Question

What is the density (in g/L) of hydrogen gas at 21.0 C and a pressure of 1640 psi?

Answer 1

You have specified a pressure of `1640*"psi"`, which is not the same as `1640*"psig"`......

Explanation:

We must first convert to kosher units.......

`1*atm-=14.7*"psi"`

And thus `1640*"psi"=(1640*"psi")/(14.7*"psi"*atm^-1)=111.6*atm.`

This is a high pressure, and we use the Ideal Gas for a first approx. If you want a better estimate, you need to find another equation of state.

WE know that `PV=nRT`, and `P/(RT)=n/V=("mass"/"molar mass")/(V)`

And so....`P/(RT)xx"molar mass"="mass"/V=rho_"gas [density](https://socratic.org/chemistry/measurement-in-chemistry/density)"`

And thus `rho=(111.6*atmxx2.015*g*mol^-1)/(0.0821*L*atm*K^-1*mol^-1xx294.15K)`

`=9.31*g*L^-1`

Note that `"Kelvin temperature"` is specified.........And given the pressure, (which is unreasonably hi), this is a very high gas density.