# Pressure induced transition of graphite to diamond -- homework help

by mh7049
Tags: diamond, graphite, homework, induced, pressure, transition
 P: 3 1. The problem statement, all variables and given/known data Process is being carried out at 25*C and requires an increase in pressure until he graphite and diamond are in equilibrium. The following data is given at 25*C dG(25*C, 1atm) = gdiamond - ggraphite = 2866 J/mol density diamond = 3.51 g/cm^3 density graphite = 2.26 g/cm^3 Estimate the pressure at which these two forms of carbon are in equilibrium at 25*C. This is a homework question for my thermo II class, I have been looking at it for a day or so now and can't come up with a solution. Based on the densities it's clear it will take a lot of pressure for the transition. Any help would be greatly appreciated, it's been 6 months since thermo I so I am a little rusty! Thanks again! 2. Relevant equations 3. The attempt at a solution
 HW Helper P: 1,273 can you explain to me your notation and also any ideas you might have as how to proceed?
 HW Helper P: 1,273 ...but failing anything else, I might recommend to estimate the pressure by saying that the (gibbs free?) energy difference (dG) you have written is that quantity which should be related to the energy transfered by compressing... something like dG = p dv where v is the specific volume, which you can easily related to the change in density which is known.
P: 3

## Pressure induced transition of graphite to diamond -- homework help

how would you relate that to the change in density?
 P: 3 so far I have that dG=deltaVdP - deltaSdT but because there is no change in temperature the change in Gibbs free energy = deltaVdP. I was also not given an initial amount of graphite or how much is to transition to diamond. I plan to assume 1g of graphite is transition to diamond. Any other suggestions would be greatly appreciated!
HW Helper
P: 1,273
 Quote by mh7049 how would you relate that to the change in density?
calculus. density and volume are related by:

$\rho = M/V$

so $\delta \rho = \delta V \frac{d \rho}{d V}$

and you can find $d\rho/d V$ from the first equation.
 P: 4 2 years later and i've come up with the same question. Same numbers and everything. I seem to get an answer, but it appears to be far too low when checking against information found elsewhere. I've assumed that the process occurs at STP thus 1 atm or 101325 Pa and 25 C. The general equation that I've got is. 2886 j/mol = (cm^3 / (3.51-2.26)g) * (12.01 g/mol) * (1 m^3 / 10^6 cm^3) * (Pf - 101325) Pf being final pressure Basically I took the given dG (Gibbs free energy) and set it equal to the change in volume which was given by the densities. I converted everything into the proper units of mols and m^3 and multiplied by change in pressure assuming that initial pressure is under STP. Additionally, I know that there is no change in temperature thus the entropy portion of the equation is zero and that we can use dG=VdP I end up with 2945 atm which seems to be about 7000 atm too low in comparison to HPHT. I've tried a few other arrangements of the equation dG=Vdp-SdT and 2866=Pd-Pg.
HW Helper
P: 1,273
 Quote by CerealSteak 2 years later and i've come up with the same question. Same numbers and everything. I seem to get an answer, but it appears to be far too low when checking against information found elsewhere. I've assumed that the process occurs at STP thus 1 atm or 101325 Pa and 25 C. The general equation that I've got is. 2886 j/mol = (cm^3 / (3.51-2.26)g) * (12.01 g/mol) * (1 m^3 / 10^6 cm^3) * (Pf - 101325) Pf being final pressure Basically I took the given dG (Gibbs free energy) and set it equal to the change in volume which was given by the densities. I converted everything into the proper units of mols and m^3 and multiplied by change in pressure assuming that initial pressure is under STP. Additionally, I know that there is no change in temperature thus the entropy portion of the equation is zero and that we can use dG=VdP I end up with 2945 atm which seems to be about 7000 atm too low in comparison to HPHT. I've tried a few other arrangements of the equation dG=Vdp-SdT and 2866=Pd-Pg.
Hi,

Looks like you are on the right track with

[TEX]
\Delta G = v \Delta P
[/TEX]
where v is the specific volume.

But you used "delta v" rather than v and I'm not sure why.

To do the problem exactly we would have to know how v changes as a function of P. But failing that we could estimate

[TEX]
\Delta P=\Delta G/v
[/TEX]

And plugging in the numbers I find
[TEX]
\Delta P = 836,000,000Pa
[/TEX]
 P: 4 Thanks a lot for your time and help. I think I'm still a little confused with your comment on knowing how v changes as a function of P. Does this mean we write the equation deltaP=deltaG/v for both diamond and graphite then combine the equations to solve for deltaP or Pf?
 P: 4 I think I got the same deltaP as you by using the equation you mentioned. deltaP = deltaG / v (diamond) = 837606994.2 But what of the specific volume of graphite? Isn't this important to the solution as we have equilibrium between the two?
P: 1
 Quote by mh7049 1. The problem statement, all variables and given/known data Process is being carried out at 25*C and requires an increase in pressure until he graphite and diamond are in equilibrium. The following data is given at 25*C dG(25*C, 1atm) = gdiamond - ggraphite = 2866 J/mol density diamond = 3.51 g/cm^3 density graphite = 2.26 g/cm^3 Estimate the pressure at which these two forms of carbon are in equilibrium at 25*C. This is a homework question for my thermo II class, I have been looking at it for a day or so now and can't come up with a solution. Based on the densities it's clear it will take a lot of pressure for the transition. Any help would be greatly appreciated, it's been 6 months since thermo I so I am a little rusty! Thanks again! 2. Relevant equations 3. The attempt at a solution
The Solution right here:
dG(T,P) = dG(1,298) + I(VdP)
= 2892.3 + (-1.881 10^-6)(P-1) x (1.013 x 10^5)
The transformation is thermodynamically spontaneous at 298 K as long as dG(P,T) is negative:
2892.3 + (-1.881 x 10^-6) (P - 1) x (1.013 x 105) < 0
P > 15146 bars
Notice:
P in Pa: 1 atm = 1.013 x 10^5 Pa
V in m3: 1 m3 = 10^6 cm3
I = Integral which shows the additional external work for compression

Best Regard
Nguyen Tri Ngyen
Viet Nam - Hue - 03/28/2012

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