Question

$6.0g$ of $He$ having average velocity ${20}^{\circ }C$ is mixed with $12.0gofN{e}^{20}$ having the same average velocity. What is the average kinetic energy per mole in the mixture?

• A. 6653J
• B. 807.84 kJ
• C. 1607.84 kJ
• D. none of these

Answer 1

The correct option is A 6653J

The number of moles of $He$ and $Ne$ are $\frac{6}{4}=1.5$ and $\frac{12}{20}=0.6$ respectively.
The average mass of the mixture is $\frac{4\left(1.5\right)+20(0.6}{1.5+0.6}=8.58g/mol$
The average velocity of $He$ is $\sqrt{\frac{8RT}{\pi M}}=\sqrt{\frac{8\times 8.314\times 293}{3.1416\times 4}}=39.39$
The average kinetic energy per mole of the mixture is $0.5m{v}^2=0.5\times 8.58\times (39.39{)}^2=6653J$.