Energy of a Gas in equilibrium with BB-radiation

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1. Nov 30, 2017

VSayantan

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. Nov 30, 2017

TSny

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?

3. Dec 1, 2017

VSayantan

You mean volume is fixed and pressure and temperature increases?

4. Dec 1, 2017

TSny

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.

5. Dec 1, 2017

VSayantan

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 $R$?

6. Dec 1, 2017

TSny

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 $T_0$?

7. Dec 1, 2017

VSayantan

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!

8. Dec 1, 2017

TSny

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 $T_0$.

9. Dec 1, 2017

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) $\alpha = \frac {4\sigma V}{c}$.

10. Dec 1, 2017

TSny

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.