Question

The ratio between kinetic energies of 16 g of $O_2$ and 28 g of $N_2$ respectively at 300 K will be?

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $4:7$

Answer 1

Answer: (B)

$1:2$

Given:-

$16g$ of $O_2$

$KE=\frac{3}{2}nRT$ $n⟶$ Number of moles

Number of moles= $\frac{16}{32}=0.5$ mole of $O_2$

$28g$ of $N_2$

$\Rightarrow n=\frac{28}{28}=1$ moles of $N_2$

$\frac{(K.E{)}_{O_2}}{(K.E{)}_{N_2}}=\frac{\frac{3}{2}\times 0.5\times R\times T}{\frac{3}{2}\times 1\times R\times T}$ $T=300K$

$=1:2$ .