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Chapter 20: Problem 42

The entropy of a macroscopic state is given by \(S=k_{B} \ln w\) where \(k_{\mathrm{B}}\) is the Boltzmann constant and \(w\) is the number of possible microscopic states. Calculate the change in entropy when \(n\) moles of an ideal gas undergo free expansion to fill the entire volume of a box after a barrier between the two halves of the box is removed.

Short Answer

Expert verified
In the given problem, we have n moles of an ideal gas undergoing a free expansion in a box. The change in entropy (ΔS) for this process is given by: ΔS = Nk_B * ln(2), where N is the number of moles and k_B is the Boltzmann constant.

Step by step solution

01

Calculate the initial microscopic states (w_initial)

In the beginning, the gas is restricted to half of the box volume, so \(V_i = \frac{V}{2}\). Using the microcanonical ensemble formula and replacing \(V_i\) with \(\frac{V}{2}\), we get: \(w_{initial} = \frac{(\frac{V}{2}^N N!)^{1/2}}{h^N (2 \pi m k_B T)^{3N/2}}\)
02

Calculate the initial entropy (S_initial)

Now we use the given entropy formula: \(S=k_B \ln w\) to find the initial entropy: \(S_{initial} = k_B \ln w_{initial} = k_B \ln \frac{(\frac{V}{2}^N N!)^{1/2}}{h^N (2 \pi m k_B T)^{3N/2}}\)
03

Calculate the final microscopic states (w_final)

After the free expansion, the gas fills the entire box volume, which is \(V\). Replacing the volume in the microcanonical ensemble formula with V, we get: \(w_{final} = \frac{(V^N N!)^{1/2}}{h^N (2 \pi m k_B T)^{3N/2}}\)
04

Calculate the final entropy (S_final)

Using the entropy formula and the obtained value for \(w_{final}\), we can calculate the final entropy: \(S_{final}=k_B \ln w_{final} = k_B \ln \frac{(V^N N!)^{1/2}}{h^N (2 \pi m k_B T)^{3N/2}}\)
05

Calculate the change in entropy (ΔS)

The change in entropy is given by \(\Delta S = S_{final} - S_{initial}\). Subtracting the initial entropy from the final entropy, we get: \(\Delta S = k_B \ln \frac{(V^N N!)^{1/2}}{h^N (2 \pi m k_B T)^{3N/2}} - k_B \ln \frac{(\frac{V}{2}^N N!)^{1/2}}{h^N (2 \pi m k_B T)^{3N/2}}\) We can simplify this expression using the logarithm properties: \(\Delta S = k_B \ln \frac{(V^N N!)^{1/2}}{(\frac{V}{2}^N N!)^{1/2}} = k_B \ln (2^N) = Nk_B \ln 2\) The change in entropy for the free expansion of n moles of an ideal gas in a box is \(\Delta S = Nk_B \ln 2\).

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Most popular questions from this chapter

A Carnot engine operates between a warmer reservoir at a temperature \(T_{1}\) and a cooler reservoir at a temperature \(T_{2}\). It is found that increasing the temperature of the warmer reservoir by a factor of 2 while keeping the same temperature for the cooler reservoir increases the efficiency of the Carnot engine by a factor of 2 as well. Find the efficiency of the engine and the ratio of the temperatures of the two reservoirs in their original form.

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 \(S_{\mathrm{A}}+S_{\mathrm{B}}\).

20.9a) The maximum efficiency of a Carnot engine is \(100 \%\) since the Carnot cycle is an ideal process. b) The Carnot cycle consists of two isothermal processes and two adiabatic processes. c) The Carnot cycle consists of two isothermal processes and two isentropic processes (constant entropy). d) The efficiency of the Carnot cycle depends solely on the temperatures of the two thermal reservoirs.

What capacity must a heat pump with a coefficient of performance of 3 have to heat a home that loses heat energy at a rate of \(12 \mathrm{~kW}\) on the coldest day of the year? a) \(3 \mathrm{~kW}\) c) \(10 \mathrm{~kW}\) e) \(40 \mathrm{~kW}\) b) \(4 \mathrm{~kW}\) d) \(30 \mathrm{~kW}\)

You are given a beaker of water. What can you do to increase its entropy? What can you do to decrease its entropy?
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