The conversion of graphite to diamond reduces its volume by X ml / mole isothermally at T = 298K . The pressure at which graphite is in equilibrium with diamonds is1.45x10⁴ bar. The standard Gibbs free energy of formation of diamond at T = 298K is 2.9KJ / mole. Then value of X is:

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answer is C.

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Detailed Solution

Using $Δ_{r} G_{(P_{2})} - Δ_{r} G_{(P_{1})} = (Δ_{r} V) (P_{2} - P_{1})$
$\begin{array}{l} P_{1} = 1 bar = 10^{5} pascal \\ P_{2} = 1.45 \times 10^{4} bar = 1.45 \times 10^{9} pascal \\ As P_{2} > P_{1} \Rightarrow (P_{2} - P_{1}) = P_{2} \\ Also Δ_{r} G_{(P_{1})} = Δ_{f} G_{(D)}^{0} - Δ_{f} G_{(G)}^{0} = 2.9 KJ - 0 = 2.9 KJ \\ \Rightarrow - 2.9 \times 10^{3} = (Δ_{r} V) (1.45 \times 10^{9}) \\ \Rightarrow Δ_{r} V = - 2 \times 10^{- 6} m^{3} / mole = - 2 ml / mole \\ \Rightarrow X = 2 \end{array}$