Chapter 5, Problem 5.136QP

The graph here represents the distribution of molecular speeds of hydrogen and neon at 200 K.

1. a Match each curve to the appropriate gas.
2. b Calculate the rms speed (in m/s) for each of the gases at 200 K.
3. c Which of the gases would you expect to have the greater effusion rate at 200 K? Justify your answer.
4. d Calculate the temperature at which the rms speed of the hydrogen gas would equal the rms speed of the neon at 200 K.

Interpretation Introduction

Interpretation:

The curve in the given graph must be matched with the appropriate gas.

Answer to Problem 5.136QP

The taller and narrow curve represents neon atoms

The flatter and wider curve represents hydrogen molecules

Explanation of Solution

Given,

The given graph represents the distribution of molecular speeds of hydrogen and neon at $200\text{ K}$

Matching of given curves with appropriate gases:

In the given graph, the taller and narrow curve whose maximum falls near $500\text{ m/s}$ represents heavier particles like neon atoms.

The flatter and wider curve whose maximum falls near $1500\text{ m/s}$ represents lighter particles like hydrogen molecules.

Conclusion

The taller and narrow curve matches with neon atoms

The flatter and wider curve matches with hydrogen molecules

Interpretation Introduction

Interpretation:

The rms speed (in $\text{m/s}$) for each of the gases at $200\text{ K}$ has to be calculated

Concept Introduction:

Root-mean-square (rms):

The root-mean-square molecular speed ( $u$) is given by the below formula.

$\begin{array}{l}u\text{=}\sqrt{\frac{3\text{RT}}{{\text{M}}_{\text{m}}}}\\ ={\left(\frac{3\text{RT}}{{\text{M}}_{\text{m}}}\right)}^{\frac{1}{2}}\end{array}$

Where,

$\text{R}$ is the molar gas constant

$\text{T}$ is the absolute temperature

${\text{M}}_{\text{m}}$ is the molar mass of the gas

Answer to Problem 5.136QP

The rms speed of neon gas at $200\text{ K}$ is $497\text{ m/s}$

The rms speed of hydrogen gas at $200\text{ K}$ is $1.57\times {10}^3\text{ m/s}$

Explanation of Solution

Given,

The given graph represents the distribution of molecular speeds of hydrogen and neon at $200\text{ K}$

Calculation of rms speed:

The rms speed of neon is calculated as follows,

$\begin{array}{l}{u}_{\text{Ne}}=\left(\frac{3\times 8.31\text{ kg}·{\text{m}}^2/({\text{s}}^2·\text{K}·\text{mol)}\times \text{200 K}}{20.18\times {10}^{-3}\text{ kg/mol}}\right)\\ =\text{497 m/s}\end{array}$

The rms speed of hydrogen gas is calculated as follows,

$\begin{array}{l}{u}_{{\text{H}}_{\text{2}}}=\left(\frac{3\times 8.31\text{ kg}·{\text{m}}^2/({\text{s}}^2·\text{K}·\text{mol)}\times \text{200 K}}{\text{2}\text{.016}\times {10}^{-3}\text{ kg/mol}}\right)\\ =\text{1}\text{.57}\times {\text{10}}^3\text{ m/s }\end{array}$

Conclusion

The rms speed of neon gas at $200\text{ K}$ is calculated as $497\text{ m/s}$

The rms speed of hydrogen gas at $200\text{ K}$ is calculated as $1.57\times {10}^3\text{ m/s}$

Interpretation Introduction

Interpretation:

The gas that has greater effusion rate at $200\text{ K}$ has to be explained

Concept Introduction:

Graham’s law of effusion:

$\text{Rate of effusion of molecules }\propto \frac{\text{1}}{\sqrt{{\text{M}}_{\text{m}}}}$

Answer to Problem 5.136QP

The gas that has greater effusion rate at $200\text{ K}$ is hydrogen gas.

Explanation of Solution

Given,

The given graph represents the distribution of molecular speeds of hydrogen and neon at $200\text{ K}$

Gas possessing greater effusion rate:

The rates of effusion are directly related to the rms speed.

Greater the rms speed, greater is the rate of effusion.

Since the rms speed of hydrogen is greater than neon, hydrogen gas will have greater effusion rate.

In the container, the fast moving molecules collide with the holes of the container more often and hence have a higher effusion probability.

Conclusion

The gas that has greater effusion rate at $200\text{ K}$ is found as hydrogen gas.

Interpretation Introduction

Interpretation:

The temperature at which the rms speed of the hydrogen gas equals the rms speed of neon gas at $200\text{ K}$ has to be calculated

Concept Introduction:

Root-mean-square (rms):

The root-mean-square molecular speed ( $u$) is given by the below formula.

$\begin{array}{l}u\text{=}\sqrt{\frac{3\text{RT}}{{\text{M}}_{\text{m}}}}\\ ={\left(\frac{3\text{RT}}{{\text{M}}_{\text{m}}}\right)}^{\frac{1}{2}}\end{array}$

Where,

$\text{R}$ is the molar gas constant

$\text{T}$ is the absolute temperature

${\text{M}}_{\text{m}}$ is the molar mass of the gas

Answer to Problem 5.136QP

The temperature at which the rms speed of the hydrogen gas equals the rms speed of neon gas at 200 K is $20.0\text{ K}$

Explanation of Solution

Given,

The given graph represents the distribution of molecular speeds of hydrogen and neon at $200\text{ K}$

Temperature calculation:

The temperature equaling the rms speed of neon is calculated from root mean square equation as follows,

$\begin{array}{l}\text{T }\text{=}\left(\frac{u^2{\text{M}}_{\text{m}}}{\text{3R}}\right)\\ =\left(\frac{{(497.06\text{ m/s)}}^{\text{2}}(20.16\times {10}^{-3}\text{ kg/mol) }}{3\times 8.31\text{ kg}·{\text{m}}^2/({\text{s}}^2·\text{K}·\text{mol)}}\right)\\ =20.0\text{ K}\end{array}$

Conclusion

The temperature at which the rms speed of the hydrogen gas equals the rms speed of neon gas at $200\text{ K}$ is calculated as $20.0\text{ K}$