Energy of a Gas in equilibrium with BB-radiation

• VSayantan
In summary: There is no potential energy, just kinetic energy.For the blackbody radiation $$U_{rad}=\sigma {T^4}$$Work done in increasing temperature from ##T_0~K## to ##{(T_0+1)}~K## of the radiation is $${(U_{rad})}_i-{(U_{rad})}_f=3RT_0-{(U_{rad})}_f$$So, ##{(U_{rad})}_f## can be determined by dividing the work done by the temperature increase.$${(U_{rad})}_f=3RT_0-{( VSayantan Homework Statement A closed, thermally-insulated box contains one mole of an ideal monatomic gas G in thermodynamic equilibrium with blackbody radiation B. The total internal energy of the system is ##U=U_{G}+U_{B}##, where ##U_{G}## and ##U_{B} (\propto T^4)## are the energies of the ideal gas and the radiation respectively. If ##U_{G}=U_{B}## at a certain temperature ##T_0~K##, then find the energy required to raised the temperature from ##T_0~K## to ##(T_0 + 1)~K##, in terms of the gas constant ##R##. Homework Equations Equation of state of an Ideal gas$$pV=nRT$$Adiabatic relation of an Ideal gas$$pV^{\gamma}=constant$$Ration of specific heats$$\gamma = 1+{\frac 2 f}$$where ##f## is the number of degrees of freedom of the gas. Work done by an Ideal gas$$W=\int p \, dV$$Pressure exerted by Blackbody radiation$$P_{rad}={\frac 1 4}\alpha T^4$$Internal energy of Blackbody radiation$$u_{rad}=3pV$$Entropy of Blackbody radiation$$S={\frac 4 3} \alpha V T^3$$The Attempt at a Solution The system is closed and thermally-insulated, so the change in energy is adiabatic. For a monatomic gas the degrees of freedom is ##3##. So, ratio of specific heats$$\frac {C_p}{C_V}=\frac 5 3$$Change in energy of an adiabatic process for an ideal gas is$$W=\int_{V_1}^{V_2} p\, dV\Rightarrow W=\int_{V_1}^{V_2} {\frac {k}{V^\gamma}}\, dV\Rightarrow W=k\int_{V_1}^{V_2} {V^{-\gamma}}\, dV\Rightarrow W={\frac {1}{\gamma - 1}}[{p_1}{V_1}-{p_2}{V_2}]$$But, with ##n=1##$${p_1}{V_1}=R{T_0}$$and ,$${p_2}{V_2}=R{(T_0+1)}$$Therefore,$$W={\frac {R}{\gamma - 1}}[{T_0}-{(T_0 +1)}]W=-{\frac {3}{2}}R$$Entropy of Blackbody radiation is$$S={\frac 4 3} \alpha V T^3$$Which gives$$T={\sqrt[3]{{\frac 3 4}{\frac {S}{\alpha}}}} V^{\frac {-1}{3}}$$Along with$$P_{rad}={\frac 1 4}\alpha T^4$$One obtains a relation$$PV^{\frac 4 3}=constant$$which is similar in form with the adiabatic relation of an ideal gas, except for the exponent. Then, the energy required for expansion is$$W_{rad}=\int_{V_1}^{V_2} p\, dV\Rightarrow W_{rad}=k\int_{V_1}^{V_2} {\frac {1}{V^{\frac {4}{3}}}}\, dV$$Which simplifies to$$W_{rad}=3[{p_1}{V_1}-{p_2}{V_2}]$$But$$U=3pV$$So,$$W_{rad}=U_i-U_f$$Also, ##U_i## is the internal energy of the Blackbody radiation at temperature ##T_0~K##, which is equal to ##U_G##. Thus,$$W_{rad}=U_G-U_f$$Now, how do I use these to expressions for energies to obtain the final result? Last edited: Hello. I'm not sure of the interpretation of the problem. To me, a "box" is something that has fixed walls; therefore, the volume would not change significantly when the temperature increases by 1 K. So, could it be that you are meant to assume that the volume stays constant while energy is added? VSayantan TSny said: Hello. I'm not sure of the interpretation of the problem. To me, a "box" is something that has fixed walls; therefore, the volume would not change significantly when the temperature increases by 1 K. So, could it be that you are meant to assume that the volume stays constant while energy is added? You mean volume is fixed and pressure and temperature increases? VSayantan said: You mean volume is fixed and pressure and temperature increases? That's how I would interpret the problem. I could be wrong. See how far you can get with just using the formula for ##U_G## in terms of ##T## for a monatomic ideal gas and using the fact that ##U_B \propto T^4## for the radiation. VSayantan TSny said: That's how I would interpret the problem. I could be wrong. See how far you can get with just using the formula for ##U_G## in terms of ##T## for a monatomic ideal gas and using the fact that ##U_B \propto T^4## for the radiation. For the gas total internal energy at temperature ##T~K## is$$U=2\times 3{N_A}\times {\frac 1 2}{k_B}T$$the factor ##2## arises because the the gas molecules have kinetic energy as well as potential energy. For the blackbody radiation$$U_{rad}=\sigma {T^4}$$Work done in increasing temperature from ##T_0~K## to ##{(T_0+1)}~K## of the gas is$$U_i-U_f=3R[T_0-(T_0+1)]$$Which is ##3R## Work done in increasing temperature from ##T_0~K## to ##{(T_0+1)}~K## of the radiation is$${(U_{rad})}_i-{(U_{rad})}_f=3RT_0-{(U_{rad})}_f$$Now how do I obtain a value of ##{(U_{rad})}_f## in terms of ##R##? VSayantan said: For the gas total internal energy at temperature ##T~K## is$$U=2\times 3{N_A}\times {\frac 1 2}{k_B}T$$the factor ##2## arises because the the gas molecules have kinetic energy as well as potential energy. For an ideal gas there is no potential energy, just kinetic energy. For the blackbody radiation$$U_{rad}=\sigma {T^4}$$OK, here ##\sigma## is some proportionality constant. Initially, the energy of the gas equals the energy of the radiation. Can you use this to determine ##\sigma## in terms of ##R## and ##T_0##? VSayantan TSny said: For an ideal gas there is no potential energy, just kinetic energy. Thanks for pointing out. So,$$U_{gas}={\frac 3 2}R$$TSny said: OK, here ##\sigma## is some proportionality constant. Initially, the energy of the gas equals the energy of the radiation. Can you use this to determine ##\sigma## in terms of ##R## and ##T_0##? Actually value of the proportionality constant ##\sigma##, the Stefan's constant, is known, given by$$\sigma = \frac {2{\pi}^5 {k_B}^4}{15{c^2}{h^3}}$$But ##k_B##, the Boltzmann constant, is ##\frac {R}{N_A}##, so,$$\sigma = \frac {2{\pi}^5 {R^4}}{15{c^2}{h^3}{N_A}^4}$$This only complicates the whole thing! The energy ##U_B## of the blackbody radiation in the box is not given by ##\sigma T^4## where ##\sigma## is the Stefan-Boltzmann constant. Rather, ##U_B = \frac{4 \sigma V}{c} T^4## where ##V## is the volume of the box and ##c## is the speed of light. For example see http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/raddens.htmlBut to work this problem you just need to know that ##U_B \propto T^4##. So, ##U_B = AT^4## for some constant ##A##. You should be able to relate the constant ##A## to ##T_0## and ##R## using the fact that ##U_B = U_G## at the temperature ##T_0##. VSayantan OK, I think I've found a way. The energy required to increase the temperature of the gas from ##T_0~K## to ##{(T_0+1)}~K## is$$\Delta {U_{gas}}={U_{gas}}_i - {U_{gas}}_f$$Which simplifies to$$\Delta {U_{gas}}={\frac 3 2}RT_0-{\frac 3 2}R{(T_0+1)}$$That is$$\Delta {U_{gas}}=-{\frac 3 2}R$$The energy required to increase the temperature of the radiation from ##T_0~K## to ##{(T_0+1)}~K## is$$\Delta {U_{rad}}={U_{rad}}_i - {U_{rad}}_f\Rightarrow {\Delta {U_{rad}}}=\alpha {[{(T_0)}^4-{(T_0+1)}^4]}\Rightarrow {\Delta {U_{rad}}}=\alpha {[{(T_0)}^2-{(T_0+1)}^2]}{[{(T_0)}^2+{(T_0+1)}^2]}\Rightarrow {\Delta {U_{rad}}}=\alpha {[{T_0}-{(T_0+1)}]}{[{(T_0)}+{(T_0+1)}]}{[{(T_0)}^2+{(T_0+1)}^2]}$$Since ##T_0 \cong T_0+1##$$ {\Delta {U_{rad}}}\approx \alpha {[{T_0}-{(T_0+1)}]}{[{2(T_0)}]}{[2{(T_0)}^2]} {\Delta {U_{rad}}}=4 \alpha {T_0}^3{[{T_0}-{(T_0+1)}]} {\Delta {U_{rad}}}=-4 \alpha {T_0}^3$$But,$${U_{rad}}_i={U_{gas}}_i\Rightarrow \alpha {T_0}^4= {\frac 3 2}RT\Rightarrow \alpha {T_0}^3= {\frac 3 2}R$$Using this, one obtains$${\Delta {U_{rad}}}=-4\times {\frac 3 2}R$$i.e.,$${\Delta {U_{rad}}}=-6R$$Therefore,$$\Delta U={\Delta {U_{rad}}}+{\Delta {U_{gas}}}\Rightarrow \Delta U=-{\frac 3 2}R - 6R$$That is$$ \Delta U=-{\frac {15} {2}}R

This looks nice, I think. (because there is an option ##7.5R## )

Thanks @TSny for helping. Relating ##\sigma## with ##T_0## was crucial!

After your last suggestion I've changed the constant ##\sigma## to ##\alpha##, where (as you correctly observed) ##\alpha = \frac {4\sigma V}{c}##.

Both the energy of the gas and the energy of the radiation should increase when the temperature increases. So, ##\Delta U## for each should be positive. Note that ##\Delta U = U_f - U_i## rather than ##\Delta U = U_i - U_f##. Otherwise, your work looks good.

Your approximation holds good as long as ##T_0 >> 1##. You can check that this is true by using your relation ##\alpha T_0^3 = \frac{3}{2} R## and solving for ##T_0## . Using the expression for ##\alpha## in terms of ##c##, ##\sigma##, and ##V##, you will find that ##T_0 >> 1## as long as the volume of the box is not unreasonably large.

VSayantan

1. What is meant by "energy of a gas in equilibrium with BB-radiation"?

The energy of a gas in equilibrium with BB-radiation refers to the balance between the energy emitted by a gas and the energy absorbed by the same gas from its surroundings, specifically from black body (BB) radiation. This equilibrium is important in understanding the thermal properties of gases and the interactions between them and their environment.

2. How is the energy of a gas in equilibrium with BB-radiation calculated?

The energy of a gas in equilibrium with BB-radiation can be calculated using the Planck's Law, which describes the intensity of radiation emitted by a black body at a given wavelength and temperature. This law takes into account the temperature of the gas and its surrounding environment, as well as the wavelength of the radiation being emitted or absorbed.

3. What factors affect the energy of a gas in equilibrium with BB-radiation?

The energy of a gas in equilibrium with BB-radiation is primarily affected by the temperature of the gas and its surrounding environment. Other factors that may influence this energy include the composition and density of the gas, as well as the wavelength of the radiation being emitted or absorbed.

4. Why is understanding the energy of a gas in equilibrium with BB-radiation important?

Understanding the energy of a gas in equilibrium with BB-radiation is important for several reasons. It helps us understand the thermal properties of gases and how they interact with their environment. This knowledge is crucial for applications in fields such as atmospheric science, astrophysics, and engineering.

5. Can the energy of a gas in equilibrium with BB-radiation be manipulated?

Yes, the energy of a gas in equilibrium with BB-radiation can be manipulated by changing the temperature of the gas or its surrounding environment, as well as by altering the composition and density of the gas. This can have significant effects on the thermal properties and behavior of the gas, making it an important factor to consider in various scientific and technological applications.

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