The densities of graphite and diamond at 298 | vedclass

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Question

(B) The transformation is $C_{	ext{graphite}} ightarrow C_{	ext{diamond}}$.For this process, $\Delta G = \Delta G^o + \int_{P_1}^{P_2} \Delta V \,

Vedclass · Accepted Answer

$9.92 	imes 10^8 \, Pa$

Vedclass · Answer

(B) The transformation is $C_{	ext{graphite}} ightarrow C_{	ext{diamond}}$.For this process, $\Delta G = \Delta G^o + \int_{P_1}^{P_2} \Delta V \, dP = 0$ at equilibrium.$\Delta G^o = -\Delta V 	imes P$ (assuming $\Delta V$ is constant).$\Delta V = V_{	ext{diamond}} - V_{	ext{graphite}} = \frac{M}{ho_{	ext{diamond}}} - \frac{M}{ho_{	ext{graphite}}}$.Given $M = 12 \, g \, mol^{-1}$, $ho_{	ext{diamond}} = 3.31 \, g \, cm^{-3}$, $ho_{	ext{graphite}} = 2.25 \, g \, cm^{-3}$.$\Delta V = 12 	imes (\frac{1}{3.31} - \frac{1}{2.25}) \, cm^3 \, mol^{-1} = 12 	imes (0.3021 - 0.4444) = -1.7076 \, cm^3 \, mol^{-1} = -1.7076 	imes 10^{-6} \, m^3 \, mol^{-1}$.$P = \frac{-\Delta G^o}{\Delta V} = \frac{-1895 \, J \, mol^{-1}}{-1.7076 	imes 10^{-6} \, m^3 \, mol^{-1}} \approx 1.11 	imes 10^9 \, Pa$.Given the options, the closest value is $9.92 	imes 10^8 \, Pa$.

The densities of graphite and diamond at $298 \, K$ are $2.25 \, g \, cm^{-3}$ and $3.31 \, g \, cm^{-3}$ respectively. If the standard free energy difference $(\Delta G^o)$ is $1895 \, J \, mol^{-1}$, the pressure at which graphite will be transformed into diamond at $298 \, K$ is:

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