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Q.
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
Kinetic Theory
Solution:
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 \longrightarrow(1)
$
For $_{2}$,
$
K \cdot E _{ O _{2}}=\frac{3}{2} n _{ O _{2}} RT \longrightarrow(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 $n _{ H _{2}}=2 n _{ O _{2}}$,
$
\frac{ K \cdot E _{ H _{2}}}{ K \cdot E _{ O _{2}}}=\frac{ n _{2 O _{2}}}{ n _{ O _{2}}}=\frac{2}{1}
$
Thus the ration is $2: 1$.