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.
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
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P = 5 x 106
Pa = 5 x 106 / 105 bar
n = 20
T = 300 K
R = 0.083 L bar K-1 mol-1
Vreal = 800 mL
Videal = nRT/P
Putting values , we have
Videal = 1004 mL
Z = Vreal/ Videal= 800/1004 = 0.796
As Z is less than 1, it means gas is more compressible under these conditions.
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