Question

The density of the vapour of a substance at 1 atm pressure and 500 K is 0.36 kg m${}^{-3}$. The vapour effuses through a small hole at a rate of 1.33 times faster than oxygen under the same condition. If the vapour behaves ideally at 1000 K, determine the average translational kinetic energy of a molecule. (in terms of J per molecule).

• A. $7.02\times {10}^{-20}$
• B. $2.07\times {10}^{-20}$
• C. $702\times {10}^{-20}$
• D. $207\times {10}^{-20}$

Answer 1

The correct option is B $2.07\times {10}^{-20}$

$E=\frac{3}{2}{K}_{B}T=\frac{3}{2}\times 1.38\times {10}^{-23}\times 1000$
$=2.07\times {10}^{-20}\text{J per molecule}$