The compressibility factor for $1$ mole of a vander Waals gas at $273 K, 8.21$ atm pressure having volume $1.092$ litre is $0.4$ . Assuming the volume of gas molecules to be negligible. The value of vander Walls constant $a$ will be ?

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

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

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

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

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Solution

The correct option is B $14.53 L^{2} m o l^{- 2} a t m$
$Z = \frac{p V}{n R T}$
$0.4 = \frac{8.21 \times V}{1 \times 0.082 \times 273}$
$V = 1.09066 L$
$(p + \frac{a}{V^{2}}) (V - b) = R T f o r 1 m o l$
Neglecting b, $(p + \frac{a}{V^{2}}) V$ $= R T$ or
$p V + \frac{a}{V} = R T$ or
$\frac{p V}{R T} + \frac{a}{V R T} = 1$ or
$a = (1 - \frac{p V}{R T}) V R T = (1 - 0.4) \times 1.09066 \times 0.082 \times 273$
$a = 14.6493 a t m L^{2} m o l^{- 2}$

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