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The compressibility factor for one mole of a van der Waals gas at 0^o C and 100 atm pressure is found to be 0.5 Assume that the volume of gas molecule is negligible. The value of van der Waals constant ‘a’ is

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1.253 atm lit² mol^–2

12.53 atm lit² mol^–2

0.125 atm lit² mol^–2

22.53 atm lit² mol^–2

answer is A.

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Detailed Solution

$z = \frac{PV}{RT} 0.5 = 100 \times \frac{V}{0.0821} \times 273 V = 0.1117 L$

When the volume of a gas molecule is negligible, van der Waal's equation becomes,

$P + \frac{a}{V^{2}} (V - 0) = RT$

Substituting the values

$A = (0.0821 \times 0.1119 \times 273) - (100 \times 0.1119 \times 0.1119) A \approx 1.253 {atmL}^{2} {mol}^{- 2}$

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