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Chapter 20: Problem 45

Suppose an atom of volume \(V_{\mathrm{A}}\) is inside a container of volume \(V\). The atom can occupy any position within this volume. For this simple model, the number of states available to the atom is given by \(V / V_{A}\). Now suppose the same atom is inside a container of volume \(2 V .\) What will be the change in entropy?

Short Answer

Expert verified
Question: Calculate the change in entropy when the volume of a container with a single atom changes from V to 2V, knowing that the number of states available to the atom is proportional to the container's volume over the atom's volume. Answer: The change in entropy, ΔS, is equal to k ln{2}, where k is the Boltzmann constant.

Step by step solution

01

Calculate Entropy of Initial State

The initial entropy of the atom, \(S_1\), can be calculated using the formula: \(S_1 = k \ln{\Omega_1}\), where \(k\) is the Boltzmann constant and \(\Omega_1\) is the number of states in container volume \(V\). The number of states in this case is given by \(\Omega_1=V/V_\mathrm{A}\).
02

Calculate Entropy of Final State

The final entropy of the atom, \(S_2\), can be calculated in the same way as in step 1, but with the container volume \(2V\). The number of states in this case is given by \(\Omega_2=(2V)/V_\mathrm{A}\). Thus, \(S_2 = k \ln{\Omega_2}\).
03

Calculate Change in Entropy

The change in entropy, \(ΔS\), is the difference between the final entropy and the initial entropy: \(ΔS = S_2 - S_1\). By substituting the expressions for \(S_1\) and \(S_2\) found in steps 1 and 2, we get: \(ΔS = k(\ln{(2V/V_\mathrm{A})} - \ln{(V/V_\mathrm{A})})\).
04

Simplify the Expression

Using the property of logarithms, we can simplify the expression for \(ΔS\): \(ΔS = k (\ln{(\frac{2V}{V_\mathrm{A}} \cdot \frac{V_\mathrm{A}}{V})}) = k \ln{2}\). The change in entropy, \(ΔS\), is equal to \(k \ln{2}\).

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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.

Consider a Carnot engine that works between thermal reservoirs with temperatures of \(1000.0 \mathrm{~K}\) and \(300.0 \mathrm{~K}\). The average power of the engine is \(1.00 \mathrm{~kJ}\) per cycle. a) What is the efficiency of this engine? b) How much energy is extracted from the warmer reservoir per cycle? c) How much energy is delivered to the cooler reservoir?

The burning of fuel transfers \(4.00 \cdot 10^{5} \mathrm{~W}\) of power into the engine of a \(2000 .-\mathrm{kg}\) vehicle. If the engine's efficiency is \(25.0 \%,\) determine the maximum speed the vehicle can achieve \(5.00 \mathrm{~s}\) after starting from rest.

Find the net change in entropy when \(100 . \mathrm{g}\) of water at \(0^{\circ} \mathrm{C}\) is added to \(100 . \mathrm{g}\) of water at \(100 .{ }^{\circ} \mathrm{C}\)

A certain refrigerator is rated as being \(32.0 \%\) as ef ficient as a Carnot refrigerator. To remove \(100 .\) J of heat from the interior at \(0^{\circ} \mathrm{C}\) and eject it to the outside at \(22^{\circ} \mathrm{C}\), how much work must the refrigerator motor do?
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