91Ó°ÊÓ

Formula to be used is$P=-(∂U/∂V{)}_{S,N}$

At constant N and S, the pressure equals the negative of the partial derivative of the total energy with respect to the volume, i.e.

$P=-{\left(\frac{∂U}{∂V}\right)}_{N,S}\left(1\right)$

To calculate the pressure, we must express the total energy in terms of entropy. The total energy is calculated as follows:

$U=\frac{8{\mathrm{Ï€}}^5(kT{)}^4}{15(hc{)}^3}V$

The entropy is given as:

$S=\frac{32{\mathrm{Ï€}}^5}{45}V{\left(\frac{kT}{hc}\right)}^{3}k$

Let

$A=\frac{8{\mathrm{Ï€}}^2{k}^4}{15(hc{)}^3}$

Total energy and entropy is:

$U=AV{T}^4\text{and}S=\frac{4}{3}AV{T}^3$

Solve entropy equation for T:

$T={\left(\frac{3S}{4AV}\right)}^{1/3}$

Substitute T in total energy:

$$\begin{gathered} U=AV{\left(\frac{3S}{4AV}\right)}^{4/3} \\ U=A{\left(\frac{3S}{4A}\right)}^{4/3}\frac{1}{V^{1/3}} \end{gathered}$$

Substitute in equation (1)

$$\begin{gathered} P=-A{\left(\frac{3S}{4A}\right)}^{4/3}\frac{∂}{∂V}\left(\frac{1}{V^{1/3}}\right) \\ P=-A{\left(\frac{3S}{4A}\right)}^{4/3}\left(-\frac{1}{3}\frac{1}{V^{4/3}}\right) \\ P=\frac{1}{3}A{\left(\frac{3S}{4AV}\right)}^{4/3} \end{gathered}$$

But,

$T={\left(\frac{3S}{4AV}\right)}^{1/3}$

Therefore,

$P=\frac{1}{3}A{T}^4$

The total energy of radiation is:

$U=\frac{8{\mathrm{Ï€}}^5(kT{)}^4}{15(hc{)}^3}V$

Substitute the values:

$$\begin{gathered} \frac{U}{V}=\frac{8{\mathrm{π}}^5{\left(\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)(1500\mathrm{K})\right)}^4}{15{\left(\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)\right)}^3} \\ =3.815×{10}^{-3}\mathrm{J}/{\mathrm{m}}^3 \end{gathered}$$

The pressure is:

$$\begin{gathered} P=\frac{1}{3}\left(3.815×{10}^{-3}\mathrm{J}/{\mathrm{m}}^3\right)=1.272×{10}^{-3}\mathrm{J}/{\mathrm{m}}^3=1.272×{10}^{-3}\mathrm{Pa} \\ \mathrm{P}=1.272×{10}^{-3}\mathrm{Pa} \end{gathered}$$

For comparison, the pressure within the kiln is the same as the pressure outside it, which is 1 atm in pascal 1.01 x 10⁵ Pa, which is about 10⁸ higher than the pressure caused by photons. The temperature at the sun's core is T = 15 x 10⁶ K, hence the energy per unit volume is:

$$\begin{gathered} \frac{U}{V}=\frac{8{\mathrm{π}}^5{\left(\left(1.38×{10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(15×{10}^6\mathrm{K}\right)\right)}^4}{15{\left(\left(6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)\right)}^3} \\ =3.815×{10}^{13}\mathrm{J}/{\mathrm{m}}^3 \end{gathered}$$

The pressure is:

role="math" localid="1647764290491" $$\begin{gathered} P=\frac{1}{3}\left(3.815×{10}^{13}\mathrm{J}/{\mathrm{m}}^3\right)=1.272×{10}^{13}\mathrm{J}/{\mathrm{m}}^3=1.272×{10}^{13}\mathrm{Pa} \\ P=1.272×{10}^{13}\mathrm{Pa} \end{gathered}$$

Number of moles per unit volume equals to:

$\frac{n}{V}=\left({10}^3\mathrm{mole}/\mathrm{kg}\right)\left({10}^5\mathrm{kg}/{\mathrm{m}}^3\right)={10}^8\mathrm{mole}/{\mathrm{m}}^3$

The pressure is:

$P=\frac{nRT}{V}=2\left({10}^8\mathrm{mole}/{\mathrm{m}}^3\right)(8.314\mathrm{J}/\mathrm{K}·\text{mole})\left(15×{10}^6\mathrm{K}\right)=2.5×{10}^{16}\mathrm{Pa}$

The factor 2 came from the fact that each ionised hydrogen atom has and electron and a proton.