A monatomic fluid in equilibrium is contained in a large insulated box of volume \(V\). The fluid is divided (conceptually) into \(m\) cells, each of which has an average number of particles \(N_{0}\), where \(N_{0}\) is large (neglect coupling between cells). Compute the variance in fluctuations of internal energy per particle \(u=U / N,\left\langle\left(\Delta u_{i}\right)^{2}\right\rangle\), in the ith cell. (Hint: Use temperature \(T\) and volume per particle \(v=V / N\) as independent variables.)

Short Answer

Expert verified
The variance in fluctuations of internal energy per particle is \(\frac{3 k_B^2 T^2}{2 N_0}\

Step by step solution

01

Understand the Given Parameters

The problem involves a monatomic fluid in equilibrium contained in a large insulated box of volume V. The fluid is conceptually divided into m cells. Each cell has an average number of particles \(N_0\) which is large. We need to compute the variance in fluctuations of internal energy per particle \(u\). Assume temperature \(T\) and volume per particle \(v=V/N\) as independent variables.
02

Define Internal Energy per Particle

The internal energy per particle for a monatomic ideal gas can be written as \(u = \frac{3}{2} k_B T\), where \(k_B\) is the Boltzmann constant and \(T\) is the temperature.
03

Express Total Internal Energy

The total internal energy for \(N_0\) particles in a cell can be given by \(U = N_0 u = N_0 \frac{3}{2} k_B T\).
04

Determine Energy Fluctuation

The fluctuation in the total internal energy \(U\) is related to the heat capacity of the system. For a monatomic ideal gas, we have \(C_V = \frac{3}{2}N_0 k_B\). The variance in the energy fluctuation can be given by \(\langle(\Delta U)^2\rangle = k_B T^2 C_V \).
05

Calculate Variance Using Heat Capacity

Substitute the expression for \(C_V\) into the variance formula: \(\langle(\Delta U)^2\rangle = k_B T^2 \frac{3}{2} N_0 k_B = \frac{3}{2} N_0 k_B^2 T^2 \).
06

Compute Variance per Particle

To find the variance per particle \(u\), divide the variance by \(N_0^2\): \(\langle(\Delta u)^2\rangle = \frac{\langle(\Delta U)^2\rangle}{N_0^2} = \frac{\frac{3}{2} N_0 k_B^2 T^2}{N_0^2} = \frac{3 k_B^2 T^2}{2 N_0}\).
07

Finalize the Answer

Thus, the variance in fluctuations of internal energy per particle \(u\) is \(\left\langle \- \langleu_i\rangle^2\right\rangle = \frac{3 k_B^2 T^2}{2 N_0}\).

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Key Concepts

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

Internal Energy Fluctuation
When studying statistical mechanics, it's important to understand how internal energy fluctuates in a system in equilibrium. Internal energy is the total energy contained within a system due to the kinetic and potential energies of its particles. In an ideal gas, particularly monatomic, these energies are influenced by temperature and the motion of the particles.

Fluctuations refer to the variations around the average value of a quantity—in this case, the internal energy. These fluctuations are especially relevant when dealing with small systems, where changes are more noticeable. For a larger number of particles, the relative size of these fluctuations decreases, but they still play a crucial role in the thermodynamic behavior of the system.

More formally, the variance in the internal energy fluctuation \textbf{per particle} can be computed to give insights into the stability and thermal properties of the gas. The variance \(\textless(\Delta u_i)^2\textgreater\) provides a measure of the expected deviation of the internal energy per particle from its mean.
Monatomic Ideal Gas
A monatomic ideal gas is a theoretical gas composed of identical, single-atom particles that do not interact with each other except through elastic collisions. This idealization helps simplify the complex interactions seen in real gases, making it easier to apply statistical mechanics principles.

The internal energy of a monatomic ideal gas is directly related to its temperature and can be expressed as:\[u = \frac{3}{2} k_B T\] where \(\textsl{k_B}\) is the Boltzmann constant and \(\textsl{T}\) is the temperature in Kelvin. This equation shows that the internal energy per particle is proportional to the temperature, implying that higher temperatures lead to greater internal energies due to increased particle movement.

In the context of our problem, understanding this relationship is critical as it lays the groundwork for calculating fluctuations and other thermodynamic properties.
Variance Calculation
The variance of a quantity measures how spread out its values are around the mean. In the context of our exercise, we seek to determine the variance in the fluctuations of internal energy per particle. Let’s consider the steps involved:

  • The total internal energy \(\textsl{U}\) for \( \textsl{N_0} \) particles in a cell is given by: \[ U = N_0 u = N_0 \frac{3}{2} k_B T \]

  • The heat capacity at constant volume for a monatomic ideal gas is: \[ C_V = \frac{3}{2} N_0 k_B \]

  • The variance in the fluctuation of total internal energy is then: \[ \langle(\Delta U)^2 \rangle = k_B T^2 C_V \]

  • Using the value of \( \textsl{C_V} \): \[ \langle(\Delta U)^2 \rangle = k_B T^2 \frac{3}{2} N_0 k_B = \frac{3}{2} N_0 k_B^2 T^2 \]\

  • To find the variance per particle, divide by \( N_0^2 \): \[ \langle(\Delta u)^2 \rangle = \frac{\langle (\Delta U)^2 \rangle}{N_0^2} = \frac{\frac{3}{2} N_0 k_B^2 T^2 }{N_0^2} = \frac{3 k_B^2 T^2}{2 N_0} \]


This final expression tells us the variance in energy fluctuations per particle is inversely proportional to the average number of particles in each cell and directly proportional to the square of the temperature.
Heat Capacity Relation
Heat capacity is a fundamental concept in thermodynamics and statistical mechanics, often denoted as \( \textsl{C} \), which measures the amount of heat needed to change a system's temperature by a certain amount. For a monatomic ideal gas at constant volume, the heat capacity \( C_V \) is given by: \[ C_V = \frac{3}{2} N_0 k_B \]

This relation reflects the linear relationship between the internal energy and the temperature. It is also essential for calculating the fluctuations in internal energy, linking temperature changes to energy changes.

In our exercise, heat capacity plays a critical role in determining the variance in energy fluctuations. The relationship: \[ \langle (\Delta U)^2 \rangle = k_B T^2 C_V \]

highlights how the heat capacity affects the internal energy's fluctuations. By substituting the value for \( C_V \), we derived the energy variance which directly leads to understanding individual particles' fluctuation behavior.

This illustrates that as the heat capacity increases, either by more particles or higher specific heat, the magnitudes of fluctuations in energy also increase. Such relations are crucial for analyzing thermal systems and predicting their behavior under different conditions.

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Most popular questions from this chapter

Compute the heat capacity at constant magnetic field \(C_{H, n}\), the susceptibilities \(X_{T}, \mathrm{n}\) and \(X_{S, n}\) ' and the thermal expansivity \(\alpha_{H, n}\) for a magnetic system, given that the mechanical equation of state is \(M=\mathrm{n} D H / T\) and the heat capacity is \(C_{M, n}=\pi c_{\prime}\) where \(M\) is the magnetization, \(H\) is the magnetic field, \(\mathrm{n}\) is the number of moles, \(D\) is a constant, \(c\) is the molar heat capacity, and \(T\) is the temperature.

Experimentally one finds that for a rubber band $$ \begin{aligned} &\left(\frac{\partial J}{\partial L}\right)_{\mathrm{T}, M}=\frac{a T}{L_{0}}\left[1+2\left(\frac{L_{0}}{L}\right)^{3}\right] \quad \text { and } \\\ &\left(\frac{\partial J}{\partial T}\right)_{L, M}=\frac{a L}{L_{0}}\left[1-\left(\frac{L_{0}}{L}\right)^{3}\right] \end{aligned} $$ where \(J\) is the tension, \(a=1.0 \times 10^{3} \mathrm{dyn} / \mathrm{K}\), and \(L_{0}=0.5 \mathrm{~m}\) is the length of the band when no tension is applied. The mass \(M\) of the rubber band is held fixed. (a) Compute \((\partial L / \partial T)_{J, M}\) and discuss its physical meaning. (b) Find the equation of state and show that \(\mathrm{d} J\) is an exact differential. (c) Assume that the heat capacity at constant length is \(C_{L}=1.0 \mathrm{~J} / \mathrm{K}\). Find the work necessary to stretch the band reversibly and adiabatically to a length of \(1 \mathrm{~m}\). Assume that when no tension is applied, the temperature of the band is \(T=290 \mathrm{~K}\). What is the change in temperature?

For a low-density gas the virial expansion can be terminated at first order in the density and the equation of state is \( P=\frac{N k_{\mathrm{B}} T}{V}\left[1+\frac{N}{V} B_{2}(T)\right] $$ where \)B_{2}(T)\( is the second virial coefficient. The heat capacity will have corrections to its ideal gas value. We can write it in the form $$ C_{V, N}=\frac{3}{2} N k_{\mathrm{B}}-\frac{N^{2} k_{\mathrm{B}}}{V} F(T) $$ (a) Find the form that \)F(T)\( must have in order for the two equations to be thermodynamically consistent. (b) Find \)C_{P, N}$. (c) Find the entropy and internal energy.

A biological molecule of unknown mass can be prepared in pure powdered form. If 15 \(\mathrm{g}\) of this powder is added to a container with \(1 \mathrm{~L}\) of water at \(T=300 \mathrm{~K}\), which is initially at atmospheric pressure, the pressure inside the container increases to \(P=1.3 \mathrm{~atm}\). (a) What is the molecular weight of the biological molecules? (b) What is the mass of each molecule expressed in atomic units?

A van der Waals gas can be cooled by free expansion. Since no work is done and no heat is added during free expansion, the internal energy remains constant. An infinitesimal change in volume \(\mathrm{d} V\) causes an infinitesimal temperature change in \(\mathrm{d} T\), where $$ \mathrm{d} T=\left(\frac{\partial T}{\partial V}\right)_{U, \mathrm{n}} \mathrm{d} V $$ (a) Compute the Joule coefficient \({ }^{(\partial T / \partial V)}_{U, \mathrm{H}}\) for a van der Waals gas (note that the heat capacity \(C_{V, n}\) is independent of volume and use \(C_{V, n}=3 / 2 \mathrm{n} R\) ) (b) Compute the change in temperature of one mole of oxygen \(\left(\mathrm{O}_{2}\right)\) and one mole of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) if they each expand from an initial volume \(V_{\mathrm{i}}=10^{-3} \mathrm{~m}^{3}\) at temperature \(T_{\mathrm{i}}=300 \mathrm{~K}\) to a final volume \(V_{\mathrm{f}}=\infty\). (For \(\mathrm{O}_{2}\) the van der Waals constant is \(a=0.1382 \mathrm{~Pa} \mathrm{~m}^{6} / \mathrm{mol}^{2}\) and for \(\mathrm{CO}_{2}\) it is \(a=0.3658 \mathrm{~Pa} \mathrm{~m}^{6} / \mathrm{mol}^{2}\).)

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