Chem 3510 001 Entropy 1: A System Is Composed Of Four Distin

Chem 3510 001 Entropy 1 A System Is Composed Of Four Distinguishable

A system comprises four distinguishable molecules (A, B, C, and D), each capable of occupying one of three available energy states with energies E1=1, E2=2, and E3=3. The questions involve calculating the number of microstates and the entropy of the system under specific total energy constraints, as well as examining the effect of expanding the energy states available to each molecule.

Specifically, the problem asks: (a) to determine the total number of microstates when the total energy of the system is 6; (b) to calculate the entropy corresponding to this total energy; (c) to determine the entropy if the total energy is 4; and (d) to assess how increasing the number of available energy states per molecule from three to four impacts the answer to part (a).

Paper For Above instruction

Understanding the foundational principles of statistical mechanics allows us to analyze the microstates and entropy of complex systems. In this scenario, four distinguishable molecules can each occupy discrete energy states, and the total energy of the system constrains the combinations of microstates. This analysis underscores the relationship between microstates, macrostate energy, and entropy, illustrating how these properties evolve with changes in system parameters.

Part (a): Determining the number of microstates when the total energy is 6

Given the molecules have energies E1=1, E2=2, and E3=3, and the system comprises four molecules A, B, C, and D, each molecule's state can be represented as a variable assuming these energy levels. Since the molecules are distinguishable, permutations of their states contribute to the total number of microstates, provided the sum of their energies equals 6.

The problem reduces to solving the equation:

e_A + e_B + e_C + e_D = 6

where each e_i ∈ {1, 2, 3}.

To find the total number of microstates, enumerate all possible combinations of the four molecules' energies that sum to 6:

• All four in E3 (3,3,3,3): sum = 12, too high, discard.
• Three in E3 and one in E1 (3,3,3,1): sum = 10, too high, discard.
• Two in E3 and two in E2 (3,3,2,2): sum = 3+3+2+2=10, discard.
• Two in E3, one in E2, and one in E1 (3,3,2,1): sum= 3+3+2+1=9, discard.
• Other combinations where the sum is exactly 6 include:

- One molecule in E3, three in E1 (3,1,1,1): sum= 3+1+1+1=6.

- Two molecules in E2 and two in E1 (2,2,1,1): sum= 2+2+1+1=6.

- One molecule in E2, three in E1 (2,1,1,1): sum= 2+1+1+1=5, discard.

- Four molecules in E1 (1,1,1,1): sum=4, discard as sum ≠ 6.

- Combinations with other distributions do not sum to 6 when considering energies {1,2,3}.

Now, systematically, count the microstates for each valid energy combination, considering the distinguishability of molecules:

Case 1: One molecule in E3, three in E1 (3,1,1,1)

• Number of arrangements: Number of permutations of four molecules where one is in E3 and three are in E1 is
• Number of permutations: 4! / 3! = 4, since choosing which molecule is in E3 among the four.
• Microstates: 4.

Case 2: Two molecules in E2 and two in E1 (2,2,1,1)

• Number of arrangements: Number of permutations for two molecules in E2 and two in E1
• Number of permutations: 4! / (2! 2!) = 6.
• Microstates: 6.

Total microstates when total energy = 6 is sum over the two cases:

Microstates_total = 4 + 6 = 10.

Part (b): Calculating the entropy of the system for total energy 6

Entropy (S) is related to the number of microstates (Ω) by Boltzmann's equation:

S = k_B * ln(Ω)

where k_B is Boltzmann's constant. Using the total microstates Ω=10, and taking k_B as a unit coefficient for simplicity:

S = ln(10) ≈ 2.3026 (in units where k_B = 1)

Part (c): Entropy when the total energy is 4

Repeat the enumeration for total energy 4:

Question reduces to: sum of energies from variables e_A, e_B, e_C, e_D ∈ {1, 2, 3} equal to 4.

Possible combinations:

• All four in E1: sum=4, energy states: (1,1,1,1).
• One in E2, three in E1: sum= 2+1+1+1=5, discard.
• Two in E2, two in E1: sum= 2+2+1+1=6, discard.
• One in E3, others in E1 or E2: sum>4, discard.
• Other combinations do not sum to 4.

Only one valid microstate: all four molecules in E1 (1,1,1,1): sum=4.

Number of microstates: 1

Thus, entropy S=ln(1)=0.

Part (d): Changing the available states from E1=1, E2=2, E3=3 to E1=1, E2=2, E3=3, and E4=5

In this case, each molecule can occupy four energy states: 1, 2, 3, or 5. The question is how this change affects the answer to part (a).

When total energy is 6, the combinations satisfying the sum condition must be re-evaluated considering the new energy levels. This expands the set of possible microstates, including cases previously impossible with only three energy states.

Calculations similar to those above show that the number of microstates increases because molecules can now occupy higher energy levels (E=5), allowing more combinations summing to 6, such as (5,1,0,0) (which is invalid as energies are 1,2,3,5), or other combinations that satisfy the new constraints.

For the specific calculation, we need to enumerate all solutions for the equation:

 e1 + e2 + e3 + e4 = 6, where each ei ∈ {1, 2, 3, 5}

Possible solutions include:

• (1,1,1,3): sum=6
• (1,1,2,2): sum=6
• (2,2,2,0): invalid, as zero energy is not available
• (1,5,0,0): invalid due to zeros and energies not in {1,2,3,5}

More systematically, the solutions are:

• Three molecules in E1 (1) and one in E3 (3): sum=1+1+1+3=6.
• Two in E2 (2) and two in E1 (1): sum=2+2+1+1=6.
• One in E2 and three in E1: 2+1+1+1=5, discard.
• One in E3 and others in E1/E2: sum >6 or

The corresponding counts of arrangements for each solution are similar as before, with permutations considered based on distinguishability:

• For (1,1,1,3): 4 permutations (as above).
• For (2,2,1,1): 6 permutations (as above).

This increases the total microstates denominator, impacting entropy calculations accordingly. This demonstrates how expanding available energy states enhances system complexity and microstate degeneracy, thereby increasing entropy in accordance with statistical theory.

In summary, the number of microstates and the entropy of the system are heavily influenced by the available energy levels, the total energy constraint, and the distinguishability of molecules. The addition of higher energy states enlarges the microstate space, elevating the entropy, which reflects increased disorder and the greater number of accessible microstates.

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