Chapter 3, Problem 3D.5E

(a)

Interpretation Introduction

Interpretation:

Time period required for ${\text{CO}}_2$ to effuse has to be determined.

Concept Introduction:

Effusion is explained as movement of the gas molecule through small hole.

Diffusion can be explained as when one gas molecule mix with another gas molecule by random motion.

Graham’s law of effusion gives relation of rate of effusion with molar mass. Rate of effusion and square root of molar mass are inversely related. The mathematical expression is as follows:

  $\frac{\text{Effusion rate of A}}{\text{Effusion rate of B}}=\sqrt{\frac{M_{\text{B}}}{M_{\text{A}}}}$

Here,

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

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

Answer to Problem 3D.5E

Time period required for ${\text{CO}}_2$ to effuse is $151\text{ s}$.

Explanation of Solution

The expression for molar masses of ${\text{CO}}_2$ and $\text{Ar}$ gases and their time for effusion is as follows:

  $\frac{{\text{Time for CO}}_2\text{ to effuse}}{\text{Time for Ar to effuse}}=\sqrt{\frac{M_{{\text{CO}}_2}}{M_{\text{Ar}}}}$        (1)

Rearrange equation (1) to calculate value of time for ${\text{CO}}_2$ to effuse.

  ${\text{Time for CO}}_2\text{ to effuse}=\left(\text{Time for Ar to effuse}\right)\left(\sqrt{\frac{M_{{\text{CO}}_2}}{M_{\text{Ar}}}}\right)$        (2)

Substitute $42.01\text{ g/mol}$ for $M_{{\text{CO}}_2}$, $39.95\text{ g/mol}$ for $M_{\text{Ar}}$, and $147\text{ s}$ for time for $\text{Ar}$ to effuse in equation (2).

  $\begin{array}{c}{\text{Time for CO}}_2\text{ to effuse}=\left(147\text{ s}\right)\left(\sqrt{\frac{42.01\text{ g/mol}}{39.95\text{ g/mol}}}\right)\\ \approx 151\text{ s}\end{array}$

Hence time period required for ${\text{CO}}_2$ to effuse is $151\text{ s}$.

(b)

Interpretation Introduction

Interpretation:

Time period required for ${\text{C}}_2{\text{H}}_4$ to effuse has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3D.5E

Time period required for ${\text{C}}_2{\text{H}}_4$ to effuse is $123\text{ s}$.

Explanation of Solution

The expression for molar masses of ${\text{C}}_2{\text{H}}_4$ and $\text{Ar}$ gases and their time for effusion is as follows:

  $\frac{{\text{Time for C}}_2{\text{H}}_4\text{ to effuse}}{\text{Time for Ar to effuse}}=\sqrt{\frac{M_{{\text{C}}_2{\text{H}}_4}}{M_{\text{Ar}}}}$        (3)

Rearrange equation (1) to calculate value of time for ${\text{C}}_2{\text{H}}_4$ to effuse.

  ${\text{Time for C}}_2{\text{H}}_4\text{ to effuse}=\left(\text{Time for Ar to effuse}\right)\left(\sqrt{\frac{M_{{\text{C}}_2{\text{H}}_4}}{M_{\text{Ar}}}}\right)$        (4)

Substitute $28.05\text{ g/mol}$ for $M_{{\text{C}}_2{\text{H}}_4}$, $39.95\text{ g/mol}$ for $M_{\text{Ar}}$, and $147\text{ s}$ for time for $\text{Ar}$ to effuse in equation (2).

  $\begin{array}{c}{\text{Time for C}}_2{\text{H}}_4\text{ to effuse}=\left(147\text{ s}\right)\left(\sqrt{\frac{28.05\text{ g/mol}}{39.95\text{ g/mol}}}\right)\\ =123\text{ s}\end{array}$

Hence time period required for ${\text{C}}_2{\text{H}}_4$ to effuse is $123\text{ s}$.

(c)

Interpretation Introduction

Interpretation:

Time period required for ${\text{H}}_2$ to effuse has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3D.5E

Time period required for ${\text{H}}_2$ to effuse is $33.0\text{ s}$.

Explanation of Solution

The expression for molar masses of ${\text{H}}_2$ and $\text{Ar}$ gases and their time for effusion is as follows:

  $\frac{{\text{Time for H}}_2\text{ to effuse}}{\text{Time for Ar to effuse}}=\sqrt{\frac{M_{{\text{H}}_2}}{M_{\text{Ar}}}}$        (5)

Rearrange equation (1) to calculate value of time for ${\text{H}}_2$ to effuse.

  ${\text{Time for H}}_2\text{ to effuse}=\left(\text{Time for Ar to effuse}\right)\left(\sqrt{\frac{M_{{\text{H}}_2}}{M_{\text{Ar}}}}\right)$        (6)

Substitute $2.02\text{ g/mol}$ for $M_{{\text{H}}_2}$, $39.95\text{ g/mol}$ for $M_{\text{Ar}}$, and $147\text{ s}$ for time for $\text{Ar}$ to effuse in equation (2).

  $\begin{array}{c}{\text{Time for H}}_2\text{ to effuse}=\left(147\text{ s}\right)\left(\sqrt{\frac{2.02\text{ g/mol}}{39.95\text{ g/mol}}}\right)\\ =33.0\text{ s}\end{array}$

Hence time period required for ${\text{H}}_2$ to effuse is $33.0\text{ s}$.

(d)

Interpretation Introduction

Interpretation:

Time period required for ${\text{SO}}_2$ to effuse has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3D.5E

Time period required for ${\text{SO}}_2$ to effuse is $186\text{ s}$.

Explanation of Solution

The expression for molar masses of ${\text{SO}}_2$ and $\text{Ar}$ gases and their time for effusion is as follows:

  $\frac{{\text{Time for SO}}_2\text{ to effuse}}{\text{Time for Ar to effuse}}=\sqrt{\frac{M_{{\text{SO}}_2}}{M_{\text{Ar}}}}$        (7)

Rearrange equation (1) to calculate value of time for ${\text{SO}}_2$ to effuse.

  ${\text{Time for SO}}_2\text{ to effuse}=\left(\text{Time for Ar to effuse}\right)\left(\sqrt{\frac{M_{{\text{SO}}_2}}{M_{\text{Ar}}}}\right)$        (8)

Substitute $64.07\text{ g/mol}$ for $M_{{\text{SO}}_2}$, $39.95\text{ g/mol}$ for $M_{\text{Ar}}$, and $147\text{ s}$ for time for $\text{Ar}$ to effuse in equation (2).

  $\begin{array}{c}{\text{Time for SO}}_2\text{ to effuse}=\left(147\text{ s}\right)\left(\sqrt{\frac{64.07\text{ g/mol}}{39.95\text{ g/mol}}}\right)\\ =186\text{ s}\end{array}$

Hence time period required for ${\text{SO}}_2$ to effuse is $186\text{ s}$.