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Using Van der Waals's equation . Calculate the constant 'a' when two mole of a gas confined in a four litre flask exerts a pressure of 11.0 atmospheres at a temperature of 300K . The value of b is 0.05 lit $mol^{-1}$

$\begin{array}{1,1}(a)\;646 atm\;litre^2\;mol^{-2}\\(b)\;6.46\;atm\;litre^2\;mol^{-2}\\(c)\;64.6 atm\;litre^2\;mol^{-2}\\(d)\;0.64atm\;litre^2\;mol^{-2}\end {array}$

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A)
Van der Waal's equation for n mole of gas is
[P + $\large\frac{n^2a}{V^2}$] [ V-nb] = nRT
Given
V = 4 litre
P = 11.0 atm
T = 300K
b = 0.05 litre$\;mol^{-1}$
n = 2
Thus
[ 11+ $\large\frac{2^2a}{4^2}][4-2\times0.05]$
$=2\times0.082\times300$
$a = 6.46 \;atm\;litre^2mol^{-1}$
Hence answer is (B)
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