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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-II

  • question_answer
    \[{{\Delta }_{f}}G{}^\circ \] at 500K for substance 'S' in liquid state and gaseous state are\[+100.7kcal\] \[mo{{l}^{-1}}\] and \[\text{+103 kcal mo}{{\text{l}}^{\text{-1}}}\], respectively. Vapour pressure of liquid 'S' at 500K is approximately equal to: \[\text{(R=2cal }{{\text{K}}^{-1}}mo{{l}^{-1}})\]  [JEE Online 15-04-2018 (II)]

    A)                     100atm                

    B) 1 atm    

    C) 10 atm 

    D)          0.1 atm

    Correct Answer: C

    Solution :

    \[\Delta G_{rxn}^{o}={{\Delta }_{f}}G{}^\circ (vapour)-{{\Delta }_{f}}G{}^\circ (liquid)\] \[\Delta G_{rxn}^{o}=103-100.7=2.3kcal/mol\] \[\Delta G_{rxn}^{o}=-RT\ln K\] \[\begin{align}   & 2.3kcal/mol\times 1000cal/kcal= \\  & -2cal/mol/K\times 500K\times \ln K \\ \end{align}\] \[\ln K=2.3\] \[K=10atm=\]Vapour pressure of liquid 'S' Vapour pressure of liquid 'S' at 500 K is approximately equal to 10 atm.


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