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Prove that Boltzmann's microscopic definition of entropy, \(S=k_{\mathrm{B}} \ln w\), implies that entropy is an additive variable: Given two systems, A and B, in specified thermodynamic states, with entropies \(S_{A}\) and \(S_{\mathrm{p}}\), respectively, show that the corresponding entropy of the combined system is \(S_{\mathrm{A}}+S_{\mathrm{B}}\).

Short Answer

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Question: Prove that entropy is an additive variable using Boltzmann's microscopic definition of entropy. Answer: Using the Boltzmann formula, \(S = k_B \ln w\), we expressed the entropies of two independent systems A and B, and their combined system (A+B). By determining the number of microstates for the combined system (A+B) as the product of the individual systems' microstates, we were able to demonstrate that the entropy of the combined system (A+B) is equal to the sum of the entropies of the individual systems, \(S_A + S_B\), which proves that entropy is an additive variable.

Step by step solution

01

Write down the Boltzmann formula for System A and System B

We first write down the Boltzmann formula for System A and System B. This will help us set up the given entropy expressions for both systems. \[ S_A=k_B \ln w_A, \quad S_B = k_B \ln w_B \]
02

Express the number of microstates for the combined system

Now, we need to determine the number of microstates for the combined System A+B. Since the systems are independent, the total number of microstates for the combined system is equal to the product of the individual systems: \[ w_{A+B} = w_A \cdot w_B \]
03

Write down the Boltzmann formula for the combined system

Using the Boltzmann formula \(S = k_B \ln w\), we can express the entropy for the combined system as follows: \[ S_{A+B} = k_B \ln w_{A+B} \] Now we can substitute the expression for the number of microstates of the combined system: \[ S_{A+B} = k_B \ln(w_A \cdot w_B) \]
04

Use logarithm properties to split the sum

Next, we take advantage of the logarithm properties to separate the sum of the individual entropies. Recall that \(\ln(ab) = \ln(a) + \ln(b)\), so the entropy of the combined system can be expressed as the sum of logarithms: \[ S_{A+B} = k_B (\ln w_A + \ln w_B) \]
05

Show that the entropy of the combined system is the sum of the entropies of the individual systems

Now, we simply need to recognize that the result obtained in Step 4 can be rewritten in terms of the entropies of the individual systems: \[ S_{A+B} = k_B \ln w_A + k_B \ln w_B = S_A + S_B \] This proves that the entropy of the combined system is equal to the sum of the entropies of the individual systems, demonstrating that entropy is an additive variable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Additivity of Entropy
Entropy is a fundamental concept in thermodynamics and statistical mechanics. Boltzmann's microscopic formula for entropy, given by \(S = k_B \ln w\), provides insight into how the entropy of a system is related to the number of microstates it can occupy. When dealing with two independent systems, A and B, each in their own thermodynamic state, the idea of additivity of entropy comes into play. If we consider that the systems are independent, the total number of ways both systems can be arranged, or their combined microstates, is calculated by multiplying the microstates of each system. This leads to a deep understanding:
  • The expression \(w_{A+B} = w_A \cdot w_B\) reflects the way probabilities in independent systems multiply.
  • Boltzmann's definition allows entropy to be represented additively when these microstates are related through a logarithmic function, thanks to the logarithm property \(\ln(ab) = \ln(a) + \ln(b)\).
Thus, for the combined system, the entropy becomes \(S_{A+B} = k_B (\ln w_A + \ln w_B) = S_A + S_B\). This sum confirms that in thermodynamics, entropy is indeed an additive quantity.
Thermodynamic Systems
In physics, a thermodynamic system is a defined portion of the universe that we study to understand the fundamental laws of energy, work, and reactions. A system's thermodynamic state is characterized by a set of state variables like pressure, volume, and temperature. These systems can be isolated, closed, or open, depending on whether they exchange energy or matter with their surroundings. Understanding thermodynamic systems is crucial when discussing entropy and microstates. The concept of a thermodynamic system becomes critical when systems are combined or interact. An isolated system, for example, can be thought of as a perfectly sealed box, where no energy or matter crosses its boundaries. In contrast, an open system exchanges both energy and matter, like a pot of boiling water. Thermodynamic systems provide the necessary framework for applying laws like the additivity of entropy. When two independent systems are combined, maintaining the independence allows us to apply the concept of additivity effortlessly, as each system's variables remain distinct and unchanged unless acted upon by external forces.
Microstates
Microstates are pivotal to understanding the concept of entropy. They refer to the specific ways the components of a system can be arranged at a particular moment. For any thermodynamic system, a microstate corresponds to a unique arrangement of all its particles and energy levels.The number of possible microstates, denoted by \(w\), is what gives a statistical measure to the system's entropy, as expressed in Boltzmann's famous relationship \(S = k_B \ln w\). Here are some key aspects:
  • A higher number of microstates equates to higher entropy since there are more ways to arrange the system's components.
  • Each distinct arrangement is equally probable, emphasizing that systems favor states of higher entropy, typical of their large number of accessible microstates.
Thus, when calculating the entropy of combined systems, the multiplication of microstates \(w_A \cdot w_B\) aligns with the system's probability multiplying rule in independent states, leading us to Boltzmann's conclusion on entropy’s additivity in independent thermodynamic systems.

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Most popular questions from this chapter

An ideal gas is enclosed in a cylinder with a movable piston at the top. The walls of the cylinder are insulated, so no heat can enter or exit. The gas initially occupies volume \(V_{1}\) and has pressure \(p_{1}\) and temperature \(T_{1}\). The piston is then moved very rapidly to a volume of \(V_{2}=3 V_{1}\). The process happens so rapidly that the enclosed gas does not do any work. Find \(p_{2}, T_{2},\) and the change in entropy of the gas.

While looking at a very small system, a scientist observes that the entropy of the system spontaneously decreases. If true, is this a Nobel-winning discovery or is it not that significant?

Which of the following processes (all constanttemperature expansions) produces the most work? a) An ideal gas consisting of 1 mole of argon at \(20^{\circ} \mathrm{C}\) expands from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\). b) An ideal gas consisting of 1 mole of argon at \(20^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\). c) An ideal gas consisting of 2 moles of argon at \(10^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\). d) An ideal gas consisting of 1 mole of argon at \(40^{\circ} \mathrm{C}\) expands from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\) e) An ideal gas consisting of 1 mole of argon at \(40^{\circ} \mathrm{C}\) expands from \(2 \mathrm{~L}\) to \(4 \mathrm{~L}\).

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Consider a room air conditioner using a Carnot cycle at maximum theoretical efficiency and operating between the temperatures of \(18^{\circ} \mathrm{C}\) (indoors) and \(35^{\circ} \mathrm{C}\) (outdoors). For each 1.00 J of heat flowing out of the room into the air conditioner: a) How much heat flows out of the air conditioner to the outdoors? b) By approximately how much does the entropy of the room decrease? c) By approximately how much does the entropy of the outdoor air increase?

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