Chapter 17, Problem 29P

(a)

To determine

The pressure at the center of sun.

Given:

Temperature at the center of the gas is $1\times {10}^7\text{K}$ .

Density at the center of sun is $1\times {10}^5\text{kg}/{\text{m}}^{\text{3}}$ .

Formula used:

Write the expression for the ideal gas.

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant and $T$ is the temperature.

Solve the above equation for $P$ .

  $P=\frac{nRT}{V}$ ........ (1)

Write the expression for the number of moles.

  $n_p=\frac{m_p}{M_p}$ ........ (2)

Here, $n_p$ is the number of moles of protons, $m_p$ is the mass of protons and $M_p$ is the molar ma of proton.

Calculation:

Substitute ${10}^5\text{kg}$ for $m_p$ and ${10}^{-3}\text{kg}$ for $M_p$ in equation (2).

  $\begin{array}{l}{n}_p=\frac{{10}^5\text{kg}}{{10}^{-3}\text{kg}}\\ n_p={10}^8\end{array}$

The number of electrons is $2\times {10}^8$ .

Substitute $2\times {10}^8\text{mol}$ for $n$ , $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ , ${10}^7\text{K}$ for $T$ and $1{\text{m}}^{\text{3}}$ for $V$ in equation (1).

  $\begin{array}{l}P=\frac{\left(2\times {10}^8\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right){10}^7\text{K}}{1{\text{m}}^{\text{3}}}\\ P=2\times {10}^{11}\text{atm}\end{array}$

Conclusion:

The pressure is $2\times {10}^{11}\text{atm}$ .

Explanation of Solution

Given:

Temperature at the center of the gas is $1\times {10}^7\text{K}$ .

Density at the center of sun is $1\times {10}^5\text{kg}/{\text{m}}^{\text{3}}$ .

Formula used:

Write the expression for the ideal gas.

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant and $T$ is the temperature.

Solve the above equation for $P$ .

  $P=\frac{nRT}{V}$ ........ (1)

Write the expression for the number of moles.

  $n_p=\frac{m_p}{M_p}$ ........ (2)

Here, $n_p$ is the number of moles of protons, $m_p$ is the mass of protons and $M_p$ is the molar ma of proton.

Calculation:

Substitute ${10}^5\text{kg}$ for $m_p$ and ${10}^{-3}\text{kg}$ for $M_p$ in equation (2).

  $\begin{array}{l}{n}_p=\frac{{10}^5\text{kg}}{{10}^{-3}\text{kg}}\\ n_p={10}^8\end{array}$

The number of electrons is $2\times {10}^8$ .

Substitute $2\times {10}^8\text{mol}$ for $n$ , $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ , ${10}^7\text{K}$ for $T$ and $1{\text{m}}^{\text{3}}$ for $V$ in equation (1).

  $\begin{array}{l}P=\frac{\left(2\times {10}^8\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right){10}^7\text{K}}{1{\text{m}}^{\text{3}}}\\ P=2\times {10}^{11}\text{atm}\end{array}$

Conclusion:

The pressure is $2\times {10}^{11}\text{atm}$ .

(b)

To determine

The root mean square speed of electron and proton at the center of the sun.

Explanation of Solution

Given:

TheTemperature at the center of the gas is $1\times {10}^7\text{K}$ .

Formula used:

Write the expression for the root mean square speed of the molecule.

  $v_{\text{rms}}=\sqrt{\frac{3kT}{M}}$ ........ (1)

Here, $v_{\text{rms}}$ is the root mean square speed of the molecule, $k$ is the Boltzmann constant, $T$ is the temperature and $M$ is the molar mass of the molecule.

Calculation:

Substitute $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ , ${10}^7\text{K}$ for $T$ and $0.001\text{kg}$ for $M$ in equation (1)

  $\begin{array}{l}{v}_{\text{rms}}=\sqrt{\frac{3\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left({10}^7\text{K}\right)}{0.001\text{kg}}}\\ v_{\text{rms}}=5\times {10}^5\text{m}/\text{s}\end{array}$

Substitute $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ , ${10}^7\text{K}$ for $T$ and $\frac{1}{2000}\left(0.001\text{kg}\right)$ for $M$ in equation (1).

  $\begin{array}{l}{v}_{\text{rms}}=\sqrt{\frac{3\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left({10}^7\text{K}\right)}{\frac{1}{2000}\left(0.001\text{kg}\right)}}\\ v_{\text{rms}}=2\times {10}^7\text{m}/\text{s}\end{array}$

Conclusion:

The rms speed of proton and electron at the center of sun is $5\times {10}^5\text{m}/\text{s}$ and $2\times {10}^7\text{m}/\text{s}$ respectively.