Critical temperature of $N_{2}, C O and C H_{4}$ are 126, 134 and 190 K respectivley. The increasing order of adsorption on the surface of activated charcoal is:

N_{2} < C H_{4} < C O

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N_{2} < C O < C H_{4}

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C H_{4} < C O < N_{2}

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C O < N_{2} < C H_{4}

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Solution

The correct option is B $N_{2} < C O < C H_{4}$
Critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquefied, no matter how much pressure is applied.
Gases having higher critical temperature are easily liquefied as van der Waals' forces are stronger near the critical temperatures (TC).
Hence, they have greater force of attraction with the charcoal and therefore they are more easily adsorbed on the surface of the charcoal.
Adsorption of gas on the surface of activated charcoal depends on the critical temperature of the gas. Critical temperature has a direct relation to adsorption - Higher the critical temperature, greater will be the adsorption.
Therefore, the order of adsorption is $N_{2} < C O < C H_{4}$

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Similar questions

N_{2}, C O

C H_{4}

N_{2}, C O

C H_{4}

126, 134

190

C H_{4}

S O_{2}

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

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

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