Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Energy of a Gas in equilibrium with BB-radiation

  1. Nov 30, 2017 #1
    1. The problem statement, all variables and given/known data
    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##.

    2. Relevant 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 $$S={\frac 4 3} \alpha V T^3$$

    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: Nov 30, 2017
  2. jcsd
  3. Nov 30, 2017 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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?
     
  4. Dec 1, 2017 #3
    You mean volume is fixed and pressure and temperature increases?
     
  5. Dec 1, 2017 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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.
     
  6. Dec 1, 2017 #5
    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##?
     
  7. Dec 1, 2017 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    For an ideal gas there is no potential energy, just kinetic energy.

    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##?
     
  8. Dec 1, 2017 #7
    Thanks for pointing out.
    So, $$U_{gas}={\frac 3 2}R$$


    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!
     
  9. Dec 1, 2017 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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.html


    But 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##.
     
  10. Dec 1, 2017 #9
    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## :-p)

    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}##.
     
  11. Dec 1, 2017 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted