Question

The densities of graphite and diamond at $298K$ are $2.25$ and $3.31gc{m}^{-3}$ , respectively, If the standard free energy difference $\left(\Delta G^0\right)$ is equal to $1898Jmo{l}^{-1}$, the pressure at which graphite will be transformed diamond at $298K$ is:

• A. $9.92\times {10}^{5}Pa$
• B. $11.09\times {10}^{8}Pa$
• C. $9.92\times {10}^{7}Pa$
• D. $9.92\times {10}^{6}Pa$

Answer 1

Answer: (B)

$11.09\times {10}^{8}Pa$

As we know that,

$\text{density}=\frac{\text{mass}}{\text{volume}}$

$\therefore d_g=\frac{12}{V_g}\Rightarrow V_g=\frac{12}{2.25}{cm}^3[\text{density of graphite}=2.25g/{cm}^3]$

$\therefore d_d=\frac{12}{V_d}\Rightarrow V_d=\frac{12}{3.31}{cm}^3[\text{density of diamond}=3.31g/{cm}^3]$

$\therefore \Delta V=(\frac{12}{3.31}-\frac{12}{2.25})\times {10}^{-3}=-1.71\times {10}^{-3}L$

As we know that,

$\Delta G=-P\Delta V=\text{work done}\left(w\right)$

Given that $\Delta G=1898J/mol$

$\therefore 1898=-(-1.71\times {10}^{-3})\times P\times 101.3[\because 1L-atm=101.3J]$

$\Rightarrow P=\frac{1898}{1.71\times {10}^{-3}\times 101.3}$

$\Rightarrow P=10.95\times {10}^{3}atm=11.09\times {10}^{8}Pa$

Hence the correct answer is $11.09\times {10}^{8}Pa$.