At the center of the Sun, the density of the gas is$\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ and the composition is essentially 35% hydrogen by mass and 65% helium by mass. (a) What is the number density of protons there? (b) What is the ratio of that proton density to the density of particles in an ideal gas at standard temperature (0°C) and pressure$\left(1.01\times {10}^5\mathrm{Pa}\right)$ ?

At the center of the Sun the density of the hydrogen is 35% from the gas density,

${\mathit{\rho }}_{\mathbf{H}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathit{\rho }$

The gas density is the number density of H multiplied by the mass of H.

${\mathit{\rho }}_{\mathbf{H}}\mathbf{=}{\mathit{n}}_{\mathbf{H}}{\mathit{m}}_{\mathbf{H}}$

Combine above two equations.

$$\begin{gathered} {\mathit{n}}_{\mathbf{H}}{\mathit{m}}_{\mathbf{H}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\mathit{\rho } \\ {\mathit{n}}_{\mathbf{H}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{\rho }}{{\mathbf{m}}_{\mathbf{H}}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right) \end{gathered}$$

Here,$\mathit{\rho }$ is density, and${\mathit{m}}_{\mathbf{H}}$ is mass of hydrogen atom.

Substitute all the known values in equation (1).

$$\begin{gathered} n_H=\frac{0.35\left(1.5\times {10}^{5}kg/m^3\right)}{1.67\times {10}^{-27}kg} \\ =3.1\times {10}^{31}protons/m^3 \end{gathered}$$

Therefore, the number of density of protons is$3.1\times {10}^{31}protons/m^3$ .

From the ideal gas law,

$$\begin{gathered} PV=NkT \\ \frac{N}{V}=\frac{P}{kT}...........\left(2\right) \end{gathered}$$

Substitute all the known values in equation (2).

$$\begin{gathered} \frac{N}{V}=\frac{1.01\times {10}^{5}Pa}{\left(1.38\times {10}^{-23}J/K\right)\left(273K\right)} \\ =2.68\times {10}^{25}protons/m^3 \end{gathered}$$

Find the ratio of proton density to the density of particles as follows.

$$\begin{gathered} \frac{n_H}{N/V}=\frac{3.1\times {10}^{31}protons/m^3}{2.68\times {10}^{25}protons/m^3} \\ =1.2\times {10}^6 \end{gathered}$$

Therefore, the required ratio is$1.2\times {10}^6$ .