A sample of gas occupies $5.40 L$ at a state where its compressibility factor is $1.08$ . The volume that the same amount of an ideal gas would occupy at the same $P$ and $T$ , in litres (two decimals), is

Numerical answer type, answer in L, tolerance 0.02.

Answer: 5 L

Z = V_{real} / V_{ideal}

, so

V_{ideal} = V_{real} / Z = 5.40/1.08 = 5.000 L

, i.e.

5.0 L

. The real gas is less compressible than ideal here.