Chapter 5, Problem 5.73P

(a)

Interpretation Introduction

Interpretation:

The mass of two samples of gas one contains ${\text{H}}_{\text{2}}$ and the other contains ${\text{O}}_{\text{2}}$ at $1\text{ atm }$ and $25°\text{C}$ are to be compared.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$

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

(b)

Interpretation Introduction

Interpretation:

The density of two samples of gas one is ${\text{H}}_{\text{2}}$ and the other is ${\text{O}}_{\text{2}}$ at $1\text{ atm }$ and $25°\text{C}$ are to be compared.

Concept introduction:

The expression to calculate the density of the gas is as follows,

$d=\frac{PM}{RT}$

Here, $P$ is the pressure, $M$ is the molecular mass, $T$ is the temperature, $d$ is the density of the gas and $R$ is the gas constant.

(c)

Interpretation Introduction

Interpretation:

The mean free path of two samples of gas one is ${\text{H}}_{\text{2}}$ and the other is ${\text{O}}_{\text{2}}$ at $1\text{ atm }$ and $25°\text{C}$ are to be compared.

Concept introduction:

The mean free path can be defined as the average distance travelled by the gas molecule during a collision. There are many factors that affect the mean free path such as pressure, temperature, density and radius of the molecule.

(d)

Interpretation Introduction

Interpretation:

The average molecular kinetic energy of two samples of gas one is ${\text{H}}_{\text{2}}$ and the other is ${\text{O}}_{\text{2}}$ at $1\text{ atm }$ and $25°\text{C}$ are to be compared.

Concept introduction:

The expression to calculate the kinetic energy is as follows:

$E_{\text{k}}=\frac{1}{2}\left(\text{mass}\right){\left(\text{speed}\right)}^2$

Here, $E_{\text{k}}$ is the kinetic energy

The kinetic energy is directly proportional to the temperature.

(e)

Interpretation Introduction

Interpretation:

The average molecular speed of two samples of gas one is ${\text{H}}_{\text{2}}$ and the other is ${\text{O}}_{\text{2}}$ at $1\text{ atm }$ and $25°\text{C}$ are to be compared.

Concept introduction:

The expression to calculate the root-mean-square velocity is as follows:

$u_{rms}=\sqrt{\frac{3RT}{M}}$

Here, $M$ is the molar mass, $T$ is the temperature and $R$ is the gas constant.

The kinetic energy is directly proportional to the temperature and inversely proportional to the molar mass.

(f)

Interpretation Introduction

Interpretation:

The time taken for the effusion of two samples of gas one is ${\text{H}}_{\text{2}}$ and the other is ${\text{O}}_{\text{2}}$ at $1\text{ atm }$ and $25°\text{C}$ are to be compared.

Concept introduction:

The mathematical expression of Graham’s law of effusion is as follows:

$\frac{\text{Rate A}}{\text{Rate B}}=\frac{u_{rms\text{ of A}}}{u_{rms\text{ of B}}}=\frac{\sqrt{M_{\text{B}}}}{\sqrt{M_{\text{A}}}}=\frac{\text{time B}}{\text{time A}}$

Here, $M_{\text{B}}$ is the molar mass of gas $\text{B}$.

$M_{\text{A}}$ is the molar mass of gas $\text{A}$.