Question

If number of molecules of $H_2$ are double than that of $O_2$, then ratio o f kinetic energy of hydrogen and that of oxygen at 300 K is

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 1 : 16

Answer 1

Answer: (C)

Mean Kinetic energy is given as, $K\cdot E=\frac{3}{2}nRT$ $n=$ no. of molecules.
$R=$ Constant.
$T=$ Temperature.
For $H_2$,
$K\cdot E_{H_2}=\frac{3}{2}{n}_{H_2}RT⟶(1)$
For ${}_2$,
$K\cdot E_{O_2}=\frac{3}{2}{n}_{O_2}RT⟶(1)$
On dividing equation (1) by equation (2) we get,
$\frac{K\cdot E_{H_2}}{K\cdot E_{O_2}}=\frac{n_{H_2}}{n_{O_2}}$
given $n_{H_2}=2n_{O_2}$,
$\frac{K\cdot E_{H_2}}{K\cdot E_{O_2}}=\frac{n_{2O_2}}{n_{O_2}}=\frac{2}{1}$
Thus the ration is $2:1$.