 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q55P

Expert-verified Found in: Page 1333 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # At the center of the Sun, the density of the gas is ${\mathbf{1}}{\mathbf{.}}{\mathbf{5}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{kg}}{\mathbf{/}}{{\mathbf{m}}}^{{\mathbf{3}}}$ 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 $\left(1.01×{10}^{5}\mathrm{Pa}\right)$ ?

a) The number of density of protons is $3.1×{10}^{31}\mathrm{protons}/{\mathrm{m}}^{3}$ .

b) The required ratio is $1.2×{10}^{6}$ .

See the step by step solution

## Step 1: Describe the expression to calculate the number density of protons

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

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

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

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

Combine above two equations.

${{\mathbit{n}}}_{{\mathbf{H}}}{{\mathbit{m}}}_{{\mathbf{H}}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{35}}{\mathbit{\rho }}\phantom{\rule{0ex}{0ex}}{{\mathbit{n}}}_{{\mathbf{H}}}{\mathbf{=}}\frac{\mathbf{0}\mathbf{.}\mathbf{35}\mathbf{\rho }}{{\mathbf{m}}_{\mathbf{H}}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}\left(1\right)$

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

## Step 2(a): Find the number density of protons

Substitute all the known values in equation (1).

${n}_{H}=\frac{0.35\left(1.5×{10}^{5}kg/{m}^{3}\right)}{1.67×{10}^{-27}kg}\phantom{\rule{0ex}{0ex}}=3.1×{10}^{31}protons/{m}^{3}$

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

## Step 3(b): Find the ratio of proton density to the density of particles

From the ideal gas law,

$PV=NkT\phantom{\rule{0ex}{0ex}}\frac{N}{V}=\frac{P}{kT}...........\left(2\right)$

Substitute all the known values in equation (2).

$\frac{N}{V}=\frac{1.01×{10}^{5}Pa}{\left(1.38×{10}^{-23}J/K\right)\left(273K\right)}\phantom{\rule{0ex}{0ex}}=2.68×{10}^{25}protons/{m}^{3}$

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

$\frac{{n}_{H}}{N/V}=\frac{3.1×{10}^{31}protons/{m}^{3}}{2.68×{10}^{25}protons/{m}^{3}}\phantom{\rule{0ex}{0ex}}=1.2×{10}^{6}$

Therefore, the required ratio is $1.2×{10}^{6}$ . ### Want to see more solutions like these? 