The standard state Gibbs free energies of formation of $\ce{C(graphite)}$ and $\ce{C(diamond)}$ at $T = \pu{298 K}$ are $\pu{0 kJ mol-1}$ and $\pu{2.9 kJ mol-1}$, respectively.

The conversion of graphite $\ce{C(graphite)}$ to diamond $\ce{C(diamond)}$ reduces its volume by $\pu{2e-6 m3 mol-1}$.

If $\ce{C(graphite)}$ is converted to $\ce{C(diamond)}$ isothermally at $T = \pu{298 K}$, the pressure at which $\ce{C(graphite)}$ is in equilibrium with $\ce{C(diamond)}$ is:

(A) $\pu{14501 bar}$
(B) $\pu{58001 bar}$
(C) $\pu{1450 bar}$
(D) $\pu{29001 bar}$

JEE Adv, 2017, related 1, 2

In the solution stated in this site, there are two assumptions made about this process.

  1. $\Delta S_r = 0$ i.e. the total entropy change of the process is zero.
  2. The internal energy change is zero due to it being an isothermal process.
  3. $\Delta H_r= P \Delta V$, that is we assume pressure is constant for the process and solve for $P$. The total pressure in the final state is apparently the initial pressure plus the pressure calculate from the ratio $\frac{\Delta H_r}{\Delta V}$.

How do we justify these two assumptions?

Firstly, how can we justify the entropy change being zero? And, the second point I don't get how we can claim $\Delta U = 0$, just because it is isothermal. I know it is true for ideal gases but how does that apply here? About the last assumption, I can not understand at all what the logic is behind finding the final pressure as initial plus the calculated from the ratio.

P.S: I know the given links answer the question completely but I want to figure out how to reason these assumptions myself.

  • 1
    $\begingroup$ Why is a diamond crystal "much more ordered" than a hexagonal crystal? Both are crystals, with atoms sitting on specific lattice sites (ignoring point defect energetics). $\endgroup$
    – Jon Custer
    Commented Aug 23, 2021 at 18:39
  • $\begingroup$ Hmmm good point I guess $\endgroup$
    – Babu
    Commented Aug 23, 2021 at 19:07

1 Answer 1


This is definitely not the approach that I would have used to solve this problem. I would have used the condition that, at the equilibrium pressure $G_g=G_d$ or $\Delta G=0$. The free energy of graphite at the equilibrium condition would be $$G_g=G_g^0+v_g\Delta P$$where $G_g^0$ is the molar free energy of graphite in the standard state (0 kJ/mol), $v_g$ is the molar volume of graphite, and $\Delta P$ is the increase in pressure relative to the standard state. Similarly, for diamond, at the equilibrium condition, $$G_d=G_d^0+(v_g-\Delta v)\Delta P$$where $G_d^0$ is the molar free energy of graphite in the standard state (2.9 kJ/mol), and $\Delta v=2\times 10^{-6}\ m^3/mol$. So setting $\Delta G =0$ gives $$0=(G_d^0-G_g^0)-(\Delta v)(\Delta P)$$

  • $\begingroup$ hmm but what about justifying the assumptions? Nice approach though. Didn't think the moles would be the slope the slope of G with varying pressure $\endgroup$
    – Babu
    Commented Aug 24, 2021 at 3:35
  • 1
    $\begingroup$ The only assumption I can think of is that the change in molar volume between graphite and diamond over the range of pressures between the standard pressure and the equilibrium pressure is constant. Is that what you were referring to? $\endgroup$ Commented Aug 24, 2021 at 12:15
  • $\begingroup$ Yes, I understand, you took the first order variation of G with respect to moles. In the quoted answer, they made some assumptions, I do understand that the way shown would not be how you would solve it but could you see if you can give explanations as to why they made the assumptions that they did? $\endgroup$
    – Babu
    Commented Aug 24, 2021 at 15:25
  • $\begingroup$ I didn't take the variation of G with respect to moles. There is no change in the number of moles. I took the variation of G with respect to pressure at constant T, which is given by dG=vdP. $\endgroup$ Commented Aug 24, 2021 at 16:37
  • $\begingroup$ I see, so it is actually that variation of G with pressure is the moles $\endgroup$
    – Babu
    Commented Aug 24, 2021 at 16:39

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