If number of molecules of $H_{2}$ are double than that of $O_{2}$ , then ratio of mean kinetic energy per molecule of hydrogen to that of oxygen at 300 K is:

2:1

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1:2

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2:3

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1:4

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Solution

The correct option is A 2:1
Mean Kinetic energy is given as, $K . E = \frac{3}{2} n R T$
$n =$ no. of molecules.
$R =$ Constant.
$T =$ Temperature.
For $H_{2}$ ,
$K . E_{H_{2}} = \frac{3}{2} n_{H_{2}} R T ⟶ (1)$
For $O_{2}$ ,
$K . E_{O_{2}} = \frac{3}{2} n_{O_{2}} R T ⟶ (1)$
On dividing equation $(1)$ by equation $(2)$ we get,
$\frac{K . E_{H_{2}}}{K . E_{O_{2}}} = \frac{n_{H_{2}}}{n_{O_{2}}}$
given $n_{H_{2}} = 2 n_{O_{2}}$ ,
$\frac{K . E_{H_{2}}}{K . E_{O_{2}}} = \frac{n_{2 O_{2}}}{n_{O_{2}}} = \frac{2}{1}$
Thus the ration is $2 : 1.$

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If number of molecules of H2 are double than that of O2, then ratio of mean kinetic energy per molecule of hydrogen to that of oxygen at 300 K is:

The correct option is A 2:1Mean Kinetic energy is given as, K.E=32nRTn= no. of molecules.R= Constant.T= Temperature.For H2,K.EH2=32nH2RT⟶(1)For O2,K.EO2=32nO2RT⟶(1)On dividing equation (1) by equation (2) we get,K.EH2K.EO2=nH2nO2given nH2=2nO2,K.EH2K.EO2=n2O2nO2=21Thus the ration is 2:1.

If number of molecules of $H_{2}$ are double than that of $O_{2}$ , then ratio of mean kinetic energy per molecule of hydrogen to that of oxygen at 300 K is: