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 \(V / V_{\mathrm{A}}\). Now suppose the same atom is inside a container of volume \(2 \mathrm{~V}\). What will be the change in entropy?

Short answer

Answer: The change in entropy when the volume of the container is doubled is ΔS = k * ln(2), where k is Boltzmann's constant.

Step-by-step solution

Write the entropy formula

The entropy of a system, S, is given by: S = k * ln(W) where k is Boltzmann's constant and W is the number of possible states.

Identify the number of states in the original container

We are given the formula for the number of states: W1 = V / V_A

Identify the number of states in the larger container

Now suppose the same atom is inside a container of volume 2V. The number of states in the new container (W2) is given by: W2 = (2V) / V_A

Compute the entropy of the original container

Using the entropy formula and the number of states for the original container, we compute the entropy (S1): S1 = k * ln(W1) S1 = k * ln(V / V_A)

Compute the entropy of the larger container

Using the entropy formula and the number of states for the larger container, we compute the entropy (S2): S2 = k * ln(W2) S2 = k * ln((2V) / V_A)

Compute the change in entropy

To find the change in entropy (ΔS), we substract the entropy of the original container (S1) from the entropy of the larger container (S2): ΔS = S2 - S1 ΔS = k * ln((2V) / V_A) - k * ln(V / V_A)

Simplify the expression for the change in entropy

Using property of logarithms, we can simplify the expression for ΔS: ΔS = k * [ln((2V) / V_A) - ln(V / V_A)] ΔS = k * ln([(2V) / V_A] / [V / V_A]) ΔS = k * ln(2)

Final answer

The change in entropy when the volume of the container is doubled is: ΔS = k * ln(2)