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KVPY Sample Paper KVPY Stream-SX Model Paper-26

  • question_answer
                The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T=298 K are \[{{\Delta }_{f}}{{G}^{0}}\] [C(graphite)] \[=0\text{ }kJmo{{l}^{-1}}\]\[{{\Delta }_{f}}{{G}^{0}}\] [C (diamond)] \[=2.9kJmo{{l}^{-1}}\] The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C (graphite)] to diamond [C (diamond)] reduces its volume by \[2\times {{10}^{-6}}{{m}^{3}}mo{{l}^{-1}}\]. If C (graphite) is converted to C (diamond) isothermally at T = 298 K, the pressure at which C (graphite) is in equilibrium with C (diamond), is [Useful information: 1 J=1 kg \[{{\operatorname{m}}^{2}}{{s}^{-2}};\] Pa = 1 kg \[{{\operatorname{m}}^{-2}}{{s}^{-2}};\]1 bar \[{{10}^{5}}\]pa]

    A) 14501 bar

    B) 58001 bar

    C) 1450 bar

    D) 29001 bar

    Correct Answer: A

    Solution :

    \[{{\operatorname{C}}_{\left( graphite \right)}}\to {{\operatorname{C}}_{\left( diamond \right)}}\left( Isothermally \right)\]
    \[{{\Delta }_{r}}G{}^\circ =\Delta G{{{}^\circ }_{\left( diamond \right)}}-\Delta G{{{}^\circ }_{\left( graphite \right)}}\]
    \[=2.9-0=2.9kJ\,mo{{l}^{-1}}\]
    Gibbs free energy is the maximum useful work, then
    \[-\Delta G={{w}_{ma\,x}}=-P\Delta V\]
    \[-2.9\times {{10}^{3}}=-P\times 2\times {{10}^{-6}}\]
    \[P=\frac{2.9\times {{10}^{3}}}{2\times {{10}^{-6}}}=1.45\times {{10}^{9}}\]
    \[Pa=1.45\times {{10}^{9}}\times {{10}^{-5}}bar\]
    \[=1.45\times {{10}^{4}}bar=14500bar\]


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