Q 103 - Organic Chemistry

Question

Consider separate 1.0-L samples of O₂(g) and He(g), both at 25^oC and the same pressure. Compare the change in momentum per impact and the number of impacts per second in the two samples.

Step-by-Step Solution

Answer

Answer: The change in momentum per impact is 2.827 times larger for O₂ than for He. The number of impacts per second is 2.827 higher for He than O₂.

Step 1: Meaning of change in momentum

For an elastic collision, the change in momentum per impact is given as:

 $\mathbf{∆}\mathbf{\left(}\mathbf{mu}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{mu}$

Where, m = mass, u = velocity

 

The above expression can also be written as:

 $\mathbf{2}\mathbf{mu}\mathbf{\propto }\mathbf{2}\sqrt{\mathbf{TM}}$

Where, T = temperature, M = molar mass

Step 2: Comparison of the Change in momentum per impact

The proportion between the gases is given below:

$$\begin{gathered} \frac{\mathbf{∆}{\mathbf{\left(}\mathbf{mu}\mathbf{\right)}}_{\mathbf{O}\mathbf{2}}}{\mathbf{∆}{\mathbf{\left(}\mathbf{mu}\mathbf{\right)}}_{\mathbf{He}}}\mathbf{=}\frac{\mathbf{2}\sqrt{{\mathbf{TM}}_{\mathbf{O}\mathbf{2}}}}{\mathbf{2}\sqrt{{\mathbf{TM}}_{\mathbf{He}}}} \\ \frac{\mathbf{∆}{\mathbf{\left(}\mathbf{mu}\mathbf{\right)}}_{\mathbf{O}\mathbf{2}}}{\mathbf{∆}{\mathbf{\left(}\mathbf{mu}\mathbf{\right)}}_{\mathbf{He}}}\mathbf{=}\frac{\sqrt{{\mathbf{M}}_{\mathbf{O}\mathbf{2}}}}{\sqrt{{\mathbf{M}}_{\mathbf{He}}}} \\ \frac{\mathbf{∆}{\mathbf{\left(}\mathbf{mu}\mathbf{\right)}}_{\mathbf{O}\mathbf{2}}}{\mathbf{∆}{\mathbf{\left(}\mathbf{mu}\mathbf{\right)}}_{\mathbf{He}}}\mathbf{=}\frac{\sqrt{\mathbf{31}\mathbf{.}\mathbf{998}{\mathbf{gmol}}^{\mathbf{-}\mathbf{1}}}}{\sqrt{\mathbf{4}\mathbf{.}\mathbf{003}{\mathbf{gmol}}^{\mathbf{-}\mathbf{1}}}} \\ \frac{\mathbf{∆}{\mathbf{\left(}\mathbf{mu}\mathbf{\right)}}_{\mathbf{O}\mathbf{2}}}{\mathbf{∆}{\mathbf{\left(}\mathbf{mu}\mathbf{\right)}}_{\mathbf{He}}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{827} \end{gathered}$$

Hence, the change in momentum per impact is 2.827 times larger for O₂ than for He.

Step 3: Comparison of the number of impacts per second

The impact rate can be calculated by using the following equation:

$\mathbf{Z}\mathbf{=}\mathbf{A}\frac{\mathbf{N}}{\mathbf{V}}\sqrt{\frac{\mathbf{RT}}{\mathbf{2}\mathbf{\pi m}}}$

The proportion between the impact rate for O₂ and He is given below:

$$\begin{gathered} \frac{{\mathbf{Z}}_{\mathbf{O}\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{He}}}\mathbf{=}\frac{\mathbf{A}{\displaystyle \frac{\mathbf{N}}{\mathbf{V}}}\sqrt{{\displaystyle \frac{\mathbf{RT}}{\mathbf{2}{\mathbf{\pi M}}_{\mathbf{O}\mathbf{2}}}}}}{\mathbf{A}{\displaystyle \frac{\mathbf{N}}{\mathbf{V}}}\sqrt{{\displaystyle \frac{\mathbf{RT}}{\mathbf{2}{\mathbf{\pi M}}_{\mathbf{He}}}}}} \\ \frac{{\mathbf{Z}}_{\mathbf{O}\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{He}}}\mathbf{=}\frac{\sqrt{{\displaystyle \frac{1}{M_{O2}}}}}{\sqrt{{\displaystyle \frac{1}{M_{\mathrm{He}}}}}} \\ \frac{{\mathbf{Z}}_{\mathbf{O}\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{He}}}\mathbf{=}\frac{\sqrt{{\displaystyle M_{\mathrm{He}}}}}{\sqrt{{\mathbf{M}}_{\mathbf{O}\mathbf{2}}}} \\ \frac{{\mathbf{Z}}_{\mathbf{O}\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{He}}}\mathbf{=}\frac{\sqrt{\mathbf{4}\mathbf{.}\mathbf{003}{\mathbf{gmol}}^{\mathbf{-}\mathbf{1}}}}{\sqrt{\mathbf{31}\mathbf{.}\mathbf{998}{\mathbf{gmol}}^{\mathbf{-}\mathbf{1}}}} \end{gathered}$$

$$\begin{gathered} \frac{{\mathbf{Z}}_{\mathbf{O}\mathbf{2}}}{{\mathbf{Z}}_{\mathbf{H}\mathbf{e}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3527} \\ \frac{{\mathbf{Z}}_{\mathbf{He}}}{{\mathbf{Z}}_{\mathbf{O}\mathbf{2}}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{827} \end{gathered}$$

Hence, the number of impacts per second is 2.827 higher for He than O₂.