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


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> 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.
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> The conversion of graphite $\ce{C(graphite)}$ to diamond $\ce{C(diamond)}$ reduces its volume by $\pu{2e-6 m3 mol-1}$.
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> 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:
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> (**A**) $\pu{14501 bar}$  
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> (**B**) $\pu{58001 bar}$  
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> (**C**) $\pu{1450 bar}$  
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> (**D**) $\pu{29001 bar}$
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> JEE Adv, 2017, related [1](https://chemistry.stackexchange.com/questions/109951/pressure-at-which-graphite-and-diamond-are-in-equilibrium), [2](https://chemistry.stackexchange.com/questions/111467/conversion-of-graphite-into-diamond?noredirect=1&lq=1)
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In the solution stated in [this site](https://www.toppr.com/ask/question/the-standard-state-gibbs-free-energies-of-formation-of-cgraphite/), 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.




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




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 $$0=(G\_d^0-G\_g^0)-(\Delta v)(\Delta P)$$


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