Question

What is the ratio of kinetic energies of 3 g of hydrogen and 4 grams of oxygen at $TK$?

• A. 12 : 1
• B. 6 : 1
• C. 1 : 6
• D. 24 : 1

Answer 1

Answer: (C)

1 : 6

$k=\frac{R}{N_A}$ for $1$ mole

for $3g$ of $H_2,k=\frac{R}{\frac{3}{2}{N}_A}$

for $4g$ of $O_2,k=\frac{R}{\frac{4N_A}{16}}$

$KE=\frac{3}{2}KT$

$\therefore \frac{K{E}_{H_2}}{K{E}_{O_2}}=\frac{\frac{4N_A}{16}}{\frac{3}{2}{N}_A}$

$=\frac{1}{4}\times \frac{2}{3}$

$=\frac{2}{12}=\frac{1}{6}$ .