Question

Ratio of average kinetic energy of $14g$ of $N_2$ at ${27}^{o}C$ to $24g$ of $O_2$ at ${227}^{o}C$ is _____.

• A. $1:5$
• B. $2:5$
• C. $1:4$
• D. $3:5$

Answer 1

Answer: (D)

$3:5$

According to the kinetic theory of gases,

$KE=\frac{3}{2}kT\propto n$(at constant Temperature)

But here temperature is changing from ${27}^{o}C$ to ${227}^i=oC$.

Thus for constant number of moles, the kinetic energy is given as :

$KE\propto T$

$\therefore \frac{K{E}_1}{K{E}_2}=\frac{T_1}{T_2}$

$\therefore \frac{KE-1}{K{E}_2}=\frac{27+273}{227+273}=\frac{300}{500}=\frac{3}{5}$

Hence $K{E}_{N_2}:K{E}_{O_2}=3:5$