Question

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

• A. $9.92 \times 10^8 \, Pa $
• B. $9.92 \times 10^7 \, Pa $
• C. $9.92 \times 10^6 \, Pa $
• D. none of these

Answer 1

Answer: (D)

As we know that,
$
\begin{array}{l}
\text { density }=\frac{\text { mass }}{\text { volume }} \\
\left.\therefore d _{ g }=\frac{12}{ V _{ g }} \Rightarrow V _{ g }=\frac{12}{2.25} cm ^{3} \text { [density of graphite }=2.25 g / cm ^{3}\right] \\
\left.\therefore d _{ d }=\frac{12}{ V _{ d }} \Rightarrow V _{ d }=\frac{12}{3.31} cm ^{3} \text { [density of diamond }=3.31 g / cm ^{3}\right] \\
\therefore \Delta V =\left(\frac{12}{3.31}-\frac{12}{2.25}\right) \times 10^{-3}=-1.71 \times 10^{-3} L
\end{array}
$
As we know that,
$
\Delta G =- P \Delta V =\text { work done }( w )
$
Given that $\Delta G =1895 J / mol$
$
\begin{array}{l}
\therefore 1895=-\left(-1.71 \times 10^{-3}\right) \times P \times 101.3[\because 1 L - atm =101.3 J ] \\
\Rightarrow P =\frac{1895}{1.71 \times 10^{-3} \times 101.3} \\
\Rightarrow P =10.93 \times 10^{3} atm =11.07 \times 10^{8} Pa
\end{array}
$
Hence the correct answer is $11.07 \times 10^{8} Pa$.