Blackbody radiation in a box of volume \(V\) and at temperature \(T\) has internal energy \(U=a V T^{4}\) and pressure \(P=1 / 3 a T^{4}\), where \(a\) is the Stefan- Boltzmann constant. (a) What is the fundamental equation for blackbody radiation (the entropy)? (b) Compute the chemical potential.

Short Answer

Expert verified
The fundamental equation for the entropy is \( S = \frac{4}{3} a V T^3 \). The chemical potential \( \text{ \( \text{ ( μ ) } } \) \) is zero.

Step by step solution

01

- Understand the Internal Energy and Pressure

Given the internal energy equation \[ U = a V T^4 \] and the pressure equation \[ P = \frac{1}{3} a T^4, \] where \( a \) is the Stefan-Boltzmann constant. Identify that these variables will be used to find entropy and chemical potential.
02

- Use Thermodynamic Relations

Recall that the first law of thermodynamics is \[ dU = TdS - PdV. \]Given this relationship, express the entropy \( S \).
03

- Specify the Entropy Equation

Recognize the total differential and solve for the entropy: \[ dU = T \frac{abla U}{abla T} dT + \frac{abla U}{abla V} dV = T dS - P dV. \] Using the provided relations, we can substitute to find: \[ dU = 4a V T^3 dT + a T^4 dV = T dS - \frac{1}{3} a T^4 dV. \]
04

- Rearrange for the Entropy

Rearranging the equation to solve for the entropy yields: \[ dS = \frac{4a V T^3}{T} dT + a T^3 \frac{2}{3} dV. \]After integrating, the result is: \[ S = \frac{4}{3} a V T^3. \]
05

- Calculate the Chemical Potential

The chemical potential \( \text{ ( μ ) } \) for photons is known to be zero: \[ \text{Since blackbody radiation consists of photons, which have zero rest mass and can't be created nor destroyed, their chemical potential is zero.} \]

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Key Concepts

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

Internal Energy
In the context of blackbody radiation, internal energy is an important concept. For a given volume, the internal energy is represented by the equation: \( U = a V T^{4} \), where
  • \( U \) is the internal energy
  • \( V \) is the volume
  • \( T \) is the temperature in Kelvin
  • \( a \) is the Stefan-Boltzmann constant
This equation shows that internal energy depends on the volume and the fourth power of the temperature. This exponential increase signifies how quickly the internal energy of blackbody radiation rises with temperature.
Entropy
Entropy is a measure of disorder or randomness in a system. For blackbody radiation, finding the entropy involves understanding the relationship between internal energy, temperature, and volume. From the first law of thermodynamics, we know: \( dU = TdS - PdV \). Given the pressure \( P \) and internal energy \( U \), you can find entropy by solving: \( dU = 4a V T^3 dT + a T^4 dV = TdS - \frac{1}{3} a T^4 dV \). After simplifying and integrating, we get the entropy equation for blackbody radiation: \( S = \frac{4}{3} a V T^3 \).
  • \( S \) represents the entropy
  • \( a \) is the Stefan-Boltzmann constant
  • \( V \) is the volume
  • \( T \) is the temperature
This shows that entropy for blackbody radiation is directly proportional to volume and the cube of the temperature.
Chemical Potential
The chemical potential (\( \mu \)) describes the change in a system's energy when the number of particles changes. For blackbody radiation, which consists of photons, the chemical potential is zero. This is because:
  • Photons are not conserved quantities—they are created or destroyed in processes without fixed numbers like in thermal radiation.
  • Photons have zero rest mass.
Therefore, the chemical potential \( \mu \) for blackbody radiation can be directly stated as zero.
Stefan-Boltzmann Constant
The Stefan-Boltzmann constant (\( a \)) is a physical constant that plays a critical role in blackbody radiation. It is denoted by the symbol \( \sigma \) and has a value: \ \( 5.670374419 \times 10^{-8} W m^{-2} K^{-4} \). This constant is key in the Stefan-Boltzmann law which states that: \[ P = \sigma T^4\] Where \( P \) represents the power radiated per unit area of a blackbody and \( T \) is the absolute temperature. This demonstrates that the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of its temperature. The same constant \( a \) reappears in the equations for internal energy and entropy, underlying its fundamental importance in thermodynamics and blackbody radiation.

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Most popular questions from this chapter

Compute the entropy, enthalpy, Helmholtz free energy, and Gibbs free energy of a paramagnetic substance and write them explicitly in terms of their natural variables when possible. Assume that the mechanical equation of state is \(m=(D H / T)\) and that the molar heat capacity at constant magnetization is \(c_{\mathrm{m}}=c\), where \(m\) is the molar magnetization, \(H\) is the magnetic field, \(D\) is a constant, \(c\) is a constant, and \(T\) is the temperature.

A stochastic process, involving three fluctuating quantities, \(x_{1}, x_{2}\), and \(x_{3}\), has a probability distribution $$ P\left(x_{1}, x_{2}, x_{3}\right)=C \exp \left[-\frac{1}{2}\left(2 x_{1}^{2}+2 x_{1} x_{2}+4 x_{2}^{2}+2 x_{1} x_{3}+2 x_{2} x_{3}+2 x_{3}^{2}\right)\right] $$ where \(C\) is the normalization constant. (a) Write probability distribution in the form \(P\left(x_{1}, x_{2}, x_{3}\right)=C \exp \left(-1 / 2 x^{T} \cdot g+x\right)\), where \(g\) is a \(3 \times 3\) symmetric matrix, \(x\) is a column matrix with matrix elements \(x_{i}, i=1,2,3\), and \(x^{T}\) is its transpose. Obtain the matrix \(\boldsymbol{g}\) and its inverse \(g^{-1}\). (b) Find the eigenvalues \(\lambda_{i}(i=1,2,3)\) and orthonormal eigenvectors of \(\boldsymbol{g}\) and obtain the \(3 \times 3\) orthogonal matrix \(\boldsymbol{O}\) that diagonalizes the matrix \(\boldsymbol{g}\) (get numbers for all of them). Using this orthogonal matrix, we can write \(x^{\mathrm{T}} \cdot g \cdot x=x^{\mathrm{T}} \cdot \boldsymbol{O}^{\mathrm{T}} \cdot \boldsymbol{O} \cdot g \cdot \boldsymbol{O}^{\mathrm{T}} \cdot \boldsymbol{O} \cdot \boldsymbol{x}=\boldsymbol{a}^{\mathrm{T}} \cdot \bar{\Lambda} \cdot \boldsymbol{a}=\sum_{i=1}^{3} \lambda_{i} a_{i}^{2}\) where \(\boldsymbol{O} \cdot g \cdot \boldsymbol{O}^{\mathrm{T}}=\bar{\Lambda}\) is a \(3 \mathrm{x}\) 3 diagonal matrix with matrix elements \((\bar{A})_{i, j}=\lambda_{i} \delta_{i, j}\) and \(\boldsymbol{O} \cdot \boldsymbol{x}=\boldsymbol{a}\) is a column matrix with elements, \(a_{i}(i=1,2,3)\). (c) Compute the normalization constant, C. (d) Compute the moments \(\left(x_{i}\right)(i=1,2,3),\left\langle x_{i} x_{j}\right\rangle(i=1,2,3, j=1,2,3)\left(x_{1}^{2} x_{2} x_{3}\right)\) and \(\left\langle x_{1} x_{2}^{2} x_{3}\right\rangle+\) (Note that Exercises \(\mathrm{A.7}\) and \(\mathrm{A} .8\) might be helpful.)

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Show that \(T \mathrm{~d} s=c_{x}(\partial T / \partial Y)_{x} \mathrm{~d} Y+c_{Y}(\partial T / \partial x)_{Y} \mathrm{~d} x\), where \(x=X / n\) is the amount of extensive variable, \(X\), per mole, \(c_{x}\) is the heat capacity per mole at constant \(x\), and \(c_{Y}\) is the heat capacity per mole at constant \(Y\).

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