A) 1.256 \[{{L}^{2}}mo{{l}^{-2}}atm\]
B) 0.256 \[{{L}^{2}}mo{{l}^{-2}}atm\]
C) 2.256 \[{{L}^{2}}mo{{l}^{-2}}atm\]
D) 0.0256 \[{{L}^{2}}mo{{l}^{-2}}atm\]
Correct Answer: A
Solution :
[a] \[Z=\frac{PV}{nRT}=0.5\] Now, \[\left[ P+\frac{{{n}^{2}}a}{{{V}^{2}}} \right][V-nb]=nRT\] \[\left[ P+\frac{{{n}^{2}}a}{{{V}^{2}}} \right][V]=nRT\](b is negligible) \[P{{V}^{2}}-nRTV+{{n}^{2}}a=0\] \[\therefore V=\frac{nRT\pm \sqrt{{{n}^{2}}{{R}^{2}}{{T}^{2}}-4{{n}^{2}}a\times b}}{2P}\] Since, V is constant at given P and T, thus discriminant is 0. \[\therefore {{n}^{2}}{{R}^{2}}{{T}^{2}}=4{{n}^{2}}aP\]or \[a=\frac{{{R}^{2}}{{T}^{2}}}{4P}\] \[=\frac{{{(0.0821)}^{2}}\times {{(273)}^{2}}}{4\times 100}\] \[=1.256Lmo{{l}^{-2}}atm\]
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