Chapter 3, Problem 3D.10E

(a)

Interpretation Introduction

Interpretation:

The value of $v_{\text{rms}}$  speed of ${\text{F}}_2$ gas at $220°\text{C}$ has to be determined.

Concept Introduction:

Root mean square speed is square root of average value of square of molecular speeds of gas molecules. It includes molecular weight and temperature, both of which affect kinetic energy of gas. It is denoted by $v_{\text{rms}}$. The expression for $v_{\text{rms}}$ is as follows:

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

Here,

$R$ is gas constant.

$T$ is absolute temperature of gas.

$M$ is molecular weight of gas.

Answer to Problem 3D.10E

Value of $v_{\text{rms}}$ of ${\text{F}}_2$ is $569\text{ m}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Conversion factor to convert $220°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =220°\text{C}+273.15\\ =493.15\text{K}\end{array}$

Substitute $493.15\text{K}$ for $T$, $38.00\text{ g/mol}$ for $M$, and $8.3145\text{ J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ in equation (1) to calculate $v_{\text{rms}}$ for ${\text{F}}_2$.

  $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\left(\frac{3\left(8.3145\text{ J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(493.15\text{K}\right)}{\left(38.00\text{ g/mol}\right)}\right)\left(\frac{{10}^3\text{ g}}{1\text{ kg}}\right)}\\ =569\text{ m}\cdot {\text{s}}^{-1}\end{array}$

Hence, $v_{\text{rms}}$ speed of ${\text{F}}_2$ is $569\text{ m}\cdot {\text{s}}^{-1}$.

(b)

Interpretation Introduction

Interpretation:

The value of $v_{\text{rms}}$  speed of ${\text{Cl}}_2$ at $220°\text{C}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3D.10E

Value of $v_{\text{rms}}$ for ${\text{Cl}}_2$ is $416\text{ m}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Conversion factor to convert $220°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =220°\text{C}+273.15\\ =493.15\text{K}\end{array}$

Substitute $493.15\text{K}$ for $T$, $70.91\text{ g/mol}$ for $M$, and $8.3145\text{ J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ in equation (1) to calculate $v_{\text{rms}}$ for ${\text{Cl}}_2$.

  $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\left(\frac{3\left(8.3145\text{ J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(493.15\text{K}\right)}{\left(70.91\text{ g/mol}\right)}\right)\left(\frac{{10}^3\text{ g}}{1\text{ kg}}\right)}\\ =416\text{ m}\cdot {\text{s}}^{-1}\end{array}$

Hence, $v_{\text{rms}}$ speed of ${\text{Cl}}_2$ is $416\text{ m}\cdot {\text{s}}^{-1}$.

(c)

Interpretation Introduction

Interpretation:

The value of $v_{\text{rms}}$  speed of ${\text{Br}}_2$ gas at $220°\text{C}$ has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3D.10E

Value of $v_{\text{rms}}$ for ${\text{Br}}_2$ is $277\text{ m}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Conversion factor to convert $220°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =220°\text{C}+273.15\\ =493.15\text{K}\end{array}$

Substitute $493.15\text{K}$ for $T$, $38.00\text{ g/mol}$ for $M$, and $8.3145\text{ J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$ in equation (1) to calculate $v_{\text{rms}}$ for ${\text{Br}}_2$.

  $\begin{array}{c}{v}_{\text{rms}}=\sqrt{\left(\frac{3\left(8.3145\text{ J}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(493.15\text{K}\right)}{\left(159.81\text{ g/mol}\right)}\right)\left(\frac{{10}^3\text{ g}}{1\text{ kg}}\right)}\\ =277\text{ m}\cdot {\text{s}}^{-1}\end{array}$

Hence, $v_{\text{rms}}$ speed of ${\text{Br}}_2$ is $277\text{ m}\cdot {\text{s}}^{-1}$.