Calculate the compressibility factor for CO₂, if one mole of it occupies 0.4 litre at 300K and 40 atm. Comment on the result.

Dear user,

Compressibility factor is given by :

$$\begin{gathered} \mathrm{Z}=\frac{\mathrm{PV}}{\mathrm{nRT}} \\ \mathrm{where} \mathrm{Z} \mathrm{is} \mathrm{compressibility} \mathrm{factor}, \mathrm{P} \mathrm{is} \mathrm{pressure}, \mathrm{V} \mathrm{is} \mathrm{volume} \\ \mathrm{n} \mathrm{is} \mathrm{number} \mathrm{of} \mathrm{moles}, \mathrm{R} \mathrm{is} \mathrm{universal} \mathrm{gas} \mathrm{constant} \mathrm{and} \mathrm{T} \mathrm{is} \\ \mathrm{temperature}. \\ \mathrm{Given}, \\ \mathrm{P}=40\mathrm{atm} \\ \mathrm{V}=0.4\mathrm{l} \\ \mathrm{n}=1\mathrm{mole} \\ \mathrm{T}=300\mathrm{K} \\ \mathrm{R}=0.0821\mathrm{L} \mathrm{atm}/\mathrm{mol} \mathrm{K} \\ \mathrm{Z}=\frac{40\times 0.4}{1\times 0.0821\times 300} \\ =0.65 \end{gathered}$$

Value of Z is less than 1 and hence nRT>PV. In order to have value of Z=1, volume of CO₂ must have been more
at same pressure and temperature or CO₂ is more compressible  than ideal gas.