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Chapter 19: Problem 90

Consider a system that consists of two standard playing dice, with the state of the system defined by the sum of the values shown on the top faces. (a) The two arrangements of top faces shown here can be viewed as two possible microstates of the system. Explain. (b) To which state does each microstate correspond? (c) How many possible states are there for the system? (d) Which state or states have the highest entropy? Explain. (e) Which state or states have the lowest entropy? Explain. (f) Calculate the absolute entropy of the two-dice system.

Short Answer

Expert verified
The two arrangements of top faces can be viewed as microstates, representing two different arrangements of the two-dice system. There are 36 possible microstates and 11 possible states (sums) in this system. The state with the highest entropy (most randomness) is the sum 7, while the states with the lowest entropy (least randomness) are the sums 2 and 12. The absolute entropy of the two-dice system is approximately \(1.29 \times 10^{-22} \, \text{J/K}\).

Step by step solution

01

(a) Explain the two arrangements as possible microstates of the system

A microstate is a specific configuration or arrangement of a system, while a macrostate is a collection of microstates with the same macroscopic properties. In this problem, the arrangement of the dice with specific numbers shown on their top faces are the microstates of the system. Since each dice face corresponds to a value from 1 to 6, there are a total of 36 possible microstates (6 possible values for each die, combined).
02

(b) Identify the state to which each microstate corresponds

Recall that the state of our system is given by the sum of the values on the top faces of the two dice. For each arrangement (microstate), add up the values of the two top faces to find the corresponding state (also known as macrostate).
03

(c) Find the number of possible states

To find the number of possible states, we count all the possible sums of the top faces of the dice. The minimum sum is 2 (when both dice show a 1), and the maximum sum is 12 (when both dice show a 6). Thus, there are 11 possible states in total (corresponding to the sums 2, 3, 4, ..., 12).
04

(d) Identify the state or states with the highest entropy

Entropy is a measure of disorder or randomness in a system. The state with the highest entropy is the one that has the most microstates associated with it since more microstates mean more possible random configurations. In the case of the two-dice system, the state with the highest entropy is the one with the sum 7, as it has the most possible microstate combinations (1+6, 2+5, 3+4, 4+3, 5+2, and 6+1). There is only one state (sum 7) with the highest entropy.
05

(e) Identify the state or states with the lowest entropy

The state with the lowest entropy has the least number of microstates associated with it (i.e., the least random arrangements possible). For the two-dice system, these are the states with the lowest sums (2 and 12). Each of these states has only one possible microstate combination (1+1 for the sum 2 and 6+6 for the sum 12). Thus, there are two states (sum 2 and sum 12) with the lowest entropy.
06

(f) Calculate the absolute entropy of the two-dice system

To calculate the absolute entropy of the system, we can use Boltzmann's entropy formula: \(S = k_B \ln W\), where \(S\) is the entropy, k_B is Boltzmann's constant (1.38 x 10^-23 J/K), and \(W\) is the number of microstates in the system. In this case, the total number of microstates (combinations of top face values) is 36. So, the absolute entropy is: \(S = (1.38 \times 10^{-23} \, \text{J/K}) \ln(36) \approx 1.29 \times 10^{-22} \, \text{J/K}\). The absolute entropy of the two-dice system is approximately \(1.29 \times 10^{-22} \, \text{J/K}\).

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Most popular questions from this chapter

(a) Is the standard free-energy change, \(\Delta G^{\circ}\), always larger than \(\Delta G ?\) (b) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0 ?\) (c) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process necessarily has a low activation barrier?

Using data from Appendix \(\mathrm{C}\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free- energy change for these reactions at \(298 \mathrm{~K}:\) (a) $\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)$ (b) $2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)$ (c) $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$

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Indicate whether each statement is true or false. (a) The second law of thermodynamics says that entropy can only be produced but cannot not be destroyed. (b) In a certain process the entropy of the system changes by $1.2 \mathrm{~J} / \mathrm{K}\( (increase) and the entropy of the surroundings changes by \)-1.2 \mathrm{~J} / \mathrm{K}$ (decrease). Thus, this process must be spontaneous. (c) In a certain process the entropy of the system changes by $1.3 \mathrm{~J} / \mathrm{K}\( (increase) and the entropy of the surroundings changes by \)-1.2 \mathrm{~J} / \mathrm{K}$ (decrease). Thus, this process must be reversible.
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