/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Consider a large system of \(N\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a large system of \(N\) indistinguishable, noninteracting molecules (perhaps in an ideal gas or a dilute solution). Find an expression for the Helmholtz free energy of this system, in terms of \(Z_{1}\), the partition function for a single molecule. (Use Stirling's approximation to eliminate the \(N !\).) Then use your result to find the chemical potential, again in terms of \(Z_{1}\).

Short Answer

Expert verified
Helmholtz free energy is \(F = NkT (\ln(N) - \ln(Z_1) - 1)\) and the chemical potential is \(\mu = kT (\ln(N) + 1 - \ln(Z_1))\).

Step by step solution

01

Understand the Problem

We have a system of indistinguishable, noninteracting molecules and need to find the Helmholtz free energy using the single molecule partition function, \(Z_1\). Stirling's approximation can be used to simplify factorial expressions.
02

Helmholtz Free Energy Expression

For a large system of \(N\) indistinguishable noninteracting molecules, the total partition function \(Z = \frac{Z_1^N}{N!}\). The Helmholtz free energy \(F\) is given by \(F = -kT \ln(Z)\), where \(k\) is the Boltzmann constant and \(T\) is the temperature.
03

Apply Stirling's Approximation

Stirling's approximation states that \(\ln(N!) \approx N \ln(N) - N\). Use this to approximate the expression: \(Z = \frac{Z_1^N}{N!}\), so \(\ln(Z) = N \ln(Z_1) - \ln(N!) = N \ln(Z_1) - (N \ln(N) - N)\).
04

Substitute in the Helmholtz Equation

Substitute the expression from Step 3 into the Helmholtz equation: \(F = -kT (N \ln(Z_1) - N \ln(N) + N)\). Simplifying gives: \(F = NkT (\ln(N) - \ln(Z_1) - 1)\).
05

Derive the Chemical Potential

The chemical potential \(\mu\) is the partial derivative of \(F\) with respect to \(N\), at constant temperature and volume. From the expression \(F = NkT (\ln(N) - \ln(Z_1) - 1)\), take the derivative: \(\mu = \left(\frac{\partial F}{\partial N} \right)_{T, V} = kT (\ln(N) + 1 - \ln(Z_1))\).

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Key Concepts

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

Partition Function
In statistical mechanics, the partition function is a crucial concept that helps us understand the statistical properties of a system at thermal equilibrium. For our problem, it plays a pivotal role in expressing the Helmholtz free energy of a system with many molecules.The partition function for a single molecule, denoted as \(Z_1\), gives us a way to capture all possible energy states of that molecule. It essentially sums over all possible states, weighted by the Boltzmann factor \(e^{-E_i/kT}\), where \(E_i\) is the energy of state \(i\), \(k\) is the Boltzmann constant, and \(T\) is the temperature.
For a system with many indistinguishable molecules, the total partition function is given by the expression \(Z = \frac{Z_1^N}{N!}\). Here, \(N!\) accounts for the indistinguishability of the molecules, ensuring that permutations of molecules do not count as unique states in our system. The partition function is then used in calculating other thermodynamic quantities such as Helmholtz free energy.
Stirling's Approximation
Stirling's approximation is a mathematical tool used to simplify the factorial of large numbers. It is particularly handy in thermodynamics when dealing with a large number of molecules, like in our problem, where \(N\) can be huge.This approximation states that \(\ln(N!) \approx N \ln(N) - N\). By using this approximation, we can make the computation of the logarithm of the factorial manageable, which is crucial in our formula for the partition function: \(Z = \frac{Z_1^N}{N!}\).
Applying Stirling's approximation allows us to rewrite the natural log of \(Z\) as \(\ln(Z) = N \ln(Z_1) - (N \ln(N) - N)\). This simplification is essential to derive the Helmholtz free energy expression, making it easier to handle analytically.
Chemical Potential
The chemical potential, denoted by \(\mu\), is a key concept in thermodynamics and chemistry, describing how the Gibbs free energy of a system changes with the addition of particles. In simple terms, it's the 'cost' in energy to add one more particle to the system.In our exercise, the chemical potential is derived from the Helmholtz free energy \(F\) by taking the partial derivative with respect to the number of molecules \(N\) while keeping temperature and volume constant. We start with the expression \(F = NkT (\ln(N) - \ln(Z_1) - 1)\), and differentiating with respect to \(N\) gives us:\[\mu = \left(\frac{\partial F}{\partial N}\right)_{T,V} = kT (\ln(N) + 1 - \ln(Z_1))\]
This result shows how the chemical potential depends on both the partition function \(Z_1\) and the number of particles \(N\), providing insights into how thermodynamic equilibrium can shift with changes in system composition.
Indistinguishable Molecules
Indistinguishable molecules are a concept that changes the way we count states in a system. It assumes that swapping two identical particles doesn't result in a new, unique configuration. This is fundamentally important when considering the statistics of particles, especially in quantum mechanics and statistical thermodynamics.In our problem scenario, where we have many indistinguishable noninteracting molecules, we account for this by dividing the partition function by \(N!\), the factorial of the number of molecules. This step ensures that each specific arrangement of molecules that only differs by a permutation isn't considered a distinct state.
This concept ties directly into the principles of quantum mechanics, where particles like atoms or molecules in a gas are indistinguishable. Therefore, the simple division by \(N!\) corrects for overcounting those configurations in the partition function, which is vital for accurate thermodynamic calculations.

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Most popular questions from this chapter

A particle near earth's surface traveling faster than about \(11 \mathrm{km} / \mathrm{s}\) has enough kinetic energy to completely escape from the earth, despite earth's gravitational pull. Molecules in the upper atmosphere that are moving faster than this will therefore escape if they do not suffer any collisions on the way out. (a) The temperature of earth's upper atmosphere is actually quite high, around \(1000 \mathrm{K} .\) Calculate the probability of a nitrogen molecule at this temperature moving faster than \(11 \mathrm{km} / \mathrm{s}\), and comment on the result. (b) Repeat the calculation for a hydrogen molecule (H \(_{2}\) ) and for a helium atom, and discuss the implications. (c) Escape speed from the moon's surface is only about \(2.4 \mathrm{km} / \mathrm{s}\). Explain why the moon has no atmosphere.

A water molecule can vibrate in various ways, but the easiest type of vibration to excite is the "flexing" mode in which the hydrogen atoms move toward and away from each other but the HO bonds do not stretch. The oscillations of this mode are approximately harmonic, with a frequency of \(4.8 \times 10^{13} \mathrm{Hz}\). As for any quantum harmonic oscillator, the energy levels are \(\frac{1}{2} h f, \frac{3}{2} h f, \frac{5}{2} h f,\) and so on. None of these levels are degenerate. (a) Calculate the probability of a water molecule being in its flexing ground state and in each of the first two excited states, assuming that it is in equilibrium with a reservoir (say the atmosphere) at \(300 \mathrm{K}\). (Hint: Calculate \(Z\) by adding up the first few Boltzmann factors, until the rest are negligible.) (b) Repeat the calculation for a water molecule in equilibrium with a reservoir at \(700 \mathrm{K}\) (perhaps in a steam turbine).

You might wonder why all the molecules in a gas in thermal equilibrium don't have exactly the same speed. After all, when two molecules collide, doesn't the faster one always lose energy and the slower one gain energy? And if so, wouldn't repeated collisions eventually bring all the molecules to some common speed? Describe an example of a billiard-ball collision in which this is not the case: The faster ball gains energy and the slower ball loses energy. Include numbers, and be sure that your collision conserves both energy and momentum.

Suppose you have 10 atoms of weberium: 4 with energy 0 eV, 3 with energy \(1 \mathrm{eV}, 2\) with energy \(4 \mathrm{eV}\), and 1 with energy \(6 \mathrm{eV}\). (a) Compute the average energy of all your atoms, by adding up all their energies and dividing by 10 . (b) Compute the probability that one of your atoms chosen at random would have energy \(E\), for each of the four values of \(E\) that occur. (c) Compute the average energy again, using the formula \(\bar{E}=\sum_{s} E(s) \mathcal{P}(s)\).

Consider a classical "degree of freedom" that is linear rather than quadratic: \(E=c|q|\) for some constant \(c .\) (An example would be the kinetic energy of a highly relativistic particle in one dimension, written in terms of its momentum.) Repeat the derivation of the equipartition theorem for this system, and show that the average energy is \(\bar{E}=k T\).

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