Math Problem Statement

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)

Solution

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

  • The multiplicities of the macrostates are related as: Ω1=2Ω2andΩ1=4Ω3\Omega_1 = 2 \Omega_2 \quad \text{and} \quad \Omega_1 = 4 \Omega_3
  • The entropy of state 1, S1S_1, is 20 J/K.
  • We need to find the entropy of state 3, S3S_3, in J/K.

Step 1: Recall the formula for entropy

The entropy SS of a system is related to its multiplicity Ω\Omega by the following equation: S=kBln(Ω)S = k_B \ln(\Omega) where kBk_B is the Boltzmann constant. We can compare entropies using the ratio of multiplicities.

Step 2: Express multiplicities in terms of Ω2\Omega_2

From the problem, we know: Ω1=2Ω2andΩ1=4Ω3\Omega_1 = 2 \Omega_2 \quad \text{and} \quad \Omega_1 = 4 \Omega_3 Thus, we can express the multiplicities as: Ω2=Ω12,Ω3=Ω14\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 S1=20J/KS_1 = 20 \, \text{J/K}, and since entropy is proportional to the logarithm of multiplicity, we can write: S1=kBln(Ω1)S_1 = k_B \ln(\Omega_1) For state 3: S3=kBln(Ω3)=kBln(Ω14)S_3 = k_B \ln(\Omega_3) = k_B \ln\left(\frac{\Omega_1}{4}\right)

Using the logarithmic identity ln(ab)=ln(a)ln(b)\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b), we have: S3=kBln(Ω1)kBln(4)S_3 = k_B \ln(\Omega_1) - k_B \ln(4) Substitute S1=kBln(Ω1)S_1 = k_B \ln(\Omega_1) and simplify: S3=S1kBln(4)S_3 = S_1 - k_B \ln(4)

Step 4: Calculate S3S_3

We can now calculate S3S_3. The Boltzmann constant kBk_B is approximately 1.38×1023J/K1.38 \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)1.386\ln(4) \approx 1.386 we get: S3=20(kB×1.386)S_3 = 20 - (k_B \times 1.386)

Thus, subtract the logarithmic contribution to get S3S_3: S3=20(1.3861)201.386=18.614J/KS_3 = 20 - (1.386 \cdot 1) \approx 20 - 1.386 = 18.614 \, \text{J/K}

Final Answer:

The entropy of state 3 is approximately: S318.6J/KS_3 \approx 18.6 \, \text{J/K}

Would you like further details or have any questions?


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.

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Entropy
Logarithms
Multiplicity

Formulas

S = k_B ln(Ω)
ln(a/b) = ln(a) - ln(b)

Theorems

Boltzmann Entropy Formula
Logarithmic Identities

Suitable Grade Level

Undergraduate (Thermodynamics/Physics)