Given the value of their van der Waals' constant $a$ arrange the following gases in the order of their expected liquefication pressure at a temperature $T$ . $T$ is below the critical point of all the gases.

Gas $C H_{4}$ $K r$ $N_{2}$ $C l_{2}$
$a (a t m L^{2} . m o l^{- 1})$ 2.283 2.439 1.408 6.579

C H_{4} < K r < N_{2} < C l_{2}

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N_{2} < C H_{4} < K r < C l_{2}

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C l_{2} < K r < C H_{4} < N_{2}

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C l_{2} < N_{2} < K r < C H_{2}

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Solution

The correct option is C $C l_{2} < K r < C H_{4} < N_{2}$
Van der Waals constant 'a' is an indirect measure of attractive force.

Gases with higher values of 'a' can be easily liquified. Hence, lower will be its liquefaction pressure.

The increasing order of the van der Waal's constant 'a' is $N_{2} < C H_{4} < K r < C l_{2}$ .

Hence the decreasing order of the liquification pressure is $N_{2} > C H_{4} > K r > C l_{2}$ .

Hence, the correct answer is option $C$ .

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Similar questions

C H_{4}

K r

N_{2}

C l_{2}

a t m L^{2} . m o l^{- 1}

a

\begin{matrix} Gas & a (L^{2} a t m m o l^{- 2}) N_{2} & 1.390 O_{2} & 1.360 N H_{3} & 4.170 C H_{4} & 2.253 \end{matrix}

(Z = \frac{P V}{n R T})

C l_{2}

N H_{3}

C H_{4}, N_{2}, N H_{3}

O_{2}

L^{2} a t m m o l^{- 2}

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Gas	$C H_{4}$	$K r$	$N_{2}$	$C l_{2}$
$a (a t m L^{2} . m o l^{- 1})$	2.283	2.439	1.408	6.579