# Why is helium an ideal gas?

Aug 9, 2016

An gas is sufficiently ideal when its compressibility factor $Z$ is close to $1$.

The compressibility factor is $Z = \frac{P V}{n R T}$, and it describes the ease or difficulty in compressing the gas:

• Is the molar volume $\overline{V} = \frac{V}{n}$ smaller than for an ideal gas? If so, $Z < 1$.
• Is the molar volume $\overline{V} = \frac{V}{n}$ larger than for an ideal gas? If so, $Z > 1$.

When $Z < 1$, the attractive forces dominate, and when $Z > 1$, the repulsive forces dominate, when it comes to the volume of $\text{1 mol}$ of the gas at STP ($\text{1 bar}$, ${0}^{\circ} \text{C}$).

For helium, $\textcolor{b l u e}{Z = 1.0005}$ at $\text{1.013 bar}$ and ${15}^{\circ} \text{C}$, so helium is close enough to ideal.

NOTE: Even if you use the Ideal Gas Law, the only thing you need to turn it into what I would call the "Real Gas Law" is the real $\overline{V}$.

The other variables, $P$ (pressure) and $T$ (temperature) are independent of the gas's identity.

Hence, if you know $Z$ (which you can look up), you know what the real (not just ideal) $\overline{V}$ is, and you've accounted for the only observable value that differentiates a real gas from an ideal gas: $\overline{V}$.