Which gas can be easily liquified?
given 'a' for $N H_{3} = 4.17, C O_{2} = 3.59, S O_{2} = 6.71, C l_{2} = 6.49$

N H_{3}

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C l_{2}

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S O_{2}

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C O_{2}

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Solution

The correct option is A $S O_{2}$
Given values for gases are:
${N H}_{3} = 4.17$ , ${C O}_{2} = 3.59$ , ${S O}_{2} = 6.71$
$u_{2} = 6.49$
Now according to the van der Waal's equation, a and b are correction constants that is they are Van der Waal's force of attraction constant.
Also, 'a' is said to be the measure of the intermolecular force of attraction.
Now, we know greater the force of intermolecular attraction, easier is its liquification.
Therefore, Easily liquifiable gas has greater value
Sulphur dioxide is easily liquifiable here.

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Similar questions

a

{N H}_{3} = 4.17, {C O}_{2} = 3.59, {S O}_{2} = 6.71, {C l}_{2} = 6.49

a

N H_{3} = 4.17, C O_{2} = 3.59, S O_{2} = 6.71 & C l_{2} = 6.49

a

N H_{3} = 4.17, C O_{2} = 3.59, S O_{2} = 6.71 & C l_{2} = 6.49

N H_{3}

H_{2}

O_{2}

C O_{2}

b a r L^{2} m o l^{- 2}

N H_{3}, H_{2}, O_{2} a n d C O_{2}

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