I came across this problem :
The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T=298 K are 0 kJmol−1 and 2.9 kJmol−1, respectively.
The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2⋅10−6 m3mol−1.
If C(graphite) is converted to C(diamond) isothermally at T=298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is
(A) 14501 bar
(B) 58001 bar
(C) 1450 bar
(D) 29001 bar
I applied : ΔG(P,T)=ΔfG∘+∫p2p1Vdp.
Since the system is at equilibrium,ΔfG∘=−∫p2p1Vdp
Now I am stuck. I have not been given any relation between Pressure and Volume. Is there any assumption i have to make to solve this integral ?
Answer
For each phase (graphite or diamond) you can write
μ=μ∘+∫PP∘VmdP at constant T.
We are asked to find the pressure P=Peq at which carbon coexists in the two phases (the Gibbs free energy is equal in both) so that
μ(diamond)=μ(graphite)
This leads to
Δμ∘=−∫PeqP∘ΔVmdP
and, since Δμ∘=ΔfG∘m, ultimately to the expression you provided:
ΔfG∘m=−∫PeqP∘ΔVmdP
The approximation you are allowed to make at this point is that the solids are incompressible, such that their Vm are constant with change of pressure. It follows that
ΔfG∘m=−ΔVm∫PeqP∘dP=−ΔVm(Peq−P∘)
which can be solved for Peq:
Peq=−ΔfG∘mΔVm+P∘
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