The critical temperatures of CO, $C H_{4}$ , HCl and $S O_{2}$ are 134K, 190K, 324K and 430K respectively. The order of extension of adsorption of these gases on charcoal is

S O_{2} > H C l > C H_{4} > C O

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C O > C H_{4} > H C l > S O_{2}

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H C l > S O_{2} > C O > C H_{4}

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C H_{4} > C O > S O_{2} > H C l

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Solution

The correct option is A $S O_{2} > H C l > C H_{4} > C O$
The gas with a higher value of critical temperature will have stronger vander waal’s forces and hence easily adsorbed on charcoal.

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Similar questions

H_{2}

S O_{2}

C H_{4}

Gas	$C O_{2}$	$S O_{2}$	$C H_{4}$	$H_{2}$
Critical temp./K	304	630	190	33

N_{2}, C O

C H_{4}

126, 134

190

Gas	$H_{2}$	$C H_{4}$	$C O_{2}$	$S O_{2}$
Critical Temperature $(K)$	$33$	$190$	$304$	$630$

N_{2}, C O and C H_{4}

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