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.

Short Answer

Expert verified
The efficiency of the Carnot engine using blackbody radiation is \( \eta = 1 - \frac{T_L}{T_H} \).

Step by step solution

01

Understand the Problem

The exercise involves blackbody radiation in a Carnot cycle. Given formulas for internal energy and pressure in terms of temperature and using them to plot a closed curve in the P-V plane for the Carnot cycle. Additionally, derive the Carnot engine efficiency using blackbody radiation as its working substance.
02

Internal Energy and Pressure Relations

Given relations are: \( U = a V T^4 \) \( P = \frac{1}{3} a T^4 \). These will be used to plot the Carnot cycle and derive the efficiency. Note that these expressions show how pressure and energy depend on volume and temperature.
03

Plotting Carnot Cycle (P-V Diagram)

A Carnot cycle has four stages: two isothermal and two adiabatic processes. - Isothermal expansion: At high temperature \( T_H \), the gas expands (\( V_1 \) to \( V_2 \)) and pressure decreases (\( P_1 \) to \( P_2 \)). Since temperature is constant, \( P V = \text{constant} \). - Adiabatic expansion: Gas expands without heat exchange, pressure and temperature fall (process \( V_2, P_2, T_H \) to \( V_3, P_3, T_L \)). - Isothermal compression: At low temperature \( T_L \), gas is compressed (volume decreases from \( V_3 \) to \( V_4 \)), pressure increases (\( P_3 \) increases to \( P_4 \)). - Adiabatic compression: Gas is compressed adiabatically, pressure and temperature rise (\( V_4, P_4, T_L \) to \( V_1, P_1, T_H \)). For each step, use the relation \( P = \frac{1}{3} a T^4 \).
04

Deriving Carnot Efficiency

Efficiency \( \eta \) for a Carnot engine is given by: \[ \eta = 1 - \frac{T_L}{T_H} \] From the given problem, since work is done and heat is exchanged typically at the high and low temperatures of the cycle, the usual Carnot efficiency derivation applies directly: - During isothermal expansion, heat absorbed \( Q_H = U (T_H, V_2) - U(T_H, V_1) \) - During isothermal compression, heat rejected \( Q_L = U(T_L, V_3) - U(T_L, V_4) \). Using these and the given formulas for pressure and energy, confirm the steps and conclude with the efficiency formula.

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Key Concepts

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

Blackbody Radiation
Blackbody radiation refers to electromagnetic radiation emitted by a body in thermal equilibrium. Such a blackbody absorbs all incidents' radiation regardless of frequency or angle. A perfect example is a theoretical construct, but it closely describes the radiation emitted by stars and planets.

In this exercise, blackbody radiation behaves similarly to a gas composed of photons with specific properties. The internal energy of this radiation is given as \( U = a V T^4 \), where \( a \) is Stefan's constant, \( V \) is volume, and \( T \) is the absolute temperature. The pressure exerted by this gas is \( P = \frac{1}{3} a T^4 \). These relationships are key to understanding how blackbody radiation operates within a Carnot cycle.

Blackbody radiation serves as an ideal working substance for thermodynamic cycles because of its predictable behavior. As we move through the cycle's stages (isothermal and adiabatic), these equations help dictate energy exchanges and pressure changes during different phases.
Thermodynamic Efficiency
Thermodynamic efficiency is a critical metric in evaluating how well an engine converts heat into work. This efficiency is represented by the symbol \( \eta \) and is inherently less than 100% due to the second law of thermodynamics.

For a Carnot engine, which operates between two heat reservoirs at different temperatures (\( T_H \) and \( T_L \)), the efficiency is given by:
\below is the formula: \texpression: \[ \eta = 1 - \frac{T_L}{T_H} \]

This formula implies that efficiency depends only on the temperatures of the hot and the cold reservoirs. It doesn't depend on the working substance, which in our case, is blackbody radiation. In our specific example, we use the relationship provided, like the heat absorbed \( Q_H \) during isothermal expansion and the heat rejected \( Q_L \) during isothermal compression, to show that the efficiency derived aligns with the general Carnot efficiency formula.

Consequently, this highlights that Carnot efficiency is universal, making it the gold standard for measuring the performance of real-world engines, even those utilizing unconventional working substances like blackbody radiation.
Isothermal and Adiabatic Processes
Isothermal and adiabatic processes are two fundamental types of thermodynamic processes that play crucial roles in the Carnot cycle.

  • Isothermal Process: During an isothermal process, the temperature of the system remains constant. For our Carnot cycle, this happens twice - during the expansion at the high temperature \( T_H \) and during the compression at the low temperature \( T_L \). Since the temperature stays constant, the product \( PV \) remains constant based on the ideal gas law. In our case, we use the relation \( P = \frac{1}{3} a T^4 \) to determine how pressure and volume change during these stages.
  • Adiabatic Process: A process is adiabatic if no heat is exchanged with the surroundings. This means that any change in the system's internal energy is due solely to work done by or on the system. There are two adiabatic processes in a Carnot cycle: an expansion phase following the high-temperature isothermal process and a compression phase preceding the low-temperature isothermal process. During these stages, both temperature and pressure change without heat exchange, driven purely by the thermodynamic properties of blackbody radiation.


Understanding these processes is crucial for plotting the Carnot cycle on a \( P-V \) diagram and calculating engine efficiency.

The combination of isothermal and adiabatic changes ensures maximum theoretical efficiency by minimizing irreversible entropy changes during the cycle.

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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.

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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 \(a=0.3658 \mathrm{~Pa} \mathrm{~m}^{6} / \mathrm{mol}^{2}\).)

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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