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

Aug 6, 2017

You have specified a pressure of $1640 \cdot \text{psi}$, which is not the same as $1640 \cdot \text{psig}$......

#### Explanation:

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

$1 \cdot a t m \equiv 14.7 \cdot \text{psi}$

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 $P V = n R T$, and $\frac{P}{R T} = \frac{n}{V} = \frac{\text{mass"/"molar mass}}{V}$

And so....$\frac{P}{R T} \times \text{molar mass"="mass"/V=rho_"gas [density](https://socratic.org/chemistry/measurement-in-chemistry/density)}$

And thus $\rho = \frac{111.6 \cdot a t m \times 2.015 \cdot g \cdot m o {l}^{-} 1}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 294.15 K}$

$= 9.31 \cdot g \cdot {L}^{-} 1$

Note that $\text{Kelvin temperature}$ is specified.........And given the pressure, (which is unreasonably hi), this is a very high gas density.