JEE Main & Advanced Chemistry States of Matter Question Bank Real gases and Vander waal's equation

  • question_answer Pressure exerted by 1 mole of methane in a 0.25 litre container at 300K using vander Waal's equation (given \[1=2.253\,atm\,{{l}^{2}}\,mo{{l}^{-2}},\,b=0.0428\,lit\,mo{{l}^{-1}})\] is [Orissa JEE 2005]

    A)                 82.82 atm           

    B)                 152.51 atm

    C)                 190.52 atm         

    D)                 70.52 atm

    Correct Answer: A

    Solution :

               \[\left( P+\frac{{{n}^{2}}a}{{{V}^{2}}} \right)\,(V-nb)=nRT\]                    \[\left( P+\frac{2.253}{0.25\times 0.25} \right)\,(0.25-0.0428)=0.0821\times 300\]                    or \[(P+36.048)(0.2072)=24.63\]                 Þ \[P+36.048=118.87\] Þ \[P=82.82\,\]atm.


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