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Calculate the change in entropy for a system in going from a condition of 5 accessible microstates to 30 accessible microstates.

Short Answer

Expert verified
The change in entropy is approximately \(2.47 \times 10^{-23} \, \text{J/K}.\)

Step by step solution

01

Understand the Formula

The change in entropy (\(\Delta S\)) for a system can be calculated using Boltzmann's entropy formula: \(S = k \, \ln(\Omega)\), where \(k\) is Boltzmann's constant (\(1.38 \times 10^{-23} \text{ J/K}\)) and \(\Omega\) is the number of accessible microstates. The change in entropy is given by \(\Delta S = k \, (\ln(\Omega_f) - \ln(\Omega_i))\), where \(\Omega_i\) and \(\Omega_f\) are the initial and final number of microstates, respectively.
02

Substitute Initial and Final Microstates

Identify \(\Omega_i = 5\) (initial microstates) and \(\Omega_f = 30\) (final microstates). Substitute these values into the entropy change formula: \(\Delta S = k \, (\ln(30) - \ln(5))\).
03

Calculate the Natural Logarithms

Calculate \(\ln(30)\) and \(\ln(5)\): \(\ln(30) \approx 3.4012\) and \(\ln(5) \approx 1.6094\). Subtract the two values: \(\ln(30) - \ln(5) \approx 3.4012 - 1.6094 = 1.7918\).
04

Calculate the Change in Entropy

Use Boltzmann's constant \(k = 1.38 \times 10^{-23} \text{ J/K}\) to calculate \(\Delta S\):\[\Delta S = 1.38 \times 10^{-23} \, \text{J/K} \, \times 1.7918 \approx 2.4727 \times 10^{-23} \, \text{J/K}.\]

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Key Concepts

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

Microstates
Microstates are a fundamental concept in statistical mechanics. They represent the different possible configurations that particles in a system can occupy, given the system's macroscopic constraints like energy. Think of a microstate as every single unique way the atoms or molecules in a system can be arranged. For instance, flipping a coin can result in two microstates: heads or tails. In more complex systems like gases or solids, the number of possible microstates can be much larger. The importance of microstates comes from their direct relation to entropy. More microstates mean more disorder or ways that a system can arrange itself while complying with the overarching constraints, which directly translates to higher entropy. When a physical system changes in a way that increases the number of its accessible microstates, its entropy will typically increase. It's crucial to grasp this concept to understand how changes in conditions, like temperature and pressure, affect the entropy of a system.
Boltzmann's entropy formula
Boltzmann's entropy formula is the mathematical relation that helps us quantify the entropy of a system based on its microstates. The formula is given by:\[ S = k \ln(\Omega) \]Here, \(S\) represents the entropy of the system, \(k\) stands for the Boltzmann's constant, and \(\Omega\) symbolizes the number of microstates. The formula shows that entropy is a function of the natural logarithm of the number of microstates, making entropy not linearly but logarithmically related to microstates.This logarithmic relationship is crucial because it nicely scales the huge number of microstates we could be dealing with to a more manageable figure of entropy, providing a clearer picture of disorder or randomness in real systems.Boltzmann's formula not only provides a way to calculate entropy but also cements the profound connection between macroscopic thermodynamic properties and the microscopic behaviors of particles in a system.
Boltzmann's constant
Boltzmann's constant is a key factor connecting the microscopic and macroscopic worlds in physics. Represented by the symbol \(k\), it has a value of approximately \(1.38 \times 10^{-23} \text{ J/K}\). For systems undergoing changes, like a gas expanding or compressing, Boltzmann's constant plays a crucial role in calculating how these molecular fluctuations result in changes in measurable quantities like entropy.In Boltzmann's entropy formula, this constant acts as the proportionality factor that bridges the statistical mechanics to thermodynamics. It translates the abstract number of microstates into actual measurable entropy units, making it possible to quantify disorder in terms we can relate to, like joules per kelvin.Understanding Boltzmann's constant helps us appreciate the utility of each component in the formula and the interrelatedness of physics at different scales, from the dance of individual electrons in an atom to the behavior of entire gas clouds.

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Most popular questions from this chapter

Identify the following processes as either spontaneous or not spontaneous. (a) Liquid water turns to ice when placed in a freezer at \(-5^{\circ} \mathrm{C}\) (b) Nitrogen gas is compressed to one half its original volume. (c) Sodium reacts with water forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{NaOH}(\mathrm{aq})\) (d) Slightly soluble \(\operatorname{CaSO}_{4}\left(K_{\mathrm{sp}}=4.5 \times 10^{-5}\right)\) dis- solves in water to form a saturated solution.

Which substance has the higher entropy? (a) dry ice (solid \(\mathrm{CO}_{2}\) ) at \(-78^{\circ} \mathrm{C}\) or \(\mathrm{CO}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C}\) (b) liquid water at \(25^{\circ} \mathrm{C}\) or liquid water at \(50^{\circ} \mathrm{C}\) (c) pure alumina, \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}),\) or ruby (ruby is \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in which some \(\mathrm{Al}^{3+}\) ions in the crystalline lattice are replaced with \(\mathrm{Cr}^{3+}\) ions) (d) one mole of \(\mathrm{N}_{2}(\mathrm{g})\) at 1 bar pressure or one mole of \(\mathrm{N}_{2}(\mathrm{g})\) at 10 bar pressure (both at \(298 \mathrm{K})\)

The ionization constant, \(K_{\mathrm{a}},\) for acetic acid is \(1.8 \times\) \(10^{-5}\) at \(25^{\circ} \mathrm{C} .\) What is the value of \(\Delta_{\mathrm{r}} \mathrm{G}^{\circ}\) for this reaction? Is this reaction product- or reactantfavored at equilibrium?

For each of the following processes, predict the algebraic sign of \(\Delta_{r} H^{\circ}, \Delta_{r} S^{\circ},\) and \(\Delta_{r} G^{\circ} .\) No calculations are necessary; use your common sense. (a) The decomposition of liquid water to give gaseous oxygen and hydrogen, a process that requires a considerable amount of energy. (b) Dynamite is a mixture of nitroglycerin, \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{N}_{3} \mathrm{O}_{9},\) and diatomaceous earth. The explo- sive decomposition of nitroglycerin gives gaseous products such as water, \(\mathrm{CO}_{2}\), and others; much heat is evolved. (c) The combustion of gasoline in the engine of your car, as exemplified by the combustion of octane. $$2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{g})+25 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 16 \mathrm{CO}_{2}(\mathrm{g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$

The normal melting point of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6},\) is \(5.5^{\circ} \mathrm{C} .\) For the process of melting, what is the sign of each of the following? (a) \(\Delta_{\mathrm{r}} H^{\circ}\) (b) \(\Delta_{\mathrm{r}} S^{\circ}\) (c) \(\Delta_{r} G^{\circ}\) at \(5.5^{\circ} \mathrm{C}\) (d) \(\Delta_{\mathrm{r}} G^{\circ}\) at \(0.0^{\circ} \mathrm{C}\) (e) \(\Delta_{r} G^{\circ}\) at \(25.0^{\circ} \mathrm{C}\)

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