Justifying assumptions about method to find equilibrium pressure for co-existence of graphite and diamond

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

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.

• 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). Aug 23 '21 at 18:39
• Hmmm good point I guess Aug 23 '21 at 19:07

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 $$0=(G_d^0-G_g^0)-(\Delta v)(\Delta P)$$