\) What will be the change in entropy?

Short answer

Question: Calculate the change in entropy when the volume of a container with a single atom changes from V to 2V, knowing that the number of states available to the atom is proportional to the container's volume over the atom's volume. Answer: The change in entropy, ΔS, is equal to k ln{2}, where k is the Boltzmann constant.

Step-by-step solution

Calculate Entropy of Initial State

The initial entropy of the atom, \(S_1\), can be calculated using the formula: \(S_1 = k \ln{\Omega_1}\), where \(k\) is the Boltzmann constant and \(\Omega_1\) is the number of states in container volume \(V\). The number of states in this case is given by \(\Omega_1=V/V_\mathrm{A}\).

Calculate Entropy of Final State

The final entropy of the atom, \(S_2\), can be calculated in the same way as in step 1, but with the container volume \(2V\). The number of states in this case is given by \(\Omega_2=(2V)/V_\mathrm{A}\). Thus, \(S_2 = k \ln{\Omega_2}\).

Calculate Change in Entropy

The change in entropy, \(ΔS\), is the difference between the final entropy and the initial entropy: \(ΔS = S_2 - S_1\). By substituting the expressions for \(S_1\) and \(S_2\) found in steps 1 and 2, we get: \(ΔS = k(\ln{(2V/V_\mathrm{A})} - \ln{(V/V_\mathrm{A})})\).

Simplify the Expression

Using the property of logarithms, we can simplify the expression for \(ΔS\): \(ΔS = k (\ln{(\frac{2V}{V_\mathrm{A}} \cdot \frac{V_\mathrm{A}}{V})}) = k \ln{2}\). The change in entropy, \(ΔS\), is equal to \(k \ln{2}\).

Key concepts

Boltzmann Constant

The Boltzmann constant, usually denoted as \(k\), serves as a bridge between microscopic and macroscopic physical phenomena. It essentially scales the entropy value from a particle count level to a more tangible quantity. This constant is crucial when dealing with variables such as temperature, energy, and entropy. The value of the Boltzmann constant is approximately \(1.38 \times 10^{-23} \text{ J/K}\).

In the context of thermodynamics and statistical mechanics, the Boltzmann constant allows us to calculate entropy, such as in the formula \(S = k \ln \Omega\). Here, \(\Omega\) represents the number of microstates, or ways a system can be arranged. Through this formula, the Boltzmann constant helps in quantifying how dispersed or spread out these states are, which essentially defines entropy.

By using the Boltzmann constant in our calculations, we connect the rigorous statistical behavior of atoms and molecules with large-scale thermodynamic properties, such as the ones described in the exercise.

Number of States

The term 'number of states' \((\Omega)\) reflects how many different configurations or arrangements a system's particles can achieve. It plays a critical role in determining the entropy of a system.

When dealing with a volume \(V\) where a single atom can position itself, the number of states can be calculated as the ratio \(V / V_A\), where \(V_A\) is the volume each atom occupies. This simple division yields the number of possible locations for the atom, essentially representing its available "states."

- In a smaller volume, \(\Omega\) is relatively small, indicating fewer configurations.
- Expanding the volume to \(2V\) effectively doubles the number of states, as seen in the exercise.

This is because our flexibility in positioning the atom increases with volume, which directly ties to an increase in entropy. More states mean the system is more disorderly, aligning with the principle that entropy tends to increase when systems are allowed to expand.

Logarithm Properties

Logarithm properties simplify complex mathematical calculations, especially in entropy-related equations. In the context of our discussion, the property that \(\ln(a) - \ln(b) = \ln(a/b)\) is particularly useful. This property allows us to condense

1. calculations
2. reduce complex expressions into simpler forms

Here’s how it helps:

Suppose we have initial and final states of a system, where their entropies are calculated as \(S_1 = k \ln \Omega_1\) and \(S_2 = k \ln \Omega_2\). To find the change in entropy, \(\Delta S\), we subtract these two:

- \(\Delta S = S_2 - S_1 = k (\ln \Omega_2 - \ln \Omega_1)\).

By applying the logarithm rule mentioned above, simplifying it becomes \(\Delta S = k \ln{(\Omega_2/\Omega_1)}\).

This significantly simplifies our work by reducing compound logarithmic terms into a single, more manageable one. In our exercise, such simplification helps us quickly determine that the change in entropy due to a doubling of volume is \(k \ln{2}\), showing the power of logarithm properties in mathematical physics.