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Hint:Critical temperature is defined as the temperature at which the gas is the first time seen as the liquid. Thomas Andrews worked on the liquefaction of $C\mathop O\nolimits_2 $. At lower temperatures, the gases can be easily liquefied. The more easily liquefiable gas is more easily adorable. At critical temperature gases have weak Vander waals forces and hence they are easily adorable.
For $C\mathop O\nolimits_2 $ critical temperature is 30.98C.
Complete answer:
A] $\mathop N\nolimits_2 $- it has less value of critical temp. and is not easily liquefied hence this option is not correct
B] $S\mathop O\nolimits_2 $– it has a very high critical temp. therefore it has weak Vander waals forces and hence can be easily liquefied so is better absorbed on activated charcoal. So, the option is correct
C] $\mathop H\nolimits_2 $- it has very low critical temp due to its small size and hence it is not easily adsorbing. So the option is incorrect.
D] $\mathop O\nolimits_2 $- oxygen has less critical temperature than Sulphur dioxide and cannot be easily liquefied so cannot be adsorbed easily and hence the option is incorrect.
Answer to this question is option B.
Note:
Critical temperature – the highest temperature at which gas is the first time observed at the liquid.
Critical volume- a volume of one mole of gas at a critical temperature.
Critical pressure – the pressure at which gas is first observed as the liquid.
For $C\mathop O\nolimits_2 $ critical pressure is 73atm.
A gas below critical temp can be liquefied by applying pressure and is called VAPOUR OF SUBSTANCES.
For $C\mathop O\nolimits_2 $ critical temperature is 30.98C.
Complete answer:
A] $\mathop N\nolimits_2 $- it has less value of critical temp. and is not easily liquefied hence this option is not correct
B] $S\mathop O\nolimits_2 $– it has a very high critical temp. therefore it has weak Vander waals forces and hence can be easily liquefied so is better absorbed on activated charcoal. So, the option is correct
C] $\mathop H\nolimits_2 $- it has very low critical temp due to its small size and hence it is not easily adsorbing. So the option is incorrect.
D] $\mathop O\nolimits_2 $- oxygen has less critical temperature than Sulphur dioxide and cannot be easily liquefied so cannot be adsorbed easily and hence the option is incorrect.
Answer to this question is option B.
Note:
Critical temperature – the highest temperature at which gas is the first time observed at the liquid.
Critical volume- a volume of one mole of gas at a critical temperature.
Critical pressure – the pressure at which gas is first observed as the liquid.
For $C\mathop O\nolimits_2 $ critical pressure is 73atm.
A gas below critical temp can be liquefied by applying pressure and is called VAPOUR OF SUBSTANCES.
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