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One way to derive Equation \(19.3\) depends on the observation that at constant \(T\) the number of ways, \(W\), of arranging \(m\) ideal-gas particles in a volume \(V\) is proportional to the volume raised to the \(m\) power: $$ W x V^{m} $$ Use this relationship and Boltzmann's relationship between entropy and number of arrangements (Equation 19.5) to derive the equation for the entropy change for the isothermal expansion or compression of \(n\) moles of an ideal gas.

Short Answer

Expert verified
To derive the equation for the entropy change during an isothermal expansion or compression of n moles of an ideal gas, we use the given relationship \(W \propto V^{m}\) and Boltzmann's relationship \(S = k_{B}\ln{W}\). By substituting the number of particles in terms of moles and Avogadro's number, we arrive at the equation \(\Delta S = nR \ln{\frac{V_{final}}{V_{initial}}}\), where ∆S represents the entropy change for the isothermal process.

Step by step solution

01

Write down the given relationship

Given the relationship \( W \propto V^{m} \), where W represents the number of ways to arrange m ideal-gas particles in a volume V.
02

Introduce a proportionality constant

Since W is proportional to the volume raised to the m power, we can replace the proportionality symbol with an equal sign and introduce a proportionality constant k: \( W = kV^{m} \).
03

Connect the relationship with entropy using Boltzmann's relationship

According to Boltzmann's relationship (Equation 19.5), the entropy S is related to the number of ways W by the equation \(S = k_{B}\ln{W}\), where \(k_{B}\) is Boltzmann's constant. So, we can express the entropy in terms of the volume V and the number of particles m as follows: \(S = k_{B} \ln{(kV^{m})}\)
04

Express the entropy in terms of moles and the Avogadro number

Since we are given n moles of the ideal gas, we can express the number of particles m in terms of moles and Avogadro's number N_A: \(m = nN_{A}\). Replace m with this expression: \(S = k_{B} \ln{(kV^{nN_{A}})}\)
05

Determine the entropy change

To find the change in entropy during an isothermal expansion or compression, we need to subtract the initial entropy from the final entropy: \(\Delta S = S_{final} - S_{initial} = k_{B} \ln{(kV_{final}^{nN_{A}})} - k_{B} \ln{(kV_{initial}^{nN_{A}})}\)
06

Simplify the expression for the entropy change

Apply the rules of logarithms to simplify the expression: \(\Delta S = k_{B} \ln{\frac{V_{final}^{nN_{A}}}{V_{initial}^{nN_{A}}}} = k_{B} \ln{(\frac{V_{final}}{V_{initial}})^{nN_{A}}}\) Since \(k_{B}N_{A} = R\), the ideal gas constant, we finally get: \(\Delta S = nR \ln{\frac{V_{final}}{V_{initial}}}\) This equation represents the entropy change for the isothermal expansion or compression of n moles of an ideal gas.

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Key Concepts

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

Boltzmann's relationship
Boltzmann's relationship is a fundamental concept in statistical mechanics that connects the macroscopic property of entropy to the microscopic parameter of the number of arrangements. Entropy is a measure of molecular disorder or randomness. According to Boltzmann, the entropy of a system, denoted by \(S\), can be determined using the formula \(S = k_{B} \ln{W}\), where:
  • \(S\) is the entropy of the system.
  • \(k_{B}\) is the Boltzmann's constant, approximately \(1.38 \times 10^{-23}\, \text{J/K}\).
  • \(W\) represents the number of microscopic ways the system can be arranged.
This relationship provides a bridge between the macroscopic laws of thermodynamics and the microscopic world of particles. When you know how many ways a system can be configured, you can determine its entropy, thus linking the properties of ideal gases and thermodynamic processes.
Isothermal processes
An isothermal process is a thermodynamic process that occurs at a constant temperature. In the context of ideal gases, such as in the isothermal expansion or compression discussed here, the temperature of the gas does not change. This is a crucial condition because temperature is a determinant of system energy.
In isothermal processes:
  • The internal energy change is zero because the temperature is constant.
  • According to the first law of thermodynamics, the heat added to the system is equal to the work done by the system: \(Q = W\).
  • Entropy change can be calculated by considering the exchange of heat with the surroundings, using the formula \(\Delta S = \frac{Q}{T}\).
The constant temperature condition allows us to use the relationship \(\Delta S = nR \ln{\frac{V_{final}}{V_{initial}}}\), derived in the step-by-step solution, which shows how entropy increases when a gas expands at constant temperature.
Ideal gas law
The ideal gas law is a fundamental equation describing the behavior of an ideal gas. This law combines several simpler gas laws and is expressed as \(PV = nRT\), where:
  • \(P\) represents the pressure of the gas.
  • \(V\) is the volume occupied by the gas.
  • \(n\) is the number of moles of gas.
  • \(R\) is the universal gas constant, approximately \(8.314 \, \text{J/(mol·K)}\).
  • \(T\) is the absolute temperature in Kelvin.
The ideal gas law provides a good approximation to the behavior of real gases under many conditions, particularly at low pressures and high temperatures. It integrates the relationships between pressure, volume, and temperature, allowing us to solve for one of these variables when the others are known.
In isothermal processes, using the ideal gas law helps to understand how volume changes affect entropy, as shown in the derivation where the entropy change for an ideal gas is linked to its volume change with \(\Delta S = nR \ln{\frac{V_{final}}{V_{initial}}}\). This equation relies on the ideal gas law to relate the volumetric changes to thermodynamic properties.

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Most popular questions from this chapter

Indicate whether each statement is true or false. (a) \(\Delta S\) for an isothermal process depends on both the temperature and the amount of heat reversibly transferred. (b) \(\Delta S\) is a state function. (c) The second law of thermodynamics says that the entropy of the system increases for all spontaneous processes.

(a) What sign for \(\Delta S\) do you expect when the pressure on \(0.600 \mathrm{~mol}\) of an ideal gas at \(350 \mathrm{~K}\) is increased isothermally from an initial pressure of \(0.750 \mathrm{~atm}\) ? (b) If the final pressure on the gas is \(1.20 \mathrm{~atm}\), calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.

Indicate whether cach statement is true or false. (a) The second law of thermodynamics says that entropy is conserved. (b) If the entropy of the system increases during a reversible process, the entropy change of the surroundings must decrease by the same amount. (c) In a certain spontaneous process the system undergoes an entropy change of \(4.2 \mathrm{~J} / \mathrm{K}\); therefore, the entropy change of the surroundings must be \(-4.2 \mathrm{~J} / \mathrm{K}\).

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta E\) zero for the process?

The following processes were all discussed in Chapter 18 , "Chemistry of the Environment." Estimate whether the entropy of the system increases or decreases during each process: (a) photodissociation of \(\mathrm{O}_{2}(g)\), (b) formation of ozone from oxygen molecules and oxygen atoms, (c) diffusion of CFCs into the stratosphere, (d) desalination of water by reverse osmosis.

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