Question

The densities of graphite and diamond at $298K$ are $2.25$ and $3.31gc{m^{ - 3}}$, respectively, if the standard free energy difference $(\Delta {G^0})$ 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

Hint: We know that In thermodynamics, the standard free energy of any substance represents the free energy change related with the formation of the substance from the elements most stable form at standard condition. We can solve this problem with the help of formula; $\Delta G = - P\Delta V = $ work done, where $\Delta G$ is difference in Gibbs free energy, $P$ is pressure and $\Delta V$ is difference in volume.

Complete step by step solution:
> We know that density $ = \dfrac{{mass}}{{volume}}$; with the help of this formula we can calculate the volume of graphite and diamond at $298K$and subtracting the volume of graphite and diamond we will get difference in volumes.
Volume of graphite = atomic weight of graphite/density of graphite $ = \dfrac{{12}}{{2.25}}$ $c{m^3}$ .
Volume of diamond = atomic weight of diamond/density of diamond $ = \dfrac{{12}}{{3.31}}c{m^3}$ .
$ \Rightarrow \Delta V = (\dfrac{{12}}{{3.31}} - \dfrac{{12}}{{2.25}}) \times {10^{ - 3}} = - 1.71 \times {10^{ - 3}}L$
> It is given in the problem that the standard free energy difference $(\Delta {G^0})$ is equal to $1898Jmo{l^{ - 1}}$ .
  As we know that $\Delta G = - P\Delta V = $ work done so we will put all the known values in the equation in order to find out the value of pressure.
$ \Rightarrow 1898 = - ( - 1.71 \times {10^{ - 3}}) \times P \times 101.3$ ; because $[1L - atm = 101.3J]$
$ \Rightarrow P = \dfrac{{1898}}{{1.71 \times {{10}^{ - 3}} \times 101.3}} = 10.93 \times {10^3}atm$
$ \Rightarrow P = 11.09 \times {10^8}Pa$
Hence option B is correct, that is the pressure at which graphite will be transformed into diamond at $298K$ is $11.09 \times {10^8}Pa$.

Note: We have approached this problem with the concept that the standard free energy difference $(\Delta {G^0})$ is equal to the maximum non reversible work that is performed by a thermodynamic system at constant pressure and temperature. So here in the problem we have to calculate volumes of diamond and graphite thus we find change in volume. With the help of equation $\Delta G = - P\Delta V = $ work done we have calculated the value of pressure.