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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: $$ W \propto V^{m} $$ 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 for the isothermal expansion or compression of n moles of an ideal gas, we follow these steps: 1. Write down the given relationships: \(W \propto V^m\) and \(S = k ln(W)\). 2. Substitute the first relationship into the second: \(S = k ln(V^m)\). 3. Simplify the expression for entropy and relate m to the number of moles: \(S = mk ln(V)\) and \(m = nN_A\). 4. Substitute the relationship between m and n into the entropy expression: \(S = (nN_A)k ln(V)\). 5. Derive the equation for entropy change: \(\Delta S = (nN_A)kln\left(\frac{V_{final}}{V_{initial}}\right)\). This gives us the equation for the entropy change for the isothermal expansion or compression of n moles of an ideal gas.

Step by step solution

01

Write down the given relationships

We are given two relationships in the problem: 1. \(W \propto V^m\), where m is the number of ideal-gas particles, and W is the number of ways to arrange these particles in a volume V. 2. Boltzmann's relationship between entropy and number of arrangements, Equation 19.5, which states: \[S = k ln(W)\] Here, S is the entropy, k is the Boltzmann constant, and W is the number of arrangements of particles.
02

Substitute the first relationship into the second

We can substitute the relationship between W and V (from Step 1) into Boltzmann's relationship between entropy and the number of arrangements: \[S = k ln(V^m)\]
03

Simplify the expression for entropy and relate m to the number of moles

Using the properties of logarithms, we can simplify the expression for entropy: \[S = k ln(V^m) = mk ln(V)\] Now, we need to relate the number of particles (m) to the number of moles (n). We know that the number of particles = number of moles × Avogadro's number: \[m = nN_A\]
04

Substitute the relationship between m and n into the entropy expression

Now, we will substitute the relationship between m and n into the entropy expression derived in Step 3: \[S = (nN_A)k ln(V)\]
05

Derive the equation for entropy change

To find the entropy change for the isothermal expansion or compression of n moles of an ideal gas, we will consider the final entropy (S_final) and the initial entropy (S_initial) for the system: The final entropy is given by: \[S_{final} = (nN_A)k ln(V_{final})\] The initial entropy is given by: \[S_{initial} = (nN_A)k ln(V_{initial})\] The entropy change (\(ΔS\)) is the difference between the final and initial entropies: \[ΔS = S_{final} - S_{initial} = (nN_A)k\left[ln(V_{final}) - ln(V_{initial})\right]\] Using the properties of logarithms, we can combine the terms inside the brackets: \[ΔS = (nN_A)kln\left(\frac{V_{final}}{V_{initial}}\right)\] This is the equation for 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 forms a core principle in statistical mechanics. It connects the disorder within a system, known as entropy, to the number of ways particles can be arranged. This relationship is mathematically expressed as:\[S = k \ln(W)\]where:
  • S is the entropy of the system.
  • k is Boltzmann's constant, which is approximately \(1.38 \times 10^{-23}\, J/K\).
  • W represents the number of distinct ways the particles in the system can be arranged.
When the arrangement possibilities increase, the entropy also rises, indicating more disorder and randomness.
Boltzmann's relationship is powerful in analyzing the microscopic states and understanding how they relate to macroscopic thermodynamic properties. By substituting a system's microstate arrangements into this expression, one can predict changes in entropy, as demonstrated in exercises involving ideal gases.
Ideal gas law
The ideal gas law is a crucial equation in physical chemistry, describing the behavior of ideal gases under various conditions. It links pressure (P), volume (V), and temperature (T) with the amount of gas (n), providing a comprehensive view of gas interactions.Mathematically, the ideal gas law is represented as:\[PV = nRT\]where:
  • P is 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 \(8.314\, J/(mol\cdot K)\).
  • T is the temperature in Kelvin.
This equation assumes that the gas particles do not interact except for elastic collisions. In the context of entropy and thermodynamic processes, the ideal gas law helps explain how gases expand or compress when heat is added or removed.
It simplifies the understanding of complex gas behaviors by providing a straightforward, albeit idealized, model to work with.
Isothermal process
An isothermal process is a thermodynamic process in which the temperature remains constant. This type of process is significant, especially in understanding how gases behave when undergoing expansion or compression without temperature change.In an isothermal process involving an ideal gas, the ideal gas law \(PV = nRT\) still applies. However, because the temperature \(T\) is constant, it implies that both pressure and volume may change, but their product will remain constant.
In practical scenarios:
  • During an isothermal expansion, a gas increases in volume as it absorbs energy in the form of heat, without an increase in temperature.
  • Conversely, during isothermal compression, the gas releases energy as it decreases in volume, again, without a temperature change.
The isothermal process is crucial for deriving expressions for entropy change. By knowing that the temperature is constant, it simplifies calculations and enables direct application of known formulas, such as the change in entropy derived using Boltzmann's relationship.
Understanding these processes allows students to predict how entropic properties shift in different thermodynamic scenarios.

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Most popular questions from this chapter

For a particular reaction, \(\Delta H=-32 \mathrm{~kJ}\) and \(\Delta S=\) \(-98 \mathrm{~J} / \mathrm{K}\). Assume that \(\Delta H\) and \(\Delta S\) do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?\) (b) If \(T\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

The volume of \(0.100 \mathrm{~mol}\) of helium gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from \(2.00 \mathrm{~L}\) to \(5.00 \mathrm{~L}\). Assuming the gas to be ideal, calculate the entropy change for the process.

(a) How can we calculate \(\Delta S\) foran isothermal process? (b) Does \(\Delta S\) for a process depend on the path taken from the initial to the final state of the system? Explain.

Carbon disulfide \(\left(\mathrm{CS}_{2}\right)\) is a toxic, highly flam mable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(l)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) : \begin{tabular}{lrl} \hline & \(\Delta H_{f}^{\circ}(\mathbf{k J} / \mathrm{mol})\) & \(\Delta G_{f}^{0}(\mathbf{k J} / \mathrm{mol})\) \\ \hline \(\mathrm{CS}_{2}(l)\) & \(89.7\) & \(65.3\) \\ \(\mathrm{CS}_{2}(g)\) & \(117.4\) & \(67.2\) \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? (b) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) bums in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C\), calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part (c). Is the reaction exothermic? Is it spontaneous at 298 K? (e) Use the data in the preceding table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(l)\). Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the preceding table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(\mathrm{l})\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(1 \mathrm{~atm}\) ?

How would each of the following changes affect the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas?

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