Chapter 13, Problem 115A

Interpretation Introduction

Interpretation:

Average speed of gas particles is to be calculated.

Concept introduction:

Average speed is given by:

$\text{v=}\sqrt{\frac{8RT}{\pi M}}$

V is average speed

R is gas constant

T is temperature

M is molar mass of the gas

Answer to Problem 115A

Highest average speed is of methane gas at 90°C and smallest is of nitrogen trifluoride at 30 °C

Explanation of Solution

Hence average speed depends on ratio T/M.

The gas molecules having more value of T/M will have more average speed and vice-versa.

$\begin{array}{l}\text{v=}\sqrt{\frac{8RT}{\pi M}}\\ \mathrm{v}\propto \sqrt{\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$M$}\right.}\end{array}$

1. For CO:

$\begin{array}{l}T=90°C\\ =90+273\\ =363K\\ M=12+16\\ =28g/mol\\ \sqrt{\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$M$}\right.}=\sqrt{\raisebox{1ex}{$363$}\!\left/ \!\raisebox{-1ex}{$28$}\right.}\\ =3.6\end{array}$

2. For NF₃

$\begin{array}{l}T=30°C\\ =30+273\\ =303K\\ M=14+19\times 3\\ =71g/mol\\ \sqrt{\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$M$}\right.}=\sqrt{\raisebox{1ex}{$303$}\!\left/ \!\raisebox{-1ex}{$71$}\right.}\\ =2.06\end{array}$

3. For methane:

$\begin{array}{l}T=90°C\\ =90+273\\ =363K\\ M=12+1\times 4\\ =16g/mol\\ \sqrt{\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$M$}\right.}=\sqrt{\raisebox{1ex}{$363$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}\\ =4.76\end{array}$

4. For CO :

$\begin{array}{l}T=30°C\\ =30+273\\ =333K\\ M=12+16\\ =28g/mol\\ \sqrt{\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$M$}\right.}=\sqrt{\raisebox{1ex}{$333$}\!\left/ \!\raisebox{-1ex}{$28$}\right.}\\ =3.44\end{array}$

Conclusion

Highest average speed is of methane gas at 90°C and smallest is of nitrogen trifluoride at 30 °C