Question

2 moles of $Ne$ gas and 5 moles of $He$ gas, both samples have an average velocity of $7\times {10}^2\text{m/s}$. Find the average translational kinetic energy per mole of the given gas mixture (in joules). Report your answer as ${}^{\prime }{X}^{\prime }$ where $X=$ (Average translational KE in joules) $\times 0.002$. The reported answer should be rounded to the nearest integer.

Answer 1

Average translational kinetic energy (per mole of gas) $=\frac{3}{2}RT$
Multiplying and dividing the previous equation by molar mass of gas $M$
= $\frac{3M}{2}\times \frac{RT}{M}$
Multiplying and dividing the previous equation by $\pi$
=$\frac{3\pi M}{16}\times \frac{8RT}{\pi M}$
$=\frac{3\pi M}{16}\times {\left(\sqrt{\frac{8RT}{\pi M}}\right)}^2$
$=\frac{3\pi M}{16}\times (u_{\text{avg}}{)}^2$
$\therefore$ For one mole of gas mixture,
$\text{Average translational kinetic energy}=\frac{3\pi M_{\text{mix}}}{16}\times (u_{\text{avg}}{)}^2$, where $u_{avg}$ is average velocity of particles and $M_{mix}$ is average molar mass of gaseous mixture in kg/mol.
$\therefore \text{Average translational kinetic energy}==\frac{3}{16}\times \frac{22}{7}\times \left[\frac{\frac{(2\times 20+5\times 4)}{7}}{1000}\right]\times (7\times {10}^2{)}^2$
$\text{Average translational kinetic energy}=2475Jmo{l}^{-1}$

$\text{Reported answer}=2475\times 0.002=4.95\approx 5$