Question

One mole of oxygen at $273K$ and one mole of sulphur dioxide at $546K$ are taken in two separate containers, then:

• A. Kinetic energy of $O_2>$ Kinetic energy of $S{O}_2$
• B. Kinetic energy of $O_2<$ Kinetic energy of $S{O}_2$
• C. Kinetic energy of both are equal
• D. None of the above

Answer 1

The correct option is B Kinetic energy of $O_2<$ Kinetic energy of $S{O}_2$

For Oxygen: It is diatomic, therefore it has 5 degrees of freedom.

$\therefore (K.E.)oxygen=\frac{5}{2}{K}_B\times T$

$=\frac{5}{2}\times 273\times K_B$

For Sulphar dioxide : It is triatomic, therefore it has 7 degrees of freedom.

$\therefore (K.E.)S{O}_2=\frac{7}{2}\times K_B\times T$

$=\frac{7}{2}\times 546\times K_B$

$\frac{(K.E.)O_2}{(K.E.)S{O}_2}=\frac{\frac{5}{2}\times 273\times K_B}{\frac{7}{2}\times 546\times K_B}$

$\frac{(K.E.)O_2}{(K.E.)S{O}_2}=\frac{5}{14}$

Hence $(K.E.)S{O}_2>(K.E.)O_2$