Question

20 mol of Chlorine gas occupies a volume of 800 mL at 300 K and 5 x 10⁶ Pa pressure. Calculate the compressibility factor of the gas. Comment on the compressibility of the gas under these conditions

Answer 1

 P = 5 x 10⁶ 

Pa = 5 x 10⁶ / 10⁵ bar 

 n = 20 

 T = 300 K 

R = 0.083 L bar K^-1 mol^-1 

 V_real = 800 mL 

 V_ideal = nRT/P 

Putting values , we have 

Vi_deal = 1004 mL 

 Z = V_real/ V_ideal= 800/1004 = 0.796 

 As Z is less than 1, it means gas is more compressible under these conditions.