Thursday, August 15, 2019

thermodynamics - Pressure at which graphite and diamond are in equilibrium


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 kJmol1 and 2.9 kJmol1, respectively.


The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2106 m3mol1.


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+PPVmdP 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=PeqPΔVmdP


and, since Δμ=ΔfGm, ultimately to the expression you provided:


ΔfGm=PeqPΔVmdP


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


ΔfGm=ΔVmPeqPdP=ΔVm(PeqP)


which can be solved for Peq:


Peq=ΔfGmΔVm+P



No comments:

Post a Comment

digital communications - Understanding the Matched Filter

I have a question about matched filtering. Does the matched filter maximise the SNR at the moment of decision only? As far as I understand, ...