A three-state system has three different macrostates that it can be in. The multiplicity of state 1 is twice that of state 2, and four times the multiplicity of state 3: Ω1 = 2Ω2, and Ω1 = 4Ω2. The entropy of state 1 is 20 J/mol x k
. What is the entropy of state 3 (in units of J/K)? (the answer is not 20)

We are tasked with finding the entropy of state 3 given the following information:

- The multiplicities of the macrostates are related as:
  \[
  \Omega_1 = 2 \Omega_2 \quad \text{and} \quad \Omega_1 = 4 \Omega_3
  \]
- The entropy of state 1, \( S_1 \), is 20 J/K.
- We need to find the entropy of state 3, \( S_3 \), in J/K.

### Step 1: Recall the formula for entropy
The entropy \( S \) of a system is related to its multiplicity \( \Omega \) by the following equation:
\[
S = k_B \ln(\Omega)
\]
where \( k_B \) is the Boltzmann constant. We can compare entropies using the ratio of multiplicities.

### Step 2: Express multiplicities in terms of \( \Omega_2 \)
From the problem, we know:
\[
\Omega_1 = 2 \Omega_2 \quad \text{and} \quad \Omega_1 = 4 \Omega_3
\]
Thus, we can express the multiplicities as:
\[
\Omega_2 = \frac{\Omega_1}{2}, \quad \Omega_3 = \frac{\Omega_1}{4}
\]

### Step 3: Relating entropies
The entropy of state 1 is given as \( S_1 = 20 \, \text{J/K} \), and since entropy is proportional to the logarithm of multiplicity, we can write:
\[
S_1 = k_B \ln(\Omega_1)
\]
For state 3:
\[
S_3 = k_B \ln(\Omega_3) = k_B \ln\left(\frac{\Omega_1}{4}\right)
\]

Using the logarithmic identity \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \), we have:
\[
S_3 = k_B \ln(\Omega_1) - k_B \ln(4)
\]
Substitute \( S_1 = k_B \ln(\Omega_1) \) and simplify:
\[
S_3 = S_1 - k_B \ln(4)
\]

### Step 4: Calculate \( S_3 \)
We can now calculate \( S_3 \). The Boltzmann constant \( k_B \) is approximately \( 1.38 \times 10^{-23} \, \text{J/K} \). However, since entropy is given in macroscopic units, we are more interested in the natural logarithmic factor. Using the fact that:
\[
\ln(4) \approx 1.386
\]
we get:
\[
S_3 = 20 - (k_B \times 1.386)
\]

Thus, subtract the logarithmic contribution to get \( S_3 \):
\[
S_3 = 20 - (1.386 \cdot 1) \approx 20 - 1.386 = 18.614 \, \text{J/K}
\]

### Final Answer:
The entropy of state 3 is approximately:
\[
S_3 \approx 18.6 \, \text{J/K}
\]

Would you like further details or have any questions?

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Here are 5 questions to expand on the topic:
1. How does entropy change with the number of microstates in a system?
2. What role does the Boltzmann constant play in the entropy equation?
3. How would the entropy change if the multiplicity of state 3 were increased?
4. How is entropy related to the second law of thermodynamics?
5. Can we calculate entropy for a system with continuous states instead of discrete ones?

**Tip:** When comparing entropies, logarithmic differences in multiplicities help quantify how systems distribute over possible microstates.

A three-state system has three different macrostates that it can be in. The multiplicity of state 1 is twice that of state 2, and four times the multiplicity of state 3: Ω1 = 2Ω2, and Ω1 = 4Ω3. The entropy of state 1 is 20 J/mol x K. What is the entropy of state 3 (in units of J/K)? (The answer is not 20.)

Learn how to calculate the entropy of state 3 in a three-state system using the relationship between multiplicity and entropy. This problem applies the Boltzmann entropy formula and logarithmic identities. Suitable for undergraduate students studying thermodynamics or physics.

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