Chapter 1, Problem 1.2PR

(a)

Interpretation Introduction

Interpretation:

The pressure halfway to the center of the sun assuming that the interior consists of ionized hydrogen atoms at the temperature of $\text{3}\text{.6MK}$ and mass density $\text{1}\text{.20}{\text{gcm}}^{\text{-3}}$ has to be calculated.

Concept introduction:

Ideal gas equation:

According to kinetic theory of gas the ideal gas is the one having almost negligible volume, no attractive or repulsive force working between the molecules.  The molecules are randomly moving and colliding with each other having elastic collisions.

Ideal gas equation can be represented as,

  $\text{PV = nRT}$

Where,

$P$ is the Pressure

$\text{V}$ is the Volume

$\text{n }$ is the number of moles

$\text{R}$ is the Gas constant

$\text{T}$ is the Temperature.

Answer to Problem 1.2PR

Pressure in midway of sun is $\text{7}{\text{.18×10}}^{\text{13}}\text{Pa}$.

Explanation of Solution

Given that the temperature in the sun is $\text{3}\text{.6MK}$ and mass density is $\text{1}\text{.20}{\text{gcm}}^{\text{-3}}$.

It has been assumed that the interior of the sun is filled with the ionized hydrogen atom.

Ideal gas equation,

  $\text{PV=nRT}$

It can also be written as,

  $\begin{array}{l}\text{PV}\text{=}{\text{Nk}}_{\text{B}}\text{T}\\ \\ \text{nR}\text{=}{\text{Nk}}_{\text{B}}\end{array}$

$\text{N}$ is the number of molecules.

${\text{k}}_{\text{B}}$ is the Boltzmann constant

The number of molecules of gas can be represented as,

  $\text{n}\text{=}\frac{\text{m}}{\text{M}}$

$\text{m}$ is the amount of substance taken and $\text{M}$ is the molar mass.

Now the mass of the ionized hydrogen atom is,

  $\text{M}\text{=}\text{μ}\text{×}{\text{m}}_{\text{p}}$

$\text{μ}$ is the average atomic weight and ${\text{m}}_{\text{P}}$ is the mass of proton.

Hence the ideal gas equation can be written as,

  $\begin{array}{l}\text{PV}\text{=}\frac{\text{m}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{m}}{\text{μ}\text{×}{\text{m}}_{\text{u}}\text{×V}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{ρ}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\left[\text{ρ}\text{=}\text{density}\text{=}\frac{\text{m}}{\text{V}}\right]\end{array}$

For, ionized hydrogen atom,

  $\begin{array}{l}\text{μ}\text{=}\frac{\text{mass}\text{in}\text{a}\text{.m}\text{.u}\text{.}}{\text{no}\text{.}\text{of}\text{species}}\\ \\ \text{μ}\text{=}\frac{\text{1}}{\text{2}}\left[\begin{array}{l}\text{no}\text{.}\text{of}\text{species=}\text{2}\\ \text{(1}{\text{H}}^{\text{+}}\text{ion}\text{and}\text{1}\text{electron)}\\ \text{mass}\text{of}\text{hydrogen}\text{is}\text{1}\text{a}\text{.m}\text{.u}\text{.}\end{array}\right]\\ \\ \text{μ}\text{=}\text{0}\text{.5}\end{array}$

Now putting all the values in the ideal gas equation,

  $\begin{array}{l}\text{P}\text{=}\frac{\text{ρ}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{1}\text{.2}\text{g}\text{cm}{}^{\text{-3}}}{\text{0}\text{.5}\text{×1}{\text{.66×10}}^{\text{-24}}\text{g}}\text{×1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\text{×3}\text{.6}\text{MK}\left[\begin{array}{l}{\text{k}}_{\text{B}}\text{=}\text{1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\\ \text{in}\text{cgs}\text{unit}\end{array}\right]\\ \\ \text{P}\text{=}\frac{\text{1}\text{.2}\text{g}\text{cm}{}^{\text{-3}}}{\text{0}\text{.5}\text{×1}{\text{.66×10}}^{\text{-24}}\text{g}}\text{×1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\text{×3}\text{.6}{\text{×10}}^{\text{6}}\text{K}\\ \\ \text{P}\text{=}\text{7}{\text{.18×10}}^{\text{14}}\text{g}{\text{cm}}^{\text{-1}}{\text{s}}^{\text{-2}}\\ \\ \text{P}\text{=}\text{7}{\text{.18×10}}^{\text{13}}\text{Pa}\left[\text{1}\text{g}{\text{cm}}^{\text{-1}}{\text{s}}^{\text{-2}}\text{=}\text{0}\text{.1}\text{Pa}\right]\end{array}$

Hence pressure in midway of sun is $\text{7}{\text{.18×10}}^{\text{13}}\text{Pa}$

(b)

Interpretation Introduction

Interpretation:

The pressure of the plasma is related to its kinetic energy density by ${\text{ρ}}_{\text{K}}\text{=}\frac{{\text{E}}_{\text{K}}}{\text{V}}$ by $\text{P}\text{=}\frac{\text{2}}{\text{3}}{\text{ρ}}_{\text{K}}$ has to be shown.

Concept introduction:

Kinetic energy density:

Kinetic energy of the gas molecules is the energy that occurs due to the random motion of the gas molecules.

It can be expressed as,

  ${\text{E}}_{\text{K}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

${\text{E}}_{\text{K}}$ is the Kinetic energy of the molecules.

$\text{m}$ is the mass of the gas molecules.

$\text{v}$ is the velocity of the gas molecules.

Kinetic energy of the molecules in a region divided by the volume of that region is the kinetic energy density of the molecules of that region.

  ${\text{ρ}}_{\text{K}}\text{=}\frac{{\text{E}}_{\text{K}}}{\text{V}}$

${\text{E}}_{\text{K}}$ is the Kinetic energy of the molecules.

${\text{ρ}}_{\text{K}}$ is the kinetic energy density

$\text{V}$ is the Volume

Equipartition of energy:

The equipartition theorem shows that in thermal equilibrium any degree of freedom which appears only quadratically in the energy has an average energy of $\frac{\text{1}}{\text{2}}{\text{k}}_{\text{B}}\text{T}$.

Explanation of Solution

According to the equipartition theorem it can be predicted that the monoatomic ideal gas has an average kinetic energy of $\frac{3}{\text{2}}{\text{k}}_{\text{B}}\text{T}$.

Hence it can be concluded that kinetic energy,

  ${\text{E}}_{\text{K}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}=\frac{3}{\text{2}}{\text{k}}_{\text{B}}\text{T}$

From this it can be concluded that,

  $\begin{array}{l}{\text{E}}_{\text{K}}\text{=}\frac{\text{3}}{\text{2}}{\text{k}}_{\text{B}}\text{T}\\ \\ {\text{E}}_{\text{K}}\text{=}\frac{\text{3}}{\text{2}}{\text{Nk}}_{\text{B}}\text{T}\text{(for}\text{N}\text{molecules)}\\ \\ {\text{Nk}}_{\text{B}}\text{T}\text{=}\frac{\text{2}}{\text{3}}{\text{E}}_{\text{K}}\end{array}$

Now from part (a) the ideal gas equation,

  $\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \\ \text{PV}\text{=}{\text{Nk}}_{\text{B}}\text{T}\end{array}$

Now combining the above two equations,

  $\begin{array}{l}\text{PV}\text{=}\frac{\text{2}}{\text{3}}{\text{E}}_{\text{K}}\\ \\ \text{P}\text{=}\frac{\text{2}}{\text{3}}\text{×}\frac{{\text{E}}_{\text{K}}}{\text{V}}\\ \\ \text{P}\text{=}\frac{\text{2}}{\text{3}}{\text{ρ}}_{\text{K}}\end{array}$

Thus it can be shown that $\text{P}\text{=}\frac{\text{2}}{\text{3}}{\text{ρ}}_{\text{K}}$.

(c)

Interpretation Introduction

Interpretation:

Kinetic energy density half way to the center of the sun has to be calculated.

Concept introduction:

Ideal gas equation:

According to kinetic theory of gas the ideal gas is the one having almost negligible volume, no attractive or repulsive force working between the molecules.  The molecules are randomly moving and colliding with each other having elastic collisions.

Ideal gas equation can be represented as,

  $\text{PV = nRT}$

Where,

$P$ is the Pressure

$\text{V}$ is the Volume

$\text{n }$ is the number of moles

$\text{R}$ is the Gas constant

$\text{T}$ is the Temperature.

Kinetic energy density:

Kinetic energy of the gas molecules is the energy that occurs due to the random motion of the gas molecules.

It can be expressed as,

  ${\text{E}}_{\text{K}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

${\text{E}}_{\text{K}}$ is the Kinetic energy of the molecules.

$\text{m}$ is the mass of the gas molecules.

$\text{v}$ is the velocity of the gas molecules.

Kinetic energy of the molecules in a region divided by the volume of that region is the kinetic energy density of the molecules of that region.

  ${\text{ρ}}_{\text{K}}\text{=}\frac{{\text{E}}_{\text{K}}}{\text{V}}$

${\text{E}}_{\text{K}}$ is the Kinetic energy of the molecules.

${\text{ρ}}_{\text{K}}$ is the kinetic energy density

$\text{V}$ is the Volume

Answer to Problem 1.2PR

The sun has kinetic energy density $\text{1}{\text{.08×10}}^{\text{14}}\text{J}{\text{m}}^{\text{-3}}$.

Explanation of Solution

Pressure to the half way to the center of the sun is $\text{7}{\text{.18×10}}^{\text{13}}\text{Pa}$.

From part (b) relation between the pressure and kinetic energy density obtained is,

  $\text{P}\text{=}\frac{\text{2}}{\text{3}}{\text{ρ}}_{\text{K}}$

Now, putting the value of pressure in this equation,

  $\begin{array}{l}{\text{ρ}}_{\text{K}}\text{=}\frac{\text{3}}{\text{2}}\text{×7}{\text{.18×10}}^{\text{13}}\text{Pa}\\ \\ {\text{ρ}}_{\text{K}}\text{=}\text{1}{\text{.08×10}}^{\text{14}}\text{J}{\text{m}}^{\text{-3}}\end{array}$

At ${\text{25}}^{\text{ο}}\text{C}$ the kinetic energy density of the earth is $1.5\times {10}^5\text{J}{\text{m}}^{\text{-3}}$

Hence at ${\text{25}}^{\text{ο}}\text{C}$ the pressure of sun,

  $\begin{array}{l}\text{P}\text{=}\frac{\text{ρ}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{1}\text{.2}\text{g}\text{cm}{}^{\text{-3}}}{\text{0}\text{.5}\text{×1}{\text{.66×10}}^{\text{-24}}\text{g}}\text{×1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\text{×(25+273)K}\\ \left[\begin{array}{l}{\text{k}}_{\text{B}}\text{=}\text{1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\\ \text{in}\text{cgs}\text{unit}\end{array}\right]\\ \\ \text{P}\text{=}5.94{\text{×10}}^{\text{10}}\text{g}{\text{cm}}^{\text{-1}}{\text{s}}^{\text{-2}}\\ \\ \text{P}\text{=}5.94{\text{×10}}^9\text{Pa}\left[\text{1}\text{g}{\text{cm}}^{\text{-1}}{\text{s}}^{\text{-2}}\text{=}\text{0}\text{.1}\text{Pa}\right]\end{array}$

Hence ${\text{25}}^{\text{ο}}\text{C}$ the kinetic energy density of sun,

  $\begin{array}{l}\text{P}\text{=}\frac{\text{2}}{\text{3}}{\text{ρ}}_{\text{K}}\\ {\text{ρ}}_{\text{K}}\text{=}\frac{\text{3}}{\text{2}}\text{×}5.94{\text{×10}}^9\text{Pa}\\ \\ {\text{ρ}}_{\text{K}}\text{=}\text{8}{\text{.91×10}}^{\text{9}}\text{J}{\text{m}}^{\text{-3}}\end{array}$

Hence the sun has much more energy density.

Hence the sun has kinetic energy density $\text{1}{\text{.08×10}}^{\text{14}}\text{J}{\text{m}}^{\text{-3}}$.

(d)

Interpretation Introduction

Interpretation:

The pressure halfway to the center of the red giant assuming that the interior consists of fully ionized carbon atoms at the temperature of $\text{3500K}$ has to be calculated.

Concept introduction:

Ideal gas equation:

According to kinetic theory of gas the ideal gas is the one having almost negligible volume, no attractive or repulsive force working between the molecules.  The molecules are randomly moving and colliding with each other having elastic collisions.

Ideal gas equation can be represented as,

  $\text{PV = nRT}$

Where,

$P$ is the Pressure

$\text{V}$ is the Volume

$\text{n }$ is the number of moles

$\text{R}$ is the Gas constant

$\text{T}$ is the Temperature.

Answer to Problem 1.2PR

The pressure in midway of red giant is $2.04{\text{×10}}^{\text{10}}\text{Pa}$.

Explanation of Solution

Given that the temperature at the halfway of the center of the red giant is $\text{3500K}$.

The red giant is filled with fully ionized carbon atom.

Ideal gas equation,

$\text{PV=nRT}$

It can also be written as,

  $\begin{array}{l}\text{PV}\text{=}{\text{Nk}}_{\text{B}}\text{T}\\ \\ \text{nR}\text{=}{\text{Nk}}_{\text{B}}\end{array}$

$\text{N}$ is the number of molecules.

${\text{k}}_{\text{B}}$ is the Boltzmann constant

The number of molecules of gas can be represented as,

  $\text{n}\text{=}\frac{\text{m}}{\text{M}}$

$\text{m}$ is the amount of substance taken and $\text{M}$ is the molar mass.

Now the mass of the fully ionized carbon atom is,

  $\text{M}\text{=}\text{μ}\text{×}{\text{m}}_{\text{p}}$

$\text{μ}$ is the average atomic weight

${\text{m}}_{\text{P}}$ is the mass of proton

Hence the ideal gas equation can be written as,

  $\begin{array}{l}\text{PV}\text{=}\frac{\text{m}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{m}}{\text{μ}\text{×}{\text{m}}_{\text{u}}\text{×V}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{ρ}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\left[\text{ρ}\text{=}\text{density}\text{=}\frac{\text{m}}{\text{V}}\right]\end{array}$

For, fully ionized carbon atom,

  $\begin{array}{l}\text{μ}\text{=}\frac{\text{mass}\text{in}\text{a}\text{.m}\text{.u}\text{.}}{\text{no}\text{.}\text{of}\text{species}}\\ \\ \text{μ}\text{=}\frac{\text{12}}{7}\left[\begin{array}{l}\text{no}\text{.}\text{of}\text{species=}7\\ \text{(1}\text{carbon}\text{ion}\text{and}\text{6}\text{electron)}\\ \text{mass}\text{of}\text{hydrogen}\text{is}\text{12}\text{a}\text{.m}\text{.u}\text{.}\end{array}\right]\\ \\ \text{μ}\text{=}1.71\end{array}$

Now putting all the values in the ideal gas equation,

$\begin{array}{l}\text{P}\text{=}\frac{\text{ρ}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{1}\text{.2}\text{g}\text{cm}{}^{\text{-3}}}{1.71\text{×1}{\text{.66×10}}^{\text{-24}}\text{g}}\text{×1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\text{×3500K}\left[\begin{array}{l}{\text{k}}_{\text{B}}\text{=}\text{1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\\ \text{in}\text{cgs}\text{unit}\end{array}\right]\\ \\ \text{P}\text{=}2.04{\text{×10}}^{\text{11}}\text{g}{\text{cm}}^{\text{-1}}{\text{s}}^{\text{-2}}\\ \\ \text{P}\text{=}2.04{\text{×10}}^{\text{10}}\text{Pa}\left[\text{1}\text{g}{\text{cm}}^{\text{-1}}{\text{s}}^{\text{-2}}\text{=}\text{0}\text{.1}\text{Pa}\right]\end{array}$

Hence pressure in midway of red giant is $2.04{\text{×10}}^{\text{10}}\text{Pa}$.

(e)

Interpretation Introduction

Interpretation:

The pressure halfway to the center of the red giant assuming that the interior consists of neutral carbon atoms at the temperature of $\text{3500K}$ has to be calculated.

Concept introduction:

Ideal gas equation:

According to kinetic theory of gas the ideal gas is the one having almost negligible volume, no attractive or repulsive force working between the molecules.  The molecules are randomly moving and colliding with each other having elastic collisions.

Ideal gas equation can be represented as,

  $\text{PV = nRT}$

Where,

$P$ is the Pressure

$\text{V}$ is the Volume

$\text{n }$ is the number of moles

$\text{R}$ is the Gas constant

$\text{T}$ is the Temperature.

Answer to Problem 1.2PR

The pressure in midway of red giant is $\text{2}{\text{.91×10}}^{\text{9}}\text{Pa}$ when it is filled with neutral carbon.

Explanation of Solution

Given that the temperature at the halfway of the center of the red giant is $\text{3500K}$.

The red giant is filled with neutral carbon atom.

Ideal gas equation,

$\text{PV=nRT}$

It can also be written as,

  $\begin{array}{l}\text{PV}\text{=}{\text{Nk}}_{\text{B}}\text{T}\\ \\ \text{nR}\text{=}{\text{Nk}}_{\text{B}}\end{array}$

$\text{N}$ is the number of molecules.

${\text{k}}_{\text{B}}$ is the Boltzmann constant

The number of molecules of gas can be represented as,

  $\text{n}\text{=}\frac{\text{m}}{\text{M}}$

$\text{m}$ is the amount of substance taken and $\text{M}$ is the molar mass.

Now the mass of the fully ionized carbon atom is,

  $\text{M}\text{=}\text{μ}\text{×}{\text{m}}_{\text{p}}$

$\text{μ}$ is the average atomic weight

${\text{m}}_{\text{P}}$ is the mass of proton

Hence the ideal gas equation can be written as,

  $\begin{array}{l}\text{PV}\text{=}\frac{\text{m}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{m}}{\text{μ}\text{×}{\text{m}}_{\text{u}}\text{×V}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{ρ}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\left[\text{ρ}\text{=}\text{density}\text{=}\frac{\text{m}}{\text{V}}\right]\end{array}$

For, neutral carbon atom,

  $\begin{array}{l}\text{μ}\text{=}\frac{\text{mass}\text{in}\text{a}\text{.m}\text{.u}\text{.}}{\text{no}\text{.}\text{of}\text{species}}\\ \\ \text{μ}\text{=}\frac{\text{12}}{\text{1}}\left[\begin{array}{l}\text{no}\text{.}\text{of}\text{species=}\text{1}\\ \text{(1}\text{carbon}\text{atom)}\\ \text{mass}\text{of}\text{hydrogen}\text{is}\text{12}\text{a}\text{.m}\text{.u}\text{.}\end{array}\right]\\ \\ \text{μ}\text{=}\text{12}\end{array}$

Now putting all the values in the ideal gas equation,

$\begin{array}{l}\text{P}\text{=}\frac{\text{ρ}}{\text{μ}\text{×}{\text{m}}_{\text{u}}}\text{×}{\text{k}}_{\text{B}}\text{T}\\ \\ \text{P}\text{=}\frac{\text{1}\text{.2}\text{g}\text{cm}{}^{\text{-3}}}{\text{12}\text{×1}{\text{.66×10}}^{\text{-24}}\text{g}}\text{×1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\text{×3500K}\left[\begin{array}{l}{\text{k}}_{\text{B}}\text{=}\text{1}{\text{.38×10}}^{\text{-16}}{\text{cm}}^{\text{2}}\text{g}{\text{s}}^{\text{-2}}{\text{K}}^{\text{-1}}\\ \text{in}\text{cgs}\text{unit}\end{array}\right]\\ \\ \text{P}\text{=}\text{2}{\text{.91×10}}^{\text{10}}\text{g}{\text{cm}}^{\text{-1}}{\text{s}}^{\text{-2}}\\ \\ \text{P}\text{=}\text{2}{\text{.91×10}}^{\text{9}}\text{Pa}\left[\text{1}\text{g}{\text{cm}}^{\text{-1}}{\text{s}}^{\text{-2}}\text{=}\text{0}\text{.1}\text{Pa}\right]\end{array}$

Hence pressure in midway of red giant is $\text{2}{\text{.91×10}}^{\text{9}}\text{Pa}$ when it is filled with neutral carbon.