Question

Eight gram each of oxygen and hydrogen at ${27}^{\circ }C$ will have the total kinetic energy in the ratio of x:y. The value of y is

Answer 1

The K.E of a gas at a given temperature is:
$K.E=\frac{3}{2}nRT$
now, $n_{O_2}=\frac{8}{32}$
$n_{H_2}=\frac{8}{2}$
$\frac{K.E_{\left(O_2\right)}}{K.E_{\left(H_2\right)}}=\frac{\left(\frac{3}{2}\right)n_{O_2}RT}{\left(\frac{3}{2}\right)n_{H_2}RT}=\frac{n_{O_2}}{n_{H_2}}=\frac{\frac{8}{32}}{\frac{8}{2}}=\frac{1}{16}$
hence, y=16