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The pressure occupied by one mole of $CO_2$ at 273K is ........ (Given a for $CO_2$ = 3.592 $dm^6atmmol^{-2}$). Assume volume occupied by $CO_2$ molecules is negligible.

$(a)\;0.99\;atm\qquad(b)\;17.85\;atm\qquad(c)\;34.96\;atm\qquad(d)\;18.20\;atm$

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$[P + \large\frac{a}{V^2}][V-b] = RT$
b is negligible
$\therefore PV + \large\frac{a}{V} = RT$
(or) $V^2P - RTV + a = 0$
$V = + RT \pm \large\frac{\sqrt{R^2T^2 - 4aP}}{2P}$
Since V has only one value
$R^2T^2 = 4aP$
$P = \large\frac{R^2T^2}{4a}$
$=\large\frac{{0.0821}^2\times273\times273}{4\times3.592}$
= 34.96 atm
Hence answer is (C)
answered Mar 31, 2014 by sharmaaparna1
 

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