# Ideal gas law, theory vs reality

I am trying to get an idea of the amount of hydrogen that I can store in some pressurized steel tanks underground.

Based on the ideal gas law PV=nRT

I did a simple calculation pressure = 200bar volume = 2000Liters temp = 25C

My results were about 16136.7 moles of hydrogen

1 mole of any gas is about 22.4 liters but that's at standard temperature and pressure.

At these elevated pressure it should be about 361,671 liters?

Have a look at the following: https://industry.airliquide.us/volume-compressed-gas-cylinder

Based on this, if you assume a gas cylinder of 47 litres (standard size) at 200 bar fill pressure, you have:

$$P_1 V_1 = P_2 V_2$$

where:

• $$P_1$$ = 200 bar (above atmospheric pressure) = 201.325 bar (absolute pressure)
• $$V_1$$ = 47 $$l$$
• $$P_2 = P_{atm}$$ = 1.01325 bar (absolute pressure)
• $$V_2$$ is what you're trying to determine

You end up with $$V_2$$ = 9338.6 $$l$$ approximately.

• The initial volume was 2,000 l, not 47 l, you made that up... – user20096 Jun 22 '19 at 0:19
• And the number of moles changes, first the OP talked about 16,136.7 and later about 1 mole. – user20096 Jun 22 '19 at 2:12
• The numbers are just there to illustrate the methodology, the OP can adapt them to his/her own problem. 47 l is a standard gas cylinder volume. – am304 Jun 22 '19 at 6:22
• thank you for the logic to solve this, best! – sean Jun 24 '19 at 13:47

Theory:

1 mole at those conditions takes up 0.123 l. Why? If 16,136.7 moles of hydrogen take up 2,000 l, then 1 mole will take up 0,123 l. You can also apply the ideal gas equation to check it:

$$V=n*R*T/p=1 [mole]*0.082 [atm*l/mol*K]*298.15 [K]/[(200*10^5/101,325) [atm]]=0.123 [l]$$

And what about reality? Well, ideal gas equation fails at such high pressures because it doesn't regard the space occupied by the mollecules of the gas (important at those high pressures becuase collisions are more frequent and repulsion forces more intense, therefore there's an increase in volume). Note: This is true for hydrogen, but not for all real gases.

Let's correct the ideal gas equation with the compressibility factor $$z$$:

Now we'll have $$p*V=z*n*R*T$$

At 20 MPa (200 bar) and 25 ºC we have $$z=1.124$$ (mean value between 1.0601 and 1.1879, highlighted down here): So let's do the math:

$$V=z*n*R*T/p=1.124*1 [mole]*0.082 [atm*l/mol*K]*298.15 [K]/[(200*10^5/101,325) [atm]]=0.139 [l]$$

And 2,246.7 l if you consider the 16,137.7 moles. So prepare a bigger tank for your hydrogen!