Energy of a Gas in equilibrium with BB-radiation

[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?

 

[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?

 

[VSayantan]

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?

 

[TSny]

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]

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 $R$?

 

[TSny]

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]

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

Thanks for pointing out.
So, $$U_{gas}={\frac 3 2}R$$

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!

 

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

 

[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.