Chapter 6, Problem 6.126QE

(a)

Interpretation Introduction

Interpretation:

The root mean square speed of samples of hydrogen and nitrogen at STP conditions has to be calculated.

Concept Introduction:

In accordance with the Kinetic molecular theory, not all of the gas particles will move at the same speed but with an average speed that is given by the expression as follows:

  $\overline{KE}=\frac{1}{2}m\overline{u^2}$

Here,

$\overline{KE}$ is the average kinetic energy.

$m$ is the mass of particle.

$u^2$ is the average of the speed.

The relation between rms speed $\left(u_{\text{rms}}\right)$, temperature, and molar mass is given by the equation,

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

Here,

$\text{R}$ is the universal gas constant.

$T$ denotes temperature in kelvins.

$M$ denotes the molar mass in $\text{kg/mol}$.

Answer to Problem 6.126QE

Root mean square speed of samples of hydrogen and nitrogen at STP is $583.48\text{ m/s}$ and $697.25\text{ m/s}$ respectively.

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Here,

$T\left(\text{K}\right)$ denotes the temperature in kelvins.

$T\left(°\text{C}\right)$ denotes the temperature in Celsius.

Substitute $\text{0 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=0°\text{C}+273\text{ K}\\ =273\text{ K}\end{array}$

The relation between rms speed, temperature, and molar mass is given as follows:

  $u_{\text{rms}}=\sqrt{\frac{3RT}{M}}$        (2)

Here,

$R$ is the universal gas constant.

$T$ denotes temperature in kelvins.

$M$ denotes the molar mass in $\text{kg/mol}$.

Substitute $8.314\text{ kg}\cdot {\text{m}}^2{\text{/s}}^2\cdot \text{mol}\cdot \text{K}$ for $R$, $273\text{ K}$ for $T$, $0.020\text{ kg/mol}$ for $M$ to calculate the rms speed of hydrogen inequation (2).

  $\begin{array}{c}{u}_{\text{rms}}=\sqrt{\frac{3\left(8.314\text{ kg}\cdot {\text{m}}^2/{\text{s}}^2\cdot \text{mol}\cdot \text{K}\right)\left(273\text{ K}\right)}{0.020\text{ kg/mol}}}\\ =583.48\text{ m/s}\end{array}$

Substitute $8.314\text{ kg}\cdot {\text{m}}^2/{\text{s}}^2\cdot \text{mol}\cdot \text{K}$ for $R$, $273\text{ K}$ for $T$, $0.014\text{ kg/mol}$ for $M$ to calculate the rms speed of nitrogen inequation (2).

  $\begin{array}{c}{u}_{\text{rms}}=\sqrt{\frac{3\left(8.314\text{ kg}\cdot {\text{m}}^2/{\text{s}}^2\cdot \text{mol}\cdot \text{K}\right)\left(273\text{ K}\right)}{0.014006\text{ kg/mol}}}\\ =697.25\text{ m/s}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The average kinetic energy of hydrogen and nitrogen at STP conditions has to be calculated.

Concept Introduction:

In accordance with the Kinetic molecular theory not all of the gas particles will move at the same speed but with an average speed that is given by the expression as follows:

  $\text{Average Kinetic energy}=\frac{1}{2}m{\left(u_{\text{rms}}\right)}^2$

Here,

$m$ is the mass of gas particles.

$u_{\text{rms}}$ is the root mean square speed.

Answer to Problem 6.126QE

The average kinetic energy of hydrogen and nitrogen at STP is $5.65340\times {10}^{-21}{\text{ kgm}}^2{\text{/s}}^2$ and $5.64704\times {10}^{-21}{\text{ kgm}}^2{\text{/s}}^2$.

Explanation of Solution

The formula to calculate the mass from the number of moles and molar mass is as follows:

  $m=\left(\frac{\text{Given number of molecules}}{\text{Avogadro's number of molecules}}\right)\left(\text{molar mass}\right)$        (3)

Substitute 1 for a given number of molecules, $6.022\times {10}^{23}$ for Avogadro’s number of molecules, $0.020\text{ g/mol}$ for molar mass in equation (3).

  $\begin{array}{c}m=\left(\frac{1}{6.022\times {10}^{23}}\text{mol}\right)\left(0.020\text{ kg/mol}\right)\\ =3.32115\times {10}^{-26}\text{ kg}\end{array}$

Substitute 1 for given number of molecules, $6.022\times {10}^{23}$ for Avogadro’s number of molecules, $0.014\text{ kg/mol}$ for molar mass in equation (3).

  $\begin{array}{c}m=\left(\frac{1}{6.022\times {10}^{23}}\text{mol}\right)\left(0.014\text{ kg/mol}\right)\\ =2.3248\times {10}^{-26}\text{ kg}\end{array}$

The average kinetic energy in terms of rms speed is given by the expression as follows:

  $\text{Average Kinetic energy}=\frac{1}{2}m{\left(u_{\text{rms}}\right)}^2$

Substitute $3.32115\times {10}^{-26}\text{ kg}$ for $m$, $583.48\text{ m/s}$ for $u_{\text{rms}}$ equation (4) to calculate the average kinetic energy of hydrogen.

  $\begin{array}{c}\text{Average Kinetic energy}=\frac{1}{2}\left(3.32115\times {10}^{-26}\text{ kg}\right){\left(583.48\text{ m/s}\right)}^2\\ =5.65340\times {10}^{-21}\text{ kg}\cdot {\text{m}}^2{\text{/s}}^2\end{array}$

Substitute $2.3248\times {10}^{-26}\text{ kg}$ for $m$, $583.48\text{ m/s}$ for ${\overline{u}}_{\text{rms}}$ equation (4) to calculate the average kinetic energy of nitrogen.

  $\begin{array}{c}\text{Average Kinetic energy}=\frac{1}{2}\left(2.3248\times {10}^{-26}\text{ kg}\right){\left(697.25\text{ m/s}\right)}^2\\ =5.64704\times {10}^{-21}\text{ kg}\cdot {\text{m}}^2{\text{/s}}^2\end{array}$