Question

At the same temperature, total $KE$ of 16 grams of oxygen is ............ that of 16 grams of $S{O}_2$.

• A. Same
• B. Two times
• C. Four times
• D. Not known

Answer 1

The correct option is B Two times

We know,

$K=N\frac{3}{2}{K}_{B}T$

Where,

$K_B$ is Boltzmann constant

$N$ is the number of moles

$K$ is kinetic energy

So kinetic energy is directly proportional to the number of moles

16 grams oxygen contain 1/2 mole of $O_2$

and 16 grams of $S{O}_2$ contain 1/4 mole of $S{O}_2$

$K_{O_2}=kineticenergyof{O}_2$

$K_{S{O}_2}=kineticenergyof|K_{S{O}_2}$

Thus, at constant temperature

$\frac{K_{O_2}}{K_{S{O}_2}}=\frac{numberofmolesof{O}_2}{numberofmolesofS{O}_2}$

$\therefore \frac{K_{O_2}}{K_{S{O}_2}}=\frac{\frac{1}{2}}{\frac{1}{4}}=2$

$\therefore K_{O_2}=2K_{S{O}_2}$

$\therefore$ Total Kinetic energy of 16 grams of oxygen is two times that of 16 grams of $S{O}_2$.