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Chapter 20: Problem 45

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

Expert verified
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

01

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 \(\Omega_1=V/V_\mathrm{A}\).
02

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, \(S_2 = k \ln{\Omega_2}\).
03

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: \(ΔS = k(\ln{(2V/V_\mathrm{A})} - \ln{(V/V_\mathrm{A})})\).
04

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 \(k \ln{2}\).

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