Q.Compressibility factor for CO$_2$ at 400 K and 71.0 bar is 0.8697. The molar volume of CO$_2$ under these conditions is

Step 1. From $Z=\dfrac{PV_m}{RT}$ (per mole), $V_m=\dfrac{ZRT}{P}$. Since P is given in bar, use $R=0.08314\ \text{dm}^3\,\text{bar}\,\text{K}^{-1}\text{mol}^{-1}$.

Step 2. Substitute $Z=0.8697$, $T=400$ K, $P=71.0$ bar: $V_m=\dfrac{0.8697\times0.08314\times400}{71.0}$. …

Rearrange Z = PV/nRT for the (per-mole) molar volume, and use R in units consi …

• Using R = 0.0821 dm$^3$ atm K$^{-1}$mol$^{-1}$ (the atm-based value) with a pressure given in bar, without converting either the pressure or R -- the two units must match. …