Chapter 13, Problem 112P

(a)

To determine

The escape velocity for an object on Earth.

Answer to Problem 112P

The escape velocity is $11.2{\text{kms}}^{-1}$.

Explanation of Solution

Write the expression for conservation of energy

$T+U=T^{\prime }+U^{\prime }$                                                                                (I)

Here, $T$, $T^{\prime }$ are the initial and final kinetic energies whereas $U$, $U^{\prime }$ are the initial and final potential energies respectively.

Write the expression for kinetic energy of a particle

$T=\frac{1}{2}m{v}^2$                                                                                      (II)

Here, $m$ is the mass of the object and $v$ is the velocity of the object.

Write the expression for gravitational potential

$U=-\frac{GMm}{R_{\text{E}}}$                                                                                 (III)

Here, $G$ is the gravitational constant, $M$ is the mass of Earth, $m$ is the mass of the object and $R_{\text{E}}$ is the radius of Earth.

At infinite distance both kinetic and potential energies are considered zero.

Substitute $0$ for $T^{\prime }$, $0$ for $U^{\prime }$ and (II), (III) in (I)

$\frac{1}{2}m{v}^2-\frac{GMm}{R_{\text{E}}}=0$

Rearrange for $v$

$v=\sqrt{\frac{2GM}{R_{\text{E}}}}$                                                                              (IV)

Substitute $6.674\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$ for $G$, $5.974\times {10}^{24}\text{kg}$ for $M$ and $6.37\times {10}^6\text{m}$ for $R_{\text{E}}$ in (IV) to find $v$.

$\begin{array}{c}v=\sqrt{\frac{2\left(6.674\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}\right)\left(5.974\times {10}^{24}\text{kg}\right)}{6.37\times {10}^6\text{m}}}\\ =11.2{\text{kms}}^{-1}\end{array}$

Thus, the escape velocity of an object is $11.2{\text{kms}}^{-1}$.

(b)

To determine

The average speed of hydrogen atom at $0°\text{C}$.

Answer to Problem 112P

The average speed of hydrogen atom is $1839{\text{ ms}}^{-1}$.

Explanation of Solution

Write the expression for average speed of an atom

$v=\sqrt{\frac{3kT}{m}}$                                                             (V)

Here, $v$ is the average speed, $k$ is the Boltzmann’s constant, $T$ is the temperature and $m$ is the mass.

Substitute $1.381\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $M$, $1.0079\text{u}$ for $m$ and $0°\text{C}$ for $T$ in (V) to find $v$

$\begin{array}{c}v=\sqrt{\frac{3\left(1.381\times {10}^{-23}\text{}{\text{JK}}^{-1}\right)\left(0°\text{C}\right)}{1.0079\text{u}}}\\ =\sqrt{\frac{3\left(1.381\times {10}^{-23}\text{}{\text{JK}}^{-1}\right)\left(0+273.15\text{K}\right)}{1.0079\text{u}\times \left(\frac{1.66\times {10}^{-27}\text{}\text{kg}}{\text{u}}\right)}}\\ =\sqrt{\frac{3\left(1.381\times {10}^{-23}\text{}{\text{kgm}}^{\text{2}}{\text{s}}^{-2}\right)\left(0+273.15\text{K}\right)}{1.0079\text{u}\times \left(\frac{1.66\times {10}^{-27}\text{}\text{kg}}{\text{u}}\right)}}\\ =1839{\text{ ms}}^{-1}\end{array}$

Thus, the average speed of hydrogen is $1839{\text{ ms}}^{-1}$.

(c)

To determine

The average speed of oxygen molecule at $0°\text{C}$.

Answer to Problem 112P

The average speed of oxygen molecule is $461.5{\text{ ms}}^{-1}$.

Explanation of Solution

Substitute $1.381\times {10}^{-23}\text{}{\text{JK}}^{-1}$ for $M$, $2\left(15.9994\text{u}\right)$ for $m$ and $0°\text{C}$ for $T$ in (V) to find $v$

$\begin{array}{c}v=\sqrt{\frac{3\left(1.381\times {10}^{-23}\text{}{\text{JK}}^{-1}\right)\left(0°\text{C}\right)}{2\left(15.9994\text{u}\right)}}\\ =\sqrt{\frac{3\left(1.381\times {10}^{-23}\text{}{\text{JK}}^{-1}\right)\left(0+273.15\text{K}\right)}{2\left(15.9994\text{u}\right)\times \left(\frac{1.66\times {10}^{-27}\text{}\text{kg}}{\text{u}}\right)}}\\ =\sqrt{\frac{3\left(1.381\times {10}^{-23}\text{}{\text{kgm}}^{\text{2}}{\text{s}}^{-2}\right)\left(0+273.15\text{K}\right)}{2\left(15.9994\text{u}\right)\times \left(\frac{1.66\times {10}^{-27}\text{}\text{kg}}{\text{u}}\right)}}\\ =461.5{\text{ ms}}^{-1}\end{array}$

Thus, the average speed of oxygen molecule is $461.5{\text{ ms}}^{-1}$.

(d)

To determine

The reason for more oxygen than hydrogen on the Earth’s atmosphere.

Answer to Problem 112P

The average speed of hydrogen is greater for hydrogen.

Explanation of Solution

Particles with velocity lesser than that of the escape velocity cannot escape Earth’s atmosphere.

From the above results, the escape velocity is about six times the average speed of the hydrogen atom. Thus, a small fraction of hydrogen atoms might possess velocity greater than the required escape velocity to cross the Earth’s atmosphere.

The average speed of oxygen molecules is lesser than that of hydrogen and hence only a smaller fraction will possess the required speed to escape Earth’s gravitational pull. Thus, oxygen is found more than hydrogen in the Earth’s atmosphere.