Q56 / 200 · NEET 2003 · Chemistry
The densities of graphite and diamond at $298 \mathrm{~K}$ are 2.25 and $3.31 \mathrm{gcm}^{-3}$, respectively. If the standard free energy difference $\left(\Delta \mathrm{G}^{\circ}\right)$ is equal to $1895 \mathrm{~J} \mathrm{~mol}^{-1}$, the pressure at which graphite will be transformed into diamond at $298 \mathrm{~K}$ is:
- A$9.92 \times 10^8 \mathrm{~Pa}$
- B$9.92 \times 10^7 \mathrm{~Pa}$
- C$9.92 \times 10^6 \mathrm{~Pa}$
- Correct answerDnone
Solution
$\Delta \mathrm{G}=-\mathrm{P} \Delta \mathrm{V}$
$$
\begin{aligned}
& 1895=P\left[\left(\frac{12}{2.25}-\frac{12}{3.31}\right) \times 10^{-6} \mathrm{~m}^3\right] \\
& P=\frac{1895 \times 10^6}{1.71}=1108.187 \times 10^6 \mathrm{Bar} \\
& =1108.1871 \times 10^{11} \mathrm{~Pa}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.