The densities of graphite and diamond at 298 K are 2.25 and 3.31 gcm^(-3) respectively. If the standard fre energydifference(DeltaG^(@)) is eqal to 1895 J mol^(-1), the pressureat which graphite will be transformed into diamond at298 K is

The densities of graphite and diamond at 298 K are 2.25 and 3.31 gcm^(

1.	The densities of graphite and diamond at 298 K are 2.25 and 3.31 gcm^(-3) respectively. If the standard fre energydifference(DeltaG^(@)) is eqal to 1895 J mol^(-1), the pressureat which graphite will be transformed into diamond at298 K is
Answer» <P>`9.92 xx 10^(8) Pa` `9.92 xx 10^(7) Pa` ` 9.92xx 10^(6) Pa` none of these Solution :`-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)]`