Question

The densities of graphite and diamond at $298K$ are $2.25$ and $3.31gc{m}^{-3},$ respectively. If the standard free energy difference $(\Delta G^{\circ })$ is equal to $1895Jmo{l}^{-1}$, the pressure at which graphite will be transformed into diamond at $298K$ 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,
$<br/>\begin{array}{l}<br/>\text{ density }=\frac{\text{ mass }}{\text{ volume }}\\ <br/>\therefore d_g=\frac{12}{V_g}\Rightarrow V_g=\frac{12}{2.25}c{m}^3\text{ [density of graphite }=2.25g/c{m}^3]\\ <br/>\therefore d_d=\frac{12}{V_d}\Rightarrow V_d=\frac{12}{3.31}c{m}^3\text{ [density of diamond }=3.31g/c{m}^3]\\ <br/>\therefore \Delta V=\left(\frac{12}{3.31}-\frac{12}{2.25}\right)\times 10^{-3}=-1.71\times 10^{-3}L<br/>\end{array}<br/>$
As we know that,
$<br/>\Delta G=-P\Delta V=\text{ work done }(w)<br/>$
Given that $\Delta G=1895J/mol$
$<br/>\begin{array}{l}<br/>\therefore 1895=-\left(-1.71\times 10^{-3}\right)\times P\times 101.3[\because 1L-atm=101.3J]\\ <br/>\Rightarrow P=\frac{1895}{1.71\times 10^{-3}\times 101.3}\\ <br/>\Rightarrow P=10.93\times 10^{3}atm=11.07\times 10^{8}Pa<br/>\end{array}<br/>$
Hence the correct answer is $11.07\times 10^{8}Pa$.