Why do you need 13.8 MJ/kg (9% of energy content) to compress hydrogen to 200 bar, but only 1.4 MJ/kg (2.5% of energy content) for methane?

I looked into compressibility factors and the compressibility factor for methane is way lower than for hydrogen (up until high pressures). Does this determine how much work is required?

  • $\begingroup$ The comparison to energy content seems unusual - you need energy to compress helium that has no energy content (for combustion that is). $\endgroup$ – Jon Custer May 4 at 16:56
  • $\begingroup$ @JonCuster Yes, it's not useful for a comparison in general. Both gases are highly popular (with H2 as raw material for CH4) in PtX processes and this is one advantage methane has over hydrogene. So it makes sense in this specific case. $\endgroup$ – Kilian Helfenbein May 5 at 18:51

The essential reason is that a kilogram of hydrogen contains 8 times as many molecules as a kilogram of methane (because the mass of a hydrogen molecule is about 1/8 of the mass of a methane molecule).

If we assume, for the sake of argument, that the compression is isothermal (constant temperature, $T$) the work needed to compress a sample of $N$ molecules of an ideal gas from pressure $p_1$ to pressure $p_2$ is

$$\text{Work}=Nk_BT \ln \left(\frac{p_2}{p_1}\right)\ \ \ \ [k_B= \text{Boltzmann's constant}]$$

So if the gases were ideal, 8 times more work would be needed per kilogram for the hydrogen, but at such high pressures the gases are far from ideal. Intermolecular forces and the finite volumes occupied by the molecules are significant and different for different gases. That would account for why the ratio of work needed is not exactly 8:1

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