The densities of graphite and diamond at `298K` are `2.25` and `3.31gcm^(-3)` , respectively. If the standard free energy difference `(DeltaG^(0))` is equal to `1895Jmol^(-1)` , the pressure at which graphite will be transformed into diamond at `298K` is

Options

A. `9.92 xx 10^(8) Pa`
B. `9.92 xx 10^(7) Pa`
C. ` 9.92xx 10^(6) Pa`
D. none of these

Answer

D

Explanation

`-DeltaG = P DeltaV`
`1895= P [ ((12)/( 2.25)- (12)/( 331))xx 10^(-6) m^(2)]`
Calculate P,
[From thermodynamics , `[ (del[DeltaG])/(delP)]_(T)=DeltaV`
or `( DeltaG _(2)-DeltaG_(1))/( P_(2)-P_(1))= DeltaV`
or`DeltaG_(2)- DeltaG_(1) = (P_(2)= P_(1))DeltaU`
Taking `P_(1)=1` bar `= 10^(5)Pa`, calculate `P_(2)]`