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Percentage Composition

KE of equal m...

Question

$K E$ of equal masses of methane, oxygen and sulphur dioxide under similar conditions will be in the ratio of:

1:1:1

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1:2:3

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4:2:1

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1:2:4

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Solution

The correct option is A 4:2:1
$K E = \frac{3 n R T}{2}$
$K E \propto n$
Thus, $K E$ of gas is inversely proportional to molecular weight of gas, when masses of gases are equal and temperature is also same.
Molecular weight of $C H_{4} = 16, O_{2} = 32, S O_{2} = 64$

$K E_{1} : K E_{2} : K E_{3} = \frac{1}{M_{1}} : \frac{1}{M_{2}} : \frac{1}{M_{3}}$

$K E_{1} : K E_{2} : K E_{3} = \frac{1}{16} : \frac{1}{32} : \frac{1}{64} = 1 : \frac{1}{2} : \frac{1}{4} = 4 : \frac{4}{2} : \frac{4}{4} = 4 : 2 : 1$
Option C is correct.

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