Question

1 mol ${\text{CO}}_{\text{2}}$ ocupies 0.4 lit at 27${}^{\circ }\text{C}$ and 40 atm. Calculate the compressibility factor:

• A. 0.65
• B. 0.85
• C. 0.15
• D. 0.6

Answer 1

The correct option is A 0.65

Compressibility factor is given by,

$z=\frac{PV}{nRT}$

Given, $p=40atmV=0.4litn=1$ $T=27+273=300K$

$z=\frac{40\times 0.4atmlit}{1\times 0.084\times 300litatm}=z=0.6496$

Compressibility factor of $C{O}_2$ is $0.65$.