- #1

VSayantan

- 59

- 4

## 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: