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JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    At very high pressures, the compressibility factor of one mole of a gas is given by :   JEE Main Online Paper (Held On 09 April 2016)

    A) \[1+\frac{pb}{RT}\]

    B) \[\frac{pb}{RT}\]

    C) \[1-\frac{b}{(VRT)}\]                     

    D) \[1-\frac{pb}{RT}\]

    Correct Answer: A

    Solution :

                    According to Vander waal's equation for one mole of gas\[\left( P+\frac{a}{{{V}^{2}}} \right)(V-b)=RT\]at high pressure \[\frac{a}{{{V}^{2}}}\]can be neglected with respect to P,                 \[\therefore \]\[P+\frac{a}{{{V}^{2}}}\simeq P\]                 \[P(V-b)=RT\]                 \[PV-Pb=RT\]                 \[PV=RT+Pb\]divided on RT on both side, \[Z=1+\frac{Pb}{RT}\]


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