The densities of graphite and diamond are `22.5` and `3.51` gm `cm^(-3)`. The `Delta_(f)G^(ɵ)` values are `0 J mol^(-1)` and `2900 J mol^(-1)` for graphite and diamond, respectively. Calculate the equilibrium pressure for the conversion of graphite into diamond at `298 K`.

The densities of graphite and diamond are `22.5` and `3.51` gm `cm^(-3

1.	The densities of graphite and diamond are `22.5` and `3.51` gm `cm^(-3)`. The `Delta_(f)G^(ɵ)` values are `0 J mol^(-1)` and `2900 J mol^(-1)` for graphite and diamond, respectively. Calculate the equilibrium pressure for the conversion of graphite into diamond at `298 K`.
Answer» We have `C_(g)rarr C_(d)` `DeltaG^(ɵ)=DeltaG_(("diamond"))-DeltaG_(("graphite"))` `rArr (2900-0)=2900 J "mol"^(-1) :. DeltaG^(ɵ)=2900 J "mol"^(-1)` `Volume=("Mass")/("density")` `:. DeltaV=(V_(d)-V_(g))=(12/3.51-12/2.25)xx10^(-6) m^(3) "mol"^(-1)` `=-1.91xx10^(-6) m^(3) "mol"^(-1)` We know that, `(DeltaG)_(T)=DeltaV del P` `:. int_(g)^(d)del (DeltaG)=int_(P_(1))^(P_(2)) DeltaV del P` `DeltaG=deltaV(P_(2)-P_(1))` `:. P_(2)=(DeltaG)/(DeltaV)+P_(1)` or `P_(2)=(2900 J "mol"^(-1))/(-1.91xx10^(-6) m^(3) "mol"^(-1))+1013225 Pa` `=1.52xx10^(9) Pa` `:.` Pressure `=1.52xx10^(9) Pa`

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The densities of graphite and diamond are `22.5` and `3.51` gm `cm^(-3)`. The `Delta_(f)G^(ɵ)` values are `0 J mol^(-1)` and `2900 J mol^(-1)` for graphite and diamond, respectively. Calculate the equilibrium pressure for the conversion of graphite into diamond at `298 K`.

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