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Chapter 3: Problem 9

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.

Short Answer

Expert verified
Solve \(C_{V, N}, C_{P, N}, S, U \) using virial expansion, balance derivative equations for \(F(T)=-T B_2(T) \) ensuring thermal consistency.

Step by step solution

01

Understand the Problem

The problem provides the virial expansion for a low-density gas and asks to find the correction to the heat capacity (\(C_{V, N}\)). It also asks for the form of \(F(T)\) to ensure thermodynamic consistency, and to find \(C_{P, N}\), entropy, and internal energy.
02

Given Equations

The provided virial expansion equation is \[ P = \frac{N k_{\text{B}} T}{V} \bigg[1 + \frac{N}{V} B_2(T)\bigg] \] and the heat capacity equation is \[ C_{V, N} = \frac{3}{2} N k_{\text{B}} - \frac{N^2 k_{\text{B}}}{V} F(T) \]
03

Thermodynamic Consistency

To ensure thermodynamic consistency, we compare the symmetric difference relations. Starting with the equation of state: \[ \frac{\text{d} P}{\text{d} T} = \frac{N k_{\text{B}}}{V} \bigg[1 + \frac{N}{V} B_2(T) + \frac{N}{V} T \frac{d B_2(T)}{d T}\bigg] \] And taking the temperature derivative of the provided specific heat capacity equation: \[ -\frac{N^2 k_{\text{B}}}{V} \frac{dF(T)}{dT}\]
04

Equate Consistency Conditions

We must match the coefficient of density terms in both equations resulting: \[ -\frac{N^2 k_{\text{B}}}{V} \frac{dF(T)}{dT} umber \frac{N^2 k_{\text{B}}}{V} T \frac{d B_2(T)}{d T} \rightarrow F(T) = -T B_2(T)\]
05

Find Heat Capacity at Constant Pressure

Using \( C_{P} - C_{V} = \frac{\text{d} P}{\text{d} T} \bigg(\frac{\text{d} V}{\text{d} T}\bigg)_P \) and provided heat capacity equation: \[ C_{P, N} = \frac{5}{2} N k_{\text{B}} - \frac{2N^2 k_{\text{B}}}{V} B_2(T)\]
06

Entropy

Entropy can be directly integrated from \(C_V(T)\) and pressure volume terms. From it: \[ S = N k_{\text{B}} \bigg[\text{ln}\frac{V}{N} +\frac{5}{2} - \frac{N B_2(T)}{V}\bigg]\]
07

Internal Energy

Internal energy calculus from heat capacities: \[ U(T) = \frac{3}{2} N k_{\text{B}} T - \frac{N^2 k_{\text{B}} T}{V} B_2(T)\]

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Key Concepts

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

Virial Expansion
To better understand thermodynamics in low-density gases, we use the virial expansion. This expansion modifies the ideal gas law to account for interactions between gas molecules.
The equation of state for a low-density gas can be written as follows:
\[ P = \frac{N k_{\mathrm{B}} T}{V} \left[1+\frac{N}{V} B_{2}(T)\right] \]
Here, \(B_{2}(T)\) is known as the second virial coefficient, which captures the first-order correction due to intermolecular forces.

  • N: The number of gas molecules
  • T: Temperature
  • V: Volume
  • P: Pressure
  • kB: Boltzmann constant
This formulation is essential for understanding gases that do not behave ideally, especially at low densities.
Heat Capacity Corrections
Heat capacity describes the amount of heat needed to change a substance's temperature. For low-density gases, corrections to the ideal gas heat capacity are necessary.
The corrected heat capacity at constant volume, \(C_{V, N}\), is given by:
\[ C_{V, N} = \frac{3}{2} N k_{\mathrm{B}} - \frac{N^2 k_{\mathrm{B}}}{V} F(T) \]
The function \(F(T)\) reflects the impact of interactions between molecules, requiring adjustment to achieve thermodynamic consistency. By comparing derivatives of pressure and heat capacity, we find that:
\[ F(T) = -T B_{2}(T) \]
This relationship ensures the thermodynamic properties of the gas remain consistent with fundamental laws.
Second Virial Coefficient
The second virial coefficient, \(B_{2}(T)\), plays a crucial role in the virial expansion.
It quantifies the deviation from ideal gas behavior due to interactions between pairs of molecules:
\[ B_{2}(T) \]
This term adjusts the pressure in the equation of state:
\[ P = \frac{N k_{\mathrm{B}} T}{V} \left[1+\frac{N}{V} B_{2}(T)\right] \]
The form of \(B_{2}(T)\) depends on temperature and the specific interactions between molecules within the gas.
For thermodynamic consistency, related heat capacity corrections must include the term \(F(T)\), linked as:
\[ F(T) = -T B_{2}(T) \]
Entropy
Entropy is a measure of the disorder or randomness in a system. For a low-density gas, it can be determined by considering the heat capacities and equation of state. For our gas, entropy \(S\) can be expressed as:
\[ S = N k_{\mathrm{B}} \bigg[\text{ln}\frac{V}{N} + \frac{5}{2} - \frac{N B_{2}(T)}{V}\bigg] \]
Here, entropy depends on the volume, number of molecules, and second virial coefficient:
  • Increased disorder when volume or temperature increases.
  • Modification by the virial coefficient \(B_{2}(T)\) due to molecular interactions.
These components help capture the non-ideal behavior of gases and how interactions affect entropy.
Internal Energy
Internal energy, \(U(T)\), encompasses the energy contained within the gas due to both molecule movement and interactions. For a low-density gas, the internal energy can be derived from heat capacities and the virial expansion:
\[ U(T) = \frac{3}{2} N k_{\mathrm{B}} T - \frac{N^2 k_{\mathrm{B}} T}{V} B_{2}(T) \]
  • First Term: Reflects kinetic energy contributions.
  • Second Term: Accounts for molecular interactions adjusted by the second virial coefficient \(B_{2}(T)\).
Understanding internal energy is crucial for studying gas behavior under various conditions, showing how temperature and molecular forces influence the system's total energy.

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Most popular questions from this chapter

Electromagnetic radiation in an evacuated vessel of volume \(V\) at equilibrium with the walls at temperature \(T\) (blackbody radiation) behaves like a gas of photons having internal energy \(U=a V T^{4}\) and pressure \(P=1 / 3 a T^{4}\), where \(a\) is Stefan's constant. (a) Plot the closed curve in the \(P-V\) plane for a Carnot cycle using blackbody radiation. (b) Derive explicitly the efficiency of a Carnot engine which uses blackbody radiation as its working substance.

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.)

A heat engine uses blackbody radiation as its operating substance. The equation of state for blackbody radiation is \(P=1 / 3 a T^{4}\) and the internal energy is \(U=a V T^{4}\), where \(a=7.566 \times 10^{-16}\) \(\mathrm{J} /\left(\mathrm{m}^{3} \mathrm{~K}^{4}\right)\) is Stefan's constant, \(P\) is pressure, \(T\) is temperature, and \(V\) is volume. The engine cycle consists of three steps. Process \(1 \rightarrow 2\) is an expansion at constant pressure \(P_{1}=P_{2} .\) Process \(2 \rightarrow 3\) is a decrease in pressure from \(P_{2}\) to \(P_{3}\) at constant volume \(V_{2}=V_{3}\). Process \(3 \rightarrow 1\) is an adiabatic contraction from volume \(V_{3}\) to \(V_{1}\). Assume that \(P_{1}=3.375 P_{3}, T_{1}=2000 \mathrm{~K}\), and \(V_{1}=10^{-3} \mathrm{~m}^{3}\). (a) Express \(V_{2}\) in terms of \(V_{1}\) and \(T_{1}=T_{2}\) in terms of \(T_{3}\) (b) Compute the work done during each part of the cycle. (c) Compute the heat absorbed during each part of the cycle. (d) What is the efficiency of this heat engine (get a number)? (e) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures.

An insulated box with fixed total volume \(V\) is partitioned into \(m\) insulated compartments, each containing an ideal gas of a different molecular species. Assume that each compartment has the same pressure but a different number of moles, a different temperature, and ( \(\left.\left.P, \mathrm{n}_{i}, T_{\mathrm{i}}, V_{\mathrm{i}}\right)_{-}\right)\)a different volume. (The thermodynamic variables for the ith compartment are If all partitions are suddenly removed and the system is allowed to reach equilibrium: (a) Find the final temperature and pressure, and the entropy of mixing. (Assume that the particles are monatomic.) (b) For the special case of \(m=2\) and parameter \(\mathrm{n}_{1}=1 \mathrm{~mol}, T_{1}=300 \mathrm{~K}, V_{1}=1 \mathrm{l}, \mathrm{n}_{2}=3 \mathrm{~mol}_{1}\) and \(V_{2}=2\) 1, obtain numerical values for all parameters in part (a).

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