Lets check the numbers from the OP's scans. Time for some thermodynamics!
We would like to find out the minimum pressure p (at room temperature) under which the phase transition from graphite (colloquially, "coal"
to diamond,
C(graphite) --> C(diamond),
takes place spontaneously. For simplicity, lets consider 12 grams of carbon since there is 1 mole of carbon, i.e. 6*10^23 C-atoms, in 12 grams of carbon (i.e. our system consists of 1 mole of carbon).
According to thermodynamics, phase changes, chemical reactions, and other physicochemical processes in these kinds of situations tend towards a state with lowest (Gibbs) free energy G.
Thus we need to evaluate the change in free energy, ΔG, when 1 mol of C undergoes the process C(graphite) --> C(diamond). If ΔG < 0 it means that the process leads to a state of smaller free energy i.e. graphite spontaneously turns into diamond.
From thermodynamic tables we can find out the ΔG for this process under normal atmospheric pressure (1 atm):
C(graphite) --> C(diamond), ΔG(1 atm) = +2.900 kJ/mol. (source: https://en.wikipedia.org/wiki/Standard_Gibbs_free_energy_of_formation )
This means that when 1 mol of C turns from "coal" into diamond at 1 atm, its free energy INCREASES by 2.9 kilojoules. Thus at 1 atm, ΔG > 0, and coal does NOT change into diamond. It turns out that when the pressure is increased the ΔG for this phase change decreases, and finally at a certain pressure p, ΔG = 0. This signifies a pressure under which graphite and diamond are at equilibrium (much like liquid water and ice are at equilibrium at 0 degrees Celsius and 1 atm). If p is increased further, then ΔG < 0 and under such pressures graphite turns into diamond spontaneously.
The lowest pressure p required to turn graphite into diamond can thus be solved from the equation ΔG(p) = 0, but for that we need to express ΔG(p) explicitly as a function of p.
The fundamental thermodynamic equation for Gibbs energy at constant T (here, T = the room temperature) tells us that:
ΔG(p) = ΔG(1 atm) + ΔV*(p - 1 atm). (1)
Here, we will set the left-hand side to 0 (since we want to find p such that ΔG(p) = 0) and ΔG(1 atm) is known from above. ΔV is the change in volume when 1 mol (i.e. 12 grams) of C turns from graphite into diamond. From literature we know the densities of gra and dia: 2.26 and 3.51 grams/cm^3, respectively (dia is denser). With the densities, we can calculate the volumes of 12 grams of gra and dia, and then calculate the desired change in volume:
ΔV = -1.891 cm^3/mol.
This means, when 1 mol of C turns from gra into dia its volume DECREASES by 1.891 cm^3 = 1.891 mL (because dia is the denser allotrope).
We now have all we need to solve the eqn. (1) for pressure p:
p = -ΔG(1 atm)/ΔV + 1 atm = 15 140 atm
Thus the pressure which changes coal into diamond spontaneously (at room temp.) needs to be LARGER than 15 140 atmospheres. The scan in the original post claims that the required pressure is 100 000 atmospheres. This is reasonable: you would need quite a bit of overpressure in order for the phase change to occur in less than "geological time-scales" (at 15 140 atm, it would take "infinitely" long).
Sound science in comics, this makes me happy!
And what about the strength needed to compress coal into diamond? In order to convert pressure into force, we need to know the contact surface area of the system which is being compressed by the force. For 1 mol of C contained within a closed fist, A = 10 cm^2, is a decent estimate for the contact surface between skin and the lump of coal. The force is:
F = p*A = 100000 atm * 10 cm^2 = 10^7 N (newtons).
Lets convert this into equivalent mass (under Earth's normal gravity):
m = F/g = 10^7 N / (10 m/s^2) = 10^6 kg = 1000 metric tons.
If A = 1 cm^2 instead, this would give m = 100 metric tons (consistent with what was said in the OP's scan). So, all in all, the figures in those scans were quite reasonable.
They even got the tetrahedral diamond crystal structure on atomic level quite right. Kudos to the writer and the artist.
