At $27^{o} C$ , the ratio of the kinetic energies of nitrogen and oxygen whose weights ratio is $7 : 8$ is:

1 : 1

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1 : 2

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2 : 1

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7 : 8

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Solution

The correct option is C $1 : 1$
$\frac{K E_{1}}{K E_{2}} = \frac{n_{1} T_{1}}{n_{2} T_{2}}$
Let weight of niitrogen gas be $7 x$ and that of oxygen gas be $8 x$
no. of moles of nitrogen = $n_{1} = \frac{7 x}{28}$
no. of moles of oxygen = $n_{2} = \frac{8 x}{32}$
Thus, at same temperature
$\frac{K E_{1}}{K E_{2}} = \frac{7 x \times 32}{8 x \times 28} = 1 : 1$

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